Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 13 Factorisation

Exercise 13(A)

Question 1. Factorise : 15x + 5
Answer:
\( 15x + 5 = 5(3x + 1) \)
In simple words: We pull out the common factor 5 from both parts of the expression, leaving 3x and 1 in the bracket.

Exam Tip: Check your final answer by multiplying the outer term back into the brackets to see if you get the original expression.

Question 2. Factorise : \( a^3 - a^2 + a \)
Answer:
\( a^3 - a^2 + a = a(a^2 - a + 1) \)
In simple words: Every term has at least one \( a \), so we take \( a \) outside the brackets.

Exam Tip: Don't forget that dividing the term \( a \) by itself leaves \( 1 \) inside the bracket, not \( 0 \).

Question 3. Factorise : \( 3x^2 + 6x^3 \)
Answer:
\( 3x^2 + 6x^3 = 3x^2(1 + 2x) \)
In simple words: We pull out \( 3x^2 \) since it is common to both terms, leaving \( 1 + 2x \) in the brackets.

Exam Tip: Pay attention to the exponents of variables - always take out the variable with the lowest exponent.

Question 4. Factorise : \( 4a^2 - 8ab \)
Answer:
\( 4a^2 - 8ab = 4a(a - 2b) \)
In simple words: We take out \( 4a \), which divides both parts, leaving \( a - 2b \) inside.

Exam Tip: Be careful to factor out both the greatest common numerical coefficient and the common literal variables.

Question 5. Factorise : \( 2x^3b^2 - 4x^5b^4 \)
Answer:
\( 2x^3b^2 - 4x^5b^4 = 2x^3b^2(1 - 2x^2b^2) \)
In simple words: The greatest common factor for the numbers is 2, for \( x \) is \( x^3 \), and for \( b \) is \( b^2 \), which we take outside.

Exam Tip: Double check variable powers when factoring out terms with multiple variables to prevent calculation errors.

Question 6. Factorise : \( 15x^4y^3 - 20x^3y \)
Answer:
\( 15x^4y^3 - 20x^3y = 5x^3y(3xy^2 - 4) \)
In simple words: We extract \( 5x^3y \) as the highest common factor from both terms.

Exam Tip: Verify that the remaining terms inside the brackets share no further common factors.

Question 7. Factorise : \( a^3b - a^2b^2 - b^3 \)
Answer:
\( a^3b - a^2b^2 - b^3 = b(a^3 - a^2b - b^2) \)
In simple words: Since only \( b \) is common to all three terms, we take it outside.

Exam Tip: Only factor out variables that are present in every single term of the expression.

Question 8. Factorise : \( 6x^2y + 9xy^2 + 4y^3 \)
Answer:
\( 6x^2y + 9xy^2 + 4y^3 = y(6x^2 + 9xy + 4y^2) \)
In simple words: Since \( y \) is the only common term across all three parts, we pull it out.

Exam Tip: Keep numerical constants inside if they do not share a common factor across all terms of the polynomial.

Question 9. Factorise : \( 17a^6b^8 - 34a^4b^6 + 51a^2b^4 \)
Answer:
\( 17a^6b^8 - 34a^4b^6 + 51a^2b^4 = 17a^2b^4(a^4b^4 - 2a^2b^2 + 3) \)
In simple words: We take out \( 17a^2b^4 \), which is the greatest common factor of all three parts.

Exam Tip: Ensure you identify the greatest common divisor for large coefficients like 17, 34, and 51.

Question 10. Factorise : \( 3x^5y - 27x^4y^2 + 12x^3y^3 \)
Answer:
\( 3x^5y - 27x^4y^2 + 12x^3y^3 = 3x^3y(x^2 - 9xy + 4y^2) \)
In simple words: We factor out \( 3x^3y \) from each of the terms.

Exam Tip: Keep signs consistent when pulling out positive common factors from terms with mixed signs.

Question 11. Factorise : \( x^2(a-b) - y^2(a-b) + z^2(a-b) \)
Answer:
\( x^2(a-b) - y^2(a-b) + z^2(a-b) = (a-b)(x^2 - y^2 + z^2) \)
In simple words: Since the whole bracket \( (a-b) \) is common to all parts, we pull it out.

Exam Tip: Expressions in parentheses can act as single common factors; treat them as a single entity.

Question 12. Factorise : \( (x+y)(a+b) + (x-y)(a+b) \)
Answer:
\( (x+y)(a+b) + (x-y)(a+b) \)
\( = (a+b)(x+y+x-y) \)
\( = (a+b)(2x) \)
\( = 2x(a+b) \)
In simple words: We factor out the common bracket \( (a+b) \) and then simplify the remaining parts to get \( 2x \).

Exam Tip: Always simplify the terms inside the second bracket after taking out the common binomial factor.

Question 13. Factorise : \( 2b(2a+b) - 3c(2a+b) \)
Answer:
\( 2b(2a+b) - 3c(2a+b) = (2a+b)(2b-3c) \)
In simple words: We pull out the common binomial factor \( (2a+b) \).

Exam Tip: When factoring binomials, write the common binomial first followed by the remaining terms in a second bracket.

Question 14. Factorise : \( 12abc - 6a^2b^2c^2 + 3a^3b^3c^3 \)
Answer:
\( 12abc - 6a^2b^2c^2 + 3a^3b^3c^3 = 3abc(4 - 2abc + a^2b^2c^2) \)
In simple words: We divide all terms by the greatest common factor \( 3abc \) and place it outside.

Exam Tip: Pay careful attention to dividing each power of the variable systematically by the factored terms.

Question 15. Factorise : \( 4x(3x-2y) - 2y(3x-2y) \)
Answer:
\( 4x(3x-2y) - 2y(3x-2y) \)
\( = (3x-2y)(4x-2y) \)
\( = (3x-2y) \times 2(2x-y) \)
\( = 2(3x-2y)(2x-y) \)
In simple words: First we factor out the common bracket \( (3x-2y) \), then we pull out the common factor of 2 from the other bracket.

Exam Tip: Always check if any remaining binomial factors can be factorised further to ensure complete factorisation.

Question 16. Factorise : \( (a+2b)(3a+b) - (a+b)(a+2b) + (a+2b)^2 \)
Answer:
\( (a+2b)(3a+b) - (a+b)(a+2b) + (a+2b)^2 \)
\( = (a+2b)(3a+b - (a+b) + (a+2b)) \)
\( = (a+2b)(3a+b - a - b + a + 2b) \)
\( = (a+2b)(3a+2b) \)
In simple words: We take the bracket \( (a+2b) \) out, and then combine the remaining terms together.

Exam Tip: When simplifying terms after factoring, be very careful to distribute negative signs across parentheses correctly.

Question 17. Factorise : \( 6xy(a^2+b^2) + 8yz(a^2+b^2) - 10xz(a^2+b^2) \)
Answer:
\( 6xy(a^2+b^2) + 8yz(a^2+b^2) - 10xz(a^2+b^2) \)
Since the greatest common factor of the coefficients 6, 8, and 10 is 2, and the binomial \( (a^2+b^2) \) is common, we have:
\( = 2(a^2+b^2)(3xy + 4yz - 5xz) \)
In simple words: We factor out the common number 2 and the common bracket \( (a^2+b^2) \).

Exam Tip: Factorise the numerical coefficients and the algebraic expressions step-by-step to avoid errors.

Exercise 13(B)

Question 1. Factorise : \( a^2 + ax + ab + bx \)
Answer:
\( a^2 + ax + ab + bx \)
\( = (a^2+ax) + (ab+bx) \)
\( = a(a+x) + b(a+x) \)
\( = (a+x)(a+b) \)
In simple words: We group the first two terms and the last two terms, factor each group, and then take out the common bracket.

Exam Tip: Grouping is successful if the binomial factor obtained in both groups is identical.

Question 2. Factorise : \( a^2 - ab - ca + bc \)
Answer:
\( a^2 - ab - ca + bc \)
\( = a(a-b) - c(a-b) \)
\( = (a-b)(a-c) \)
In simple words: We group the terms in pairs and factorise each pair to find a common bracket \( (a-b) \).

Exam Tip: Watch out for the negative sign when factoring out \( -c \); it changes the sign of the term inside the bracket.

Question 3. Factorise : \( ab - 2b + a^2 - 2a \)
Answer:
\( ab - 2b + a^2 - 2a \)
\( = b(a-2) + a(a-2) \)
\( = (a-2)(b+a) \)
In simple words: Grouping the first two and the last two terms lets us factor out \( (a-2) \).

Exam Tip: Reordering terms may sometimes make grouping easier, though here direct pairing works directly.

Question 4. Factorise : \( a^3 - a^2 + a - 1 \)
Answer:
\( a^3 - a^2 + a - 1 \)
\( = a^2(a-1) + 1(a-1) \)
\( = (a-1)(a^2+1) \)
In simple words: We group the terms as pairs, factoring out \( a^2 \) from the first and \( 1 \) from the second.

Exam Tip: Remember that when a group has no visible common factor, you can always factor out \( 1 \).

Question 5. Factorise : \( 2a - 4b - xa + 2bx \)
Answer:
\( 2a - 4b - xa + 2bx \)
\( = 2(a-2b) - x(a-2b) \)
\( = (a-2b)(2-x) \)
In simple words: We pair the terms to extract \( 2 \) and \( -x \), which reveals the common bracket \( (a-2b) \).

Exam Tip: Factoring out a negative sign transforms \( +2bx \) into \( -2b \) inside the parenthesis.

Question 6. Factorise : \( xy - ay - ax + a^2 + bx - ab \)
Answer:
\( xy - ay - ax + a^2 + bx - ab \)
\( = y(x-a) - a(x-a) + b(x-a) \)
\( = (x-a)(y-a+b) \)
In simple words: We pair the six terms into three groups of two, factoring each to get the common bracket \( (x-a) \).

Exam Tip: Expressions with six terms are typically grouped in pairs of three or three pairs of two.

Question 7. Factorise : \( 3x^5 - 6x^4 - 2x^3 + 4x^2 + x - 2 \)
Answer:
\( 3x^5 - 6x^4 - 2x^3 + 4x^2 + x - 2 \)
\( = 3x^4(x-2) - 2x^2(x-2) + 1(x-2) \)
\( = (x-2)(3x^4-2x^2+1) \)
In simple words: We group these six terms into three sets of two, revealing \( (x-2) \) as the common bracket.

Exam Tip: Ensure the remaining trinomial cannot be simplified further before completing your solution.

Question 8. Factorise : \( -x^2y - x + 3xy + 3 \)
Answer:
\( -x^2y - x + 3xy + 3 \)
On rearranging and grouping:
\( = 3 - x + 3xy - x^2y \)
\( = 1(3-x) + xy(3-x) \)
\( = (3-x)(1+xy) \)
\( = (xy+1)(3-x) \)
In simple words: We rearrange the terms first so we can easily group them and find the common bracket \( (3-x) \).

Exam Tip: Do not hesitate to rearrange terms if grouping them in their original order does not produce a common binomial.

Question 9. Factorise : \( 6a^2 - 3a^2b - bc^2 + 2c^2 \)
Answer:
\( 6a^2 - 3a^2b - bc^2 + 2c^2 \)
On rearranging the terms:
\( = 6a^2 - 3a^2b + 2c^2 - bc^2 \)
\( = 3a^2(2-b) + c^2(2-b) \)
\( = (2-b)(3a^2+c^2) \)
In simple words: We rearrange and group the terms in pairs, factoring out \( 3a^2 \) and \( c^2 \) to get the common factor \( (2-b) \).

Exam Tip: Pay attention to the order of terms inside brackets to ensure they match exactly, e.g., \( (2-b) \).

Question 10. Factorise : \( 3a^2b - 12a^2 - 9b + 36 \)
Answer:
\( 3a^2b - 12a^2 - 9b + 36 \)
Grouping the terms:
\( = 3a^2(b-4) - 9(b-4) \)
\( = (b-4)(3a^2-9) \)
\( = 3(b-4)(a^2-3) \)
In simple words: We group the terms to find the common bracket \( (b-4) \), and then factor out 3 from the remaining binomial.

Exam Tip: Always check if numerical terms can be factored out from any binomial factors in your final expression.

Question 11. Factorise : \( x^2 - (a-3)x - 3a \)
Answer:
\( x^2 - (a-3)x - 3a \)
\( = x^2 - ax + 3x - 3a \)
\( = x(x-a) + 3(x-a) \)
\( = (x-a)(x+3) \)
In simple words: First, we multiply out the bracket, then group the terms in pairs to factorise.

Exam Tip: Expanding parentheses first is often necessary before you can group terms successfully.

Question 12. Factorise : \( x^2 - (b-2)x - 2b \)
Answer:
\( x^2 - (b-2)x - 2b \)
\( = x^2 - bx + 2x - 2b \)
\( = x(x-b) + 2(x-b) \)
\( = (x-b)(x+2) \)
In simple words: Expand the middle term first, then group the four terms in pairs to find the common factor \( (x-b) \).

Exam Tip: Watch the signs carefully when distributing the negative sign over the binomial bracket.

Question 13. Factorise : \( a(b-c) - d(c-b) \)
Answer:
\( a(b-c) - d(c-b) \)
\( = a(b-c) + d(b-c) \)
\( = (b-c)(a+d) \)
In simple words: We rewrite \( (c-b) \) as \( -(b-c) \), changing the sign outside, which gives us the common bracket \( (b-c) \).

Exam Tip: Use the identity \( (y-x) = -(x-y) \) to create identical binomial factors.

Question 14. Factorise : \( ab^2 - (a-c)b - c \)
Answer:
\( ab^2 - (a-c)b - c \)
\( = ab^2 - ab + bc - c \)
\( = ab(b-1) + c(b-1) \)
\( = (b-1)(ab+c) \)
In simple words: We expand the middle term first, then group the terms in pairs to find the common bracket \( (b-1) \).

Exam Tip: Keep track of the literal coefficients when grouping and pulling out common factors.

Question 15. Factorise : \( (a^2-b^2)c + (b^2-c^2)a \)
Answer:
\( (a^2-b^2)c + (b^2-c^2)a \)
\( = a^2c - b^2c + ab^2 - ac^2 \)
\( = a^2c - ac^2 + ab^2 - b^2c \)
\( = ac(a-c) + b^2(a-c) \)
\( = (a-c)(ac+b^2) \)
In simple words: We multiply the terms out, rearrange them, and group them to find the common bracket \( (a-c) \).

Exam Tip: When direct grouping does not work, fully expand the expression and regroup the terms.

Question 16. Factorise : \( a^3 - a^2 - ab + a + b - 1 \)
Answer:
\( a^3 - a^2 - ab + a + b - 1 \)
Rearranging the terms:
\( = a^3 - a^2 - ab + b + a - 1 \)
\( = a^2(a-1) - b(a-1) + 1(a-1) \)
\( = (a-1)(a^2-b+1) \)
In simple words: We rearrange the terms and group them in pairs of three, factoring out the common bracket \( (a-1) \).

Exam Tip: Grouping can be done in pairs of two for a six-term expression. Be consistent with signs.

Question 17. Factorise : \( ab(c^2+d^2) - a^2cd - b^2cd \)
Answer:
\( ab(c^2+d^2) - a^2cd - b^2cd \)
\( = abc^2 + abd^2 - a^2cd - b^2cd \)
Rearranging the terms:
\( = abc^2 - a^2cd - b^2cd + abd^2 \)
\( = ac(bc - ad) - bd(bc - ad) \)
\( = (bc-ad)(ac-bd) \)
In simple words: Expand the first term, rearrange, and then group in pairs to find the common factor \( (bc-ad) \).

Exam Tip: Rearranging terms is key to solving this; group terms with common variables together.

Question 18. Factorise : \( 2ab^2 - aby + 2cby - cy^2 \)
Answer:
\( 2ab^2 - aby + 2cby - cy^2 \)
Rearranging the terms:
\( = 2ab^2 + 2cby - aby - cy^2 \)
\( = 2b(ab + cy) - y(ab + cy) \)
\( = (ab+cy)(2b-y) \)
In simple words: We rearrange the terms and group them to extract \( 2b \) and \( -y \), leaving the common bracket \( (ab+cy) \).

Exam Tip: Rearranging terms so that coefficients with common factors (like 2) are together makes grouping more straightforward.

Question 19. Factorise : \( ax + 2bx + 3cx - 3a - 6b - 9c \)
Answer:
\( ax + 2bx + 3cx - 3a - 6b - 9c \)
Grouping the terms:
\( = x(a + 2b + 3c) - 3(a + 2b + 3c) \)
\( = (a+2b+3c)(x-3) \)
In simple words: We group the first three terms together and the last three terms together, allowing us to factor out \( (a+2b+3c) \).

Exam Tip: Look for common factors in groups of three when dealing with six terms.

Question 20. Factorise : \( 2ab^2c - 2a + 3b^3c - 3b - 4b^2c^2 + 4c \)
Answer:
\( 2ab^2c - 2a + 3b^3c - 3b - 4b^2c^2 + 4c \)
Grouping the terms in pairs:
\( = 2a(b^2c - 1) + 3b(b^2c - 1) - 4c(b^2c - 1) \)
\( = (b^2c-1)(2a+3b-4c) \)
In simple words: We group the six terms into three pairs, which reveals the common bracket \( (b^2c-1) \).

Exam Tip: Be careful when factoring out \( -4c \) from the last pair as it flips the sign inside the parenthesis.

Exercise 13(C)

Question 1. Factorise : 16 - 9x^2
Answer:
\( 16 - 9x^2 = (4)^2 - (3x)^2 = (4 + 3x)(4 - 3x) \)
In simple words: We write the terms as squares, \( 4^2 \) and \( (3x)^2 \), and then apply the difference of squares rule.

Exam Tip: Always identify the terms as perfect squares before using the identity \( a^2 - b^2 \).

Question 2. Factorise : 1 - 100a^2
Answer:
\( 1 - 100a^2 = (1)^2 - (10a)^2 = (1 + 10a)(1 - 10a) \)
In simple words: We write the terms as \( 1^2 \) and \( (10a)^2 \) and apply the difference of squares formula.

Exam Tip: Treat 1 as \( 1^2 \) to easily apply the difference of squares formula.

Question 3. Factorise : 4x^2 - 81y^2
Answer:
\( 4x^2 - 81y^2 = (2x)^2 - (9y)^2 = (2x + 9y)(2x - 9y) \)
In simple words: We convert the terms into perfect squares and then apply the difference of squares rule.

Exam Tip: Remember to square the numerical coefficients as well as the variables inside the brackets.

Question 4. Factorise : \( \frac{4}{25} - 25b^2 \)
Answer:
\( \frac{4}{25} - 25b^2 = \left(\frac{2}{5}\right)^2 - (5b)^2 = \left(\frac{2}{5} + 5b\right)\left(\frac{2}{5} - 5b\right) \)
In simple words: We write the fraction and the other term as squares, then use the difference of squares formula.

Exam Tip: Fractions follow the exact same squaring rules as integers; find the square root of both numerator and denominator.

Question 5. Factorise : \( (a+2b)^2 - a^2 \)
Answer:
\( (a+2b)^2 - a^2 = (a+2b)^2 - (a)^2 \)
\( = (a+2b+a)(a+2b-a) \)
\( = (2a+2b)(2b) \)
\( = 2(a+b)(2b) \)
\( = 4b(a+b) \)
In simple words: Using the difference of squares, we add and subtract the terms and then simplify the final expression.

Exam Tip: Simplify the resulting brackets fully by factoring out common numerical factors.

Question 6. Factorise : \( (5a-3b)^2 - 16b^2 \)
Answer:
\( (5a-3b)^2 - 16b^2 = (5a-3b)^2 - (4b)^2 \)
\( = (5a-3b+4b)(5a-3b-4b) \)
\( = (5a+b)(5a-7b) \)
In simple words: We write \( 16b^2 \) as \( (4b)^2 \), apply the identity, and simplify inside the brackets.

Exam Tip: Ensure you combine like terms (such as \( -3b \) and \( 4b \)) inside the parentheses after applying the formula.

Question 7. Factorise : \( a^4 - (a^2-3b^2)^2 \)
Answer:
\( a^4 - (a^2-3b^2)^2 \)
\( = (a^2)^2 - (a^2-3b^2)^2 \)
\( = (a^2 + a^2 - 3b^2)(a^2 - (a^2 - 3b^2)) \)
\( = (2a^2 - 3b^2)(a^2 - a^2 + 3b^2) \)
\( = (2a^2 - 3b^2)(3b^2) \)
\( = 3b^2(2a^2 - 3b^2) \)
In simple words: We express \( a^4 \) as \( (a^2)^2 \) and apply the difference of squares formula, then simplify the brackets.

Exam Tip: Watch out for sign changes when subtracting a binomial inside the bracket.

Question 8. Factorise : \( (5a-2b)^2 - (2a-b)^2 \)
Answer:
\( (5a-2b)^2 - (2a-b)^2 \)
\( = (5a-2b + 2a-b)(5a-2b - (2a-b)) \)
\( = (7a-3b)(5a-2b-2a+b) \)
\( = (7a-3b)(3a-b) \)
In simple words: We apply the formula \( x^2 - y^2 = (x+y)(x-y) \) to both bracketed terms and combine like terms.

Exam Tip: Expanding the brackets first is unnecessary and takes longer; use the difference of squares identity directly.

Question 9. Factorise : \( 1 - 25(a+b)^2 \)
Answer:
\( 1 - 25(a+b)^2 \)
\( = (1)^2 - [5(a+b)]^2 \)
\( = [1 + 5(a+b)][1 - 5(a+b)] \)
\( = (1 + 5a + 5b)(1 - 5a - 5b) \)
In simple words: We write the expression as a difference of two squares and distribute the 5 inside each bracket.

Exam Tip: Don't forget to multiply the factor outside the parenthesis with both terms inside the bracket.

Question 10. Factorise : \( 4(2a+b)^2 - (a-b)^2 \)
Answer:
\( 4(2a+b)^2 - (a-b)^2 \)
\( = [2(2a+b)]^2 - (a-b)^2 \)
\( = [2(2a+b) + (a-b)][2(2a+b) - (a-b)] \)
\( = (4a+2b+a-b)(4a+2b-a+b) \)
\( = (5a+b)(3a+3b) \)
\( = 3(5a+b)(a+b) \)
In simple words: We write the terms as perfect squares, apply the formula, simplify the brackets, and pull out any common factors.

Exam Tip: Always factor out constants like 3 from \( (3a+3b) \) to write the answer in its simplest factored form.

Question 11. Factorise : \( 25(2x+y)^2 - 16(x-y)^2 \)
Answer:
\( 25(2x+y)^2 - 16(x-y)^2 \)
\( = [5(2x+y)]^2 - [4(x-y)]^2 \)
\( = (10x+5y)^2 - (4x-4y)^2 \)
\( = (10x+5y + 4x-4y)(10x+5y - (4x-4y)) \)
\( = (14x+y)(10x+5y-4x+4y) \)
\( = (14x+y)(6x+9y) \)
\( = 3(14x+y)(2x+3y) \)
In simple words: We express both terms as squares, apply the difference of squares identity, and then factor out 3 from the second term.

Exam Tip: Simplify and factorise completely; looking for common factors at the very end is a critical step.

Question 12. Factorise : \( 49(x-y)^2 - 9(2x+y)^2 \)
Answer:
\( 49(x-y)^2 - 9(2x+y)^2 \)
\( = [7(x-y)]^2 - [3(2x+y)]^2 \)
\( = (7x-7y)^2 - (6x+3y)^2 \)
\( = (7x-7y + 6x+3y)(7x-7y - (6x+3y)) \)
\( = (13x-4y)(7x-7y-6x-3y) \)
\( = (13x-4y)(x-10y) \)
In simple words: Express each term as a perfect square, use the difference of squares formula, and group the remaining like terms.

Exam Tip: Be extremely careful when distributing negative signs through brackets during subtraction.

Question 13. Evaluate : \( \left(6\frac{2}{3}\right)^2 - \left(2\frac{1}{3}\right)^2 \)
Answer:
\( \left(6\frac{2}{3}\right)^2 - \left(2\frac{1}{3}\right)^2 \)
Converting the mixed fractions into improper fractions:
\( = \left(\frac{20}{3}\right)^2 - \left(\frac{7}{3}\right)^2 \)
Using the formula \( a^2 - b^2 = (a+b)(a-b) \):
\( = \left(\frac{20}{3} + \frac{7}{3}\right)\left(\frac{20}{3} - \frac{7}{3}\right) \)
\( = \left(\frac{27}{3}\right)\left(\frac{13}{3}\right) \)
\( = 9 \times \frac{13}{3} \)
\( = 3 \times 13 \)
\( = 39 \)
In simple words: Convert mixed numbers to improper fractions, use the difference of squares formula, and simplify the fraction multiplication.

Exam Tip: Working with improper fractions and using algebraic identities is much faster and less error-prone than squaring mixed numbers directly.

Question 14. Evaluate : \( \left(7\frac{3}{10}\right)^2 - \left(2\frac{1}{10}\right)^2 \)
Answer:
\( \left(7\frac{3}{10}\right)^2 - \left(2\frac{1}{10}\right)^2 \)
Converting the mixed numbers to improper fractions:
\( = \left(\frac{73}{10}\right)^2 - \left(\frac{21}{10}\right)^2 \)
Using the difference of squares formula:
\( = \left(\frac{73}{10} + \frac{21}{10}\right)\left(\frac{73}{10} - \frac{21}{10}\right) \)
\( = \left(\frac{94}{10}\right)\left(\frac{52}{10}\right) \)
\( = \left(\frac{47}{5}\right)\left(\frac{26}{5}\right) \)
\( = \frac{1222}{25} \)
\( = 48\frac{22}{25} \)
In simple words: Convert mixed numbers to improper fractions, apply the difference of squares identity, and then convert the improper fraction back to a mixed number.

Exam Tip: Simplify fractions to their lowest terms before multiplying to make calculation easier.

Question 15. Evaluate : \( (0.7)^2 - (0.3)^2 \)
Answer:
\( (0.7)^2 - (0.3)^2 \)
Using the difference of squares:
\( = (0.7+0.3)(0.7-0.3) \)
\( = 1 \times 0.4 \)
\( = 0.4 \)
In simple words: We add 0.7 and 0.3 to get 1, subtract 0.3 from 0.7 to get 0.4, and multiply the results.

Exam Tip: Decimal calculations become extremely simple when using the difference of squares, as one of the terms often sums to 1 or a whole number.

Question 16. Evaluate : \( (4.5)^2 - (1.5)^2 \)
Answer:
\( (4.5)^2 - (1.5)^2 \)
Using the difference of squares:
\( = (4.5+1.5)(4.5-1.5) \)
\( = 6 \times 3 \)
\( = 18 \)
In simple words: Adding the decimals gives 6 and subtracting them gives 3. Multiplying them gives 18.

Exam Tip: Always apply the difference of squares identity to evaluate differences of squares of decimals to save time.

Question 17. Factorise : \( 75(x + y)^2 - 48(x - y)^2 \)
Answer:
We begin by taking out the common factor \( 3 \) from both terms:
\( 75(x + y)^2 - 48(x - y)^2 \)
\( = 3[25(x + y)^2 - 16(x - y)^2] \)
We can rewrite the terms inside the bracket as squares:
\( = 3[\{5(x + y)\}^2 - \{4(x - y)\}^2] \)
By applying the algebraic identity \( a^2 - b^2 = (a + b)(a - b) \), we obtain:
\( = 3[5(x + y) + 4(x - y)][5(x + y) - 4(x - y)] \)
Now, expand the terms inside the brackets:
\( = 3[5x + 5y + 4x - 4y][5x + 5y - 4x + 4y] \)
Simplifying the expressions gives:
\( = 3(9x + y)(x + 9y) \)
In simple words: First, pull out the common number 3. Then, use the difference of squares rule to break the expression into two simpler parts and simplify them.
Exam Tip: Don't forget to look for a common factor first, as it simplifies the expression before applying algebraic identities.

Question 18. Factorise : \( a^2 + 4a + 4 - b^2 \)
Answer:
We can group the first three terms to form a perfect square trinomial:
\( a^2 + 4a + 4 - b^2 \)
\( = (a^2 + 2 \times a \times 2 + 2^2) - b^2 \)
Since \( (x + y)^2 = x^2 + 2xy + y^2 \), we can write:
\( = (a + 2)^2 - b^2 \)
Now, using the difference of squares formula \( x^2 - y^2 = (x + y)(x - y) \), we get:
\( = (a + 2 + b)(a + 2 - b) \)
Rearranging the terms in a standard order:
\( = (a + b + 2)(a - b + 2) \)
In simple words: Group the first three terms to make a perfect square, \( (a + 2)^2 \). Then, use the difference of squares identity to finish factorising.
Exam Tip: Rearranging the final factors in alphabetical order is a good practice that makes your final answer look neat and professional.

Question 19. Factorise : \( a^2 - b^2 - 2b - 1 \)
Answer:
Group the last three terms by factoring out a minus sign to get a perfect square:
\( a^2 - b^2 - 2b - 1 \)
\( = a^2 - (b^2 + 2b + 1) \)
Since the terms in the parentheses represent a perfect square, \( (b + 1)^2 \), we have:
\( = a^2 - (b + 1)^2 \)
Using the difference of squares formula \( x^2 - y^2 = (x + y)(x - y) \):
\( = [a + (b + 1)][a - (b + 1)] \)
\( = (a + b + 1)(a - b - 1) \)
In simple words: Put the last three terms in a bracket with a minus sign outside. This turns them into a perfect square, \( (b + 1)^2 \). Finally, use the difference of squares formula.
Exam Tip: Be very careful with signs when putting terms inside brackets preceded by a negative sign; all signs must change.

Question 20. Factorise : \( x^2 + 6x + 9 - 4y^2 \)
Answer:
We can group the first three terms as they form a perfect square trinomial:
\( x^2 + 6x + 9 - 4y^2 \)
\( = (x^2 + 2 \times x \times 3 + 3^2) - (2y)^2 \)
Applying the perfect square identity \( (a+b)^2 = a^2 + 2ab + b^2 \), we write:
\( = (x + 3)^2 - (2y)^2 \)
Now, applying the difference of squares formula \( a^2 - b^2 = (a + b)(a - b) \), we get:
\( = (x + 3 + 2y)(x + 3 - 2y) \)
Rearranging the variables to the front:
\( = (x + 2y + 3)(x - 2y + 3) \)
In simple words: The first three terms can be written together as \( (x + 3)^2 \). This leaves us with a difference of two squares, which we can split into two brackets and then simplify.
Exam Tip: Recognizing that \( 4y^2 \) can be written as \( (2y)^2 \) is key to correctly applying the difference of two squares identity.

Exercise 13(D)

Question 1. Factorise : \( x^2 + 6x + 8 \)
Answer:
We can factorise this quadratic trinomial by splitting the middle term. We look for two numbers that multiply to \( 8 \) and add up to \( 6 \). These numbers are \( 4 \) and \( 2 \):
\( x^2 + 6x + 8 \)
\( = x^2 + 4x + 2x + 8 \)
Group the terms and factor out the common factors:
\( = x(x + 4) + 2(x + 4) \)
Factor out the common binomial \( (x + 4) \):
\( = (x + 4)(x + 2) \)
In simple words: Find two numbers that multiply to 8 and add up to 6, which are 4 and 2. Split the middle term and group them to find the common factors.
Exam Tip: When splitting the middle term, always verify that the product of the two split coefficients equals the constant term.

Question 2. Factorise : \( x^2 + 4x + 3 \)
Answer:
To split the middle term, we find two numbers whose product is \( 3 \) and sum is \( 4 \). These are \( 3 \) and \( 1 \):
\( x^2 + 4x + 3 \)
\( = x^2 + 3x + x + 3 \)
Factor by grouping:
\( = x(x + 3) + 1(x + 3) \)
Taking out the common bracket \( (x + 3) \):
\( = (x + 3)(x + 1) \)
In simple words: Split \( 4x \) into \( 3x \) and \( 1x \), because 3 and 1 multiply to 3. Then group the terms to factorise.
Exam Tip: If no number is visibly common in the second group, remember that \( 1 \) is always a common factor.

Question 3. Factorise : \( a^2 + 5a + 6 \)
Answer:
We find two numbers that multiply to \( 6 \) and add to \( 5 \). These are \( 3 \) and \( 2 \):
\( a^2 + 5a + 6 \)
\( = a^2 + 3a + 2a + 6 \)
Grouping the terms and factoring them:
\( = a(a + 3) + 2(a + 3) \)
Taking the common factor \( (a + 3) \) out:
\( = (a + 3)(a + 2) \)
In simple words: Split \( 5a \) into \( 3a + 2a \), since 3 and 2 add up to 5 and multiply to 6. Then factorise by grouping.
Exam Tip: Ensure that the binomials in the parentheses match exactly before factoring them out.

Question 4. Factorise : \( a^2 - 5a + 6 \)
Answer:
We need two numbers that multiply to \( 6 \) and add up to \( -5 \). These numbers are \( -3 \) and \( -2 \):
\( a^2 - 5a + 6 \)
\( = a^2 - 3a - 2a + 6 \)
Factoring by grouping:
\( = a(a - 3) - 2(a - 3) \)
Taking out the common binomial \( (a - 3) \):
\( = (a - 3)(a - 2) \)
In simple words: Use \( -3 \) and \( -2 \) because they add up to \( -5 \) and multiply to \( 6 \). Group the terms and factorise.
Exam Tip: Pay attention to the sign when factoring out a negative number; here, factoring out \( -2 \) changes the sign of \( +6 \) to \( -3 \).

Question 5. Factorise : \( a^2 + 5a - 6 \)
Answer:
We search for two numbers whose product is \( -6 \) and sum is \( 5 \). These are \( 6 \) and \( -1 \):
\( a^2 + 5a - 6 \)
\( = a^2 + 6a - a - 6 \)
Factor by grouping:
\( = a(a + 6) - 1(a + 6) \)
Taking out the common factor \( (a + 6) \):
\( = (a + 6)(a - 1) \)
In simple words: Choose \( 6 \) and \( -1 \) because their sum is \( 5 \) and product is \( -6 \). Split the middle term and factor.
Exam Tip: A negative constant term like \( -6 \) means the two factors must have opposite signs.

Question 6. Factorise : \( x^2 + 5xy + 4y^2 \)
Answer:
We can split the middle term \( 5xy \) into \( 4xy \) and \( xy \) because their coefficients multiply to \( 4 \) and add up to \( 5 \):
\( x^2 + 5xy + 4y^2 \)
\( = x^2 + 4xy + xy + 4y^2 \)
Grouping and factoring out the common terms:
\( = x(x + 4y) + y(x + 4y) \)
Factoring out the common bracket \( (x + 4y) \):
\( = (x + 4y)(x + y) \)
In simple words: Split \( 5xy \) into \( 4xy \) and \( 1xy \). Then group the first two and last two terms to pull out the common parts.
Exam Tip: Treat the variable \( y \) as part of the coefficient when factoring trinomials with two variables.

Question 7. Factorise : \( a^2 - 3a - 40 \)
Answer:
We find two numbers that multiply to \( -40 \) and add up to \( -3 \). These are \( -8 \) and \( 5 \):
\( a^2 - 3a - 40 \)
\( = a^2 - 8a + 5a - 40 \)
Grouping and factoring:
\( = a(a - 8) + 5(a - 8) \)
Factoring out the common binomial \( (a - 8) \):
\( = (a - 8)(a + 5) \)
In simple words: Use \( -8 \) and \( 5 \) since they multiply to \( -40 \) and add to \( -3 \). Group the terms to complete the factorisation.
Exam Tip: If the product is negative and the sum is negative, the larger of the two numbers must be negative.

Question 8. Factorise : \( x^2 - x - 72 \)
Answer:
We look for two numbers that have a product of \( -72 \) and a sum of \( -1 \). These are \( -9 \) and \( 8 \):
\( x^2 - x - 72 \)
\( = x^2 - 9x + 8x - 72 \)
Grouping the terms:
\( = x(x - 9) + 8(x - 9) \)
Factoring out \( (x - 9) \):
\( = (x - 9)(x + 8) \)
In simple words: Split the middle term into \( -9x + 8x \). Group the terms and factor out the common brackets.
Exam Tip: Always double-check your arithmetic when finding factors of larger numbers like \( 72 \).

Question 9. Factorise : \( x^2 - 10xy + 24y^2 \)
Answer:
We need two numbers that multiply to \( 24 \) and add to \( -10 \). These numbers are \( -6 \) and \( -4 \):
\( x^2 - 10xy + 24y^2 \)
\( = x^2 - 6xy - 4xy + 24y^2 \)
Factoring by grouping:
\( = x(x - 6y) - 4y(x - 6y) \)
Taking out the common term \( (x - 6y) \):
\( = (x - 6y)(x - 4y) \)
In simple words: Split \( -10xy \) into \( -6xy - 4xy \). Group and take out the common factors to finish.
Exam Tip: When both the middle term is negative and the last term is positive, both split terms must be negative.

Question 10. Factorise : \( 2a^2 + 7a + 6 \)
Answer:
Here, the coefficient of \( a^2 \) is \( 2 \). We look for two numbers that multiply to \( 2 \times 6 = 12 \) and add up to \( 7 \). These numbers are \( 4 \) and \( 3 \):
\( 2a^2 + 7a + 6 \)
\( = 2a^2 + 4a + 3a + 6 \)
Factoring by grouping:
\( = 2a(a + 2) + 3(a + 2) \)
Taking out the common binomial factor \( (a + 2) \):
\( = (a + 2)(2a + 3) \)
In simple words: Since there is a 2 in front of \( a^2 \), multiply it by 6 to get 12. Find two numbers that multiply to 12 and add to 7, which are 4 and 3. Split the middle term and factor.
Exam Tip: Remember to multiply the leading coefficient by the constant term to find the target product for splitting the middle term.

Question 11. Factorise : \( 3a^2 - 5a + 2 \)
Answer:
We need two numbers that multiply to \( 3 \times 2 = 6 \) and add to \( -5 \). These are \( -3 \) and \( -2 \):
\( 3a^2 - 5a + 2 \)
\( = 3a^2 - 3a - 2a + 2 \)
Grouping the terms:
\( = 3a(a - 1) - 2(a - 1) \)
Factoring out the common bracket \( (a - 1) \):
\( = (a - 1)(3a - 2) \)
In simple words: Split \( -5a \) into \( -3a - 2a \). Group them to factorise out the common brackets.
Exam Tip: Always make sure that the signs inside both binomial brackets are identical before grouping.

Question 12. Factorise : \( 7b^2 - 8b + 1 \)
Answer:
We look for two numbers that multiply to \( 7 \times 1 = 7 \) and add to \( -8 \). These are \( -7 \) and \( -1 \):
\( 7b^2 - 8b + 1 \)
\( = 7b^2 - 7b - b + 1 \)
Grouping and factoring:
\( = 7b(b - 1) - 1(b - 1) \)
Factoring out the common binomial \( (b - 1) \):
\( = (b - 1)(7b - 1) \)
In simple words: Split \( -8b \) into \( -7b - 1b \). Then group and pull out the common factor of \( (b-1) \).
Exam Tip: Be careful to include the \( -1 \) when factoring out the negative sign from the second group.

Question 13. Factorise : \( 2a^2 - 17ab + 26b^2 \)
Answer:
Here we need two numbers that multiply to \( 2 \times 26 = 52 \) and add up to \( -17 \). These numbers are \( -13 \) and \( -4 \):
\( 2a^2 - 17ab + 26b^2 \)
\( = 2a^2 - 13ab - 4ab + 26b^2 \)
Factoring by grouping:
\( = a(2a - 13b) - 2b(2a - 13b) \)
Factoring out the common expression \( (2a - 13b) \):
\( = (2a - 13b)(a - 2b) \)
In simple words: Split the middle term into \( -13ab - 4ab \). Group the terms to find the common brackets and complete the factorisation.
Exam Tip: With variables in both terms, keep track of \( a \) and \( b \) in each step of grouping.

Question 14. Factorise : \( 2x^2 + xy - 6y^2 \)
Answer:
We need two numbers that multiply to \( 2 \times (-6) = -12 \) and add up to \( 1 \). These are \( 4 \) and \( -3 \):
\( 2x^2 + xy - 6y^2 \)
\( = 2x^2 + 4xy - 3xy - 6y^2 \)
Grouping the terms and factoring them:
\( = 2x(x + 2y) - 3y(x + 2y) \)
Factor out the common bracket \( (x + 2y) \):
\( = (x + 2y)(2x - 3y) \)
In simple words: Split \( xy \) into \( 4xy - 3xy \). Group the terms to pull out the common parts from each pair.
Exam Tip: Since the constant term is negative, the split factors must have opposite signs.

Question 15. Factorise : \( 4c^2 + 3c - 10 \)
Answer:
We look for two numbers that multiply to \( 4 \times (-10) = -40 \) and add to \( 3 \). These are \( 8 \) and \( -5 \):
\( 4c^2 + 3c - 10 \)
\( = 4c^2 + 8c - 5c - 10 \)
Factoring by grouping:
\( = 4c(c + 2) - 5(c + 2) \)
Taking out the common binomial factor \( (c + 2) \):
\( = (c + 2)(4c - 5) \)
In simple words: Find two numbers that multiply to \( -40 \) and add to 3, which are 8 and -5. Split \( 3c \), group, and factor.
Exam Tip: Verify that the product of the first term's coefficient and the last term matches the product of your split numbers before continuing.

Question 16. Factorise : \( 14x^2 + x - 3 \)
Answer:
We need two numbers that multiply to \( 14 \times (-3) = -42 \) and add up to \( 1 \). These are \( 7 \) and \( -6 \):
\( 14x^2 + x - 3 \)
\( = 14x^2 + 7x - 6x - 3 \)
Factor by grouping:
\( = 7x(2x + 1) - 3(2x + 1) \)
Taking out the common factor \( (2x + 1) \):
\( = (2x + 1)(7x - 3) \)
In simple words: Split \( x \) into \( 7x - 6x \). Group the terms to find the common brackets and complete the factorisation.
Exam Tip: Look for the largest common factor in each group, such as \( 7x \) in the first group, to simplify easily.

Question 17. Factorise : \( 6 + 7b - 3b^2 \)
Answer:
We need two numbers that multiply to \( 6 \times (-3) = -18 \) and add to \( 7 \). These are \( 9 \) and \( -2 \):
\( 6 + 7b - 3b^2 \)
\( = 6 + 9b - 2b - 3b^2 \)
Grouping the terms and factoring:
\( = 3(2 + 3b) - b(2 + 3b) \)
Factor out the common binomial \( (2 + 3b) \):
\( = (2 + 3b)(3 - b) \)
In simple words: Split \( 7b \) into \( 9b - 2b \). Group the terms and pull out the common parts to factorise.
Exam Tip: Don't worry if the constant term is at the front; the splitting method works exactly the same way.

Question 18. Factorise : \( 5 + 7x - 6x^2 \)
Answer:
We search for two numbers that multiply to \( 5 \times (-6) = -30 \) and add up to \( 7 \). These are \( 10 \) and \( -3 \):
\( 5 + 7x - 6x^2 \)
\( = 5 + 10x - 3x - 6x^2 \)
Factoring by grouping:
\( = 5(1 + 2x) - 3x(1 + 2x) \)
Taking out the common bracket \( (1 + 2x) \):
\( = (1 + 2x)(5 - 3x) \)
In simple words: Split \( 7x \) into \( 10x - 3x \). Group the first two and the last two terms to factorise.
Exam Tip: Be careful when pulling out a negative sign along with a variable, like \( -3x \), to ensure the signs inside the bracket are correct.

Question 19. Factorise : \( 4 + y - 14y^2 \)
Answer:
We need two numbers that multiply to \( 4 \times (-14) = -56 \) and add to \( 1 \). These are \( 8 \) and \( -7 \):
\( 4 + y - 14y^2 \)
\( = 4 + 8y - 7y - 14y^2 \)
Grouping the terms:
\( = 4(1 + 2y) - 7y(1 + 2y) \)
Factor out the common binomial \( (1 + 2y) \):
\( = (1 + 2y)(4 - 7y) \)
In simple words: Split \( y \) into \( 8y - 7y \). Group them to factor out the common parts.
Exam Tip: Always make sure that both binomial groups are identical before taking them out as a common factor.

Question 20. Factorise : \( 5 + 3a - 14a^2 \)
Answer:
We search for two numbers that multiply to \( 5 \times (-14) = -70 \) and add up to \( 3 \). These numbers are \( 10 \) and \( -7 \):
\( 5 + 3a - 14a^2 \)
\( = 5 + 10a - 7a - 14a^2 \)
Factoring by grouping:
\( = 5(1 + 2a) - 7a(1 + 2a) \)
Taking out the common binomial \( (1 + 2a) \):
\( = (1 + 2a)(5 - 7a) \)
In simple words: Split \( 3a \) into \( 10a - 7a \). Group the terms to complete the factorisation.
Exam Tip: Keep track of the signs during grouping, especially when factoring out a negative term like \( -7a \).

Question 21. Factorise : \( (2a + b)^2 + 5(2a + b) + 6 \)
Answer:
To make factoring easier, we can substitute a single variable for the common binomial. Let \( 2a + b = x \):
The expression becomes:
\( x^2 + 5x + 6 \)
We split the middle term of this quadratic trinomial:
\( = x^2 + 3x + 2x + 6 \)
Group the terms and factor out the common elements:
\( = x(x + 3) + 2(x + 3) \)
Factor out the common bracket:
\( = (x + 3)(x + 2) \)
Now, replace \( x \) with its original value \( 2a + b \):
\( = (2a + b + 3)(2a + b + 2) \)
In simple words: Let \( 2a + b \) be represented by \( x \) to simplify the expression to \( x^2 + 5x + 6 \). Factorise this, then put \( 2a + b \) back in place of \( x \).
Exam Tip: Using substitution is a great way to simplify complex-looking expressions and avoid mistakes during factorisation.

Question 22. Factorise : \( 1 - (2x + 3y) - 6(2x + 3y)^2 \)
Answer:
Let \( 2x + 3y = a \), which gives \( (2x + 3y)^2 = a^2 \):
Substituting this into the expression:
\( 1 - a - 6a^2 \)
Split the middle term \( -a \) into \( -3a + 2a \):
\( = 1 - 3a + 2a - 6a^2 \)
Factor by grouping:
\( = 1(1 - 3a) + 2a(1 - 3a) \)
\( = (1 - 3a)(1 + 2a) \)
Now, substitute \( 2x + 3y \) back for \( a \):
\( = [1 - 3(2x + 3y)][1 + 2(2x + 3y)] \)
Expand the terms inside the brackets:
\( = (1 - 6x - 9y)(1 + 4x + 6y) \)
In simple words: Replace \( 2x + 3y \) with \( a \). Factorise \( 1 - a - 6a^2 \) to get \( (1 - 3a)(1 + 2a) \). Put \( 2x + 3y \) back and multiply it out.
Exam Tip: Be careful to distribute the negative sign when multiplying by \( -3 \) in the final substitution step.

Question 23. Factorise : \( (x - 2y)^2 - 12(x - 2y) + 32 \)
Answer:
Let \( x - 2y = a \):
The expression becomes:
\( a^2 - 12a + 32 \)
We need two numbers that multiply to \( 32 \) and add to \( -12 \). These are \( -8 \) and \( -4 \):
\( = a^2 - 8a - 4a + 32 \)
Grouping the terms:
\( = a(a - 8) - 4(a - 8) \)
\( = (a - 8)(a - 4) \)
Substitute \( x - 2y \) back in place of \( a \):
\( = (x - 2y - 8)(x - 2y - 4) \)
In simple words: Use a helper letter \( a \) instead of \( x - 2y \). Factorise the simpler expression using \( -8 \) and \( -4 \), then put the original terms back.
Exam Tip: Make sure both brackets in the intermediate step have the same sign before combining them.

Question 24. Factorise : \( 8 + 6(a + b) - 5(a + b)^2 \)
Answer:
Let us substitute \( a + b = x \):
The expression becomes:
\( 8 + 6x - 5x^2 \)
We look for two numbers that multiply to \( 8 \times (-5) = -40 \) and add up to \( 6 \). These numbers are \( 10 \) and \( -4 \):
\( = 8 + 10x - 4x - 5x^2 \)
Factoring by grouping:
\( = 2(4 + 5x) - x(4 + 5x) \)
\( = (4 + 5x)(2 - x) \)
Now, substitute \( a + b \) back for \( x \):
\( = [4 + 5(a + b)][2 - (a + b)] \)
Expand the brackets:
\( = [4 + 5a + 5b][2 - a - b] \)
In simple words: Represent \( a + b \) with \( x \). Factorise \( 8 + 6x - 5x^2 \) to get \( (4 + 5x)(2 - x) \). Then, substitute \( a + b \) back and simplify.
Exam Tip: Pay close attention to signs when substituting back inside the negative bracket; \( 2 - (a + b) \) becomes \( 2 - a - b \).

Question 25. Factorise : \( 2(x + 2y)^2 - 5(x + 2y) + 2 \)
Answer:
Let \( x + 2y = a \):
The expression simplifies to:
\( 2a^2 - 5a + 2 \)
We split the middle term into \( -a - 4a \) because \( -1 \times -4 = 4 \) and \( -1 + -4 = -5 \):
\( = 2a^2 - a - 4a + 2 \)
Grouping the terms and factoring:
\( = a(2a - 1) - 2(2a - 1) \)
\( = (2a - 1)(a - 2) \)
Now substitute \( x + 2y \) back for \( a \):
\( = [2(x + 2y) - 1][(x + 2y) - 2] \)
Expanding the terms inside the brackets:
\( = (2x + 4y - 1)(x + 2y - 2) \)
In simple words: Replace \( x + 2y \) with \( a \), factorise the trinomial, and then place \( x + 2y \) back into the result.
Exam Tip: Remember to multiply the entire term \( (x + 2y) \) by \( 2 \) when substituting back into the first factor.

Exercise 13(E)

Question 1. In each case find whether the trinomial is a perfect square or not:
(i) \( x^2 + 14x + 49 \)
(ii) \( a^2 - 10a + 25 \)
(iii) \( 4x^2 + 4x + 1 \)
(iv) \( 9b^2 + 12b + 16 \)
(v) \( 16x^2 - 16xy + y^2 \)
(vi) \( x^2 - 4x + 16 \)
Answer:
(i) We can write \( x^2 + 14x + 49 \) as:
\( (x)^2 + 2 \times x \times 7 + (7)^2 \)
This matches the perfect square trinomial form \( a^2 + 2ab + b^2 = (a+b)^2 \).
\( \implies (x + 7)^2 \)
Thus, the trinomial \( x^2 + 14x + 49 \) is a perfect square.

(ii) We can write \( a^2 - 10a + 25 \) as:
\( (a)^2 - 2 \times a \times 5 + (5)^2 \)
This matches the form \( a^2 - 2ab + b^2 = (a-b)^2 \).
\( \implies (a - 5)^2 \)
Thus, the trinomial \( a^2 - 10a + 25 \) is a perfect square.

(iii) We can write \( 4x^2 + 4x + 1 \) as:
\( (2x)^2 + 2 \times 2x \times 1 + (1)^2 \)
This matches the form \( a^2 + 2ab + b^2 = (a+b)^2 \).
\( \implies (2x + 1)^2 \)
Thus, the trinomial \( 4x^2 + 4x + 1 \) is a perfect square.

(iv) We can write \( 9b^2 + 12b + 16 \) as:
\( (3b)^2 + 3b \times 4 + (4)^2 \)
Let \( 3b = x \) and \( 4 = y \). The expression has the form \( x^2 + xy + y^2 \). Because it lacks the factor of \( 2 \) in the middle term, it does not match the perfect square form \( x^2 + 2xy + y^2 \).
Thus, the trinomial is not a perfect square.

(v) We can write \( 16x^2 - 16xy + y^2 \) as:
\( (4x)^2 - 4(4x)(y) + (y)^2 \)
Let \( 4x = a \) and \( y = b \). The expression takes the form \( a^2 - 4ab + b^2 \). Since the middle term has a factor of \( 4 \) instead of \( 2 \), it does not match the perfect square form \( a^2 - 2ab + b^2 \).
Thus, the trinomial is not a perfect square.

(vi) We can write \( x^2 - 4x + 16 \) as:
\( (x)^2 - (x)(4) + (4)^2 \)
Let \( x = a \) and \( 4 = b \). The expression is in the form \( a^2 - ab + b^2 \). It lacks the factor of \( 2 \) in the middle term, so it cannot be expressed as \( a^2 - 2ab + b^2 \).
Thus, the trinomial is not a perfect square.
In simple words: To check if a trinomial is a perfect square, see if the middle term is exactly twice the product of the square roots of the first and last terms. If it is, then it's a perfect square; otherwise, it's not.
Exam Tip: Always identify the square roots of the first and last terms, multiply them together, and then double that product to see if it matches the middle term.

Question 2. Factorise completely : \( 2 - 8x^2 \)
Answer:
First, we factor out the common term \( 2 \):
\( 2 - 8x^2 = 2(1 - 4x^2) \)
The terms inside the bracket can be written as a difference of squares:
\( = 2[(1)^2 - (2x)^2] \)
Applying the identity \( a^2 - b^2 = (a+b)(a-b) \), we get:
\( = 2(1 + 2x)(1 - 2x) \)
In simple words: First, pull out the common factor 2. Then, rewrite the inside part as \( 1^2 - (2x)^2 \) and apply the difference of squares identity.
Exam Tip: "Factorise completely" means you should first look for a common numerical factor before applying algebraic identities.

Question 3. Factorise completely : \( 8x^2y - 18y^3 \)
Answer:
We begin by extracting the greatest common factor, which is \( 2y \):
\( 8x^2y - 18y^3 = 2y(4x^2 - 9y^2) \)
We can rewrite the expression inside the parentheses as a difference of two squares:
\( = 2y[(2x)^2 - (3y)^2] \)
Using the difference of squares identity \( a^2 - b^2 = (a+b)(a-b) \), we obtain:
\( = 2y(2x + 3y)(2x - 3y) \)
In simple words: Pull out the common factor \( 2y \) first. This leaves a difference of two squares, \( (2x)^2 - (3y)^2 \), which can be factored easily.
Exam Tip: Don't forget the common factor \( 2y \) in your final step; it must remain part of the final factorised expression.

Question 4. Factorise completely : \( ax^2 - ay^2 \)
Answer:
Take out the common factor \( a \) from both terms:
\( ax^2 - ay^2 = a(x^2 - y^2) \)
Apply the difference of squares identity \( x^2 - y^2 = (x + y)(x - y) \):
\( = a(x + y)(x - y) \)
In simple words: Factor out the letter \( a \) first. Then, use the difference of squares rule on the remaining part.
Exam Tip: Identifying the common variable first makes applying standard identities straightforward.

Question 5. Factorise completely : \( 25x^3 - x \)
Answer:
First, factor out the common term \( x \):
\( 25x^3 - x = x(25x^2 - 1) \)
The binomial inside the parentheses can be written as a difference of two squares:
\( = x[(5x)^2 - 1^2] \)
Applying the difference of squares identity, we get:
\( = x(5x + 1)(5x - 1) \)
In simple words: Pull out \( x \) first, leaving \( 25x^2 - 1 \). Rewrite this as \( (5x)^2 - 1^2 \) and factorise.
Exam Tip: Remember that \( 1 \) is a perfect square, so \( 25x^2 - 1 \) can always be factored using the difference of squares.

Question 6. Factorise completely : \( a^4 - b^4 \)
Answer:
We can write the terms as squares of squares:
\( a^4 - b^4 = (a^2)^2 - (b^2)^2 \)
Applying the difference of squares identity, we get:
\( = (a^2 + b^2)(a^2 - b^2) \)
The second term is also a difference of squares and can be factored further:
\( = (a^2 + b^2)(a + b)(a - b) \)
In simple words: Rewrite \( a^4 - b^4 \) as \( (a^2)^2 - (b^2)^2 \) and factorise once. Then, factorise the negative term \( a^2 - b^2 \) again.
Exam Tip: Check each factor of your result to see if any can be factorised further, as is often the case with higher powers like \( a^4 \).

Question 7. Factorise completely : \( 16x^4 - 81y^4 \)
Answer:
We express the terms as squares of squares:
\( 16x^4 - 81y^4 = (4x^2)^2 - (9y^2)^2 \)
Using the difference of squares identity:
\( = (4x^2 + 9y^2)(4x^2 - 9y^2) \)
The second factor is another difference of squares that can be simplified further:
\( = (4x^2 + 9y^2)[(2x)^2 - (3y)^2] \)
Applying the difference of squares identity to the second bracket:
\( = (4x^2 + 9y^2)(2x + 3y)(2x - 3y) \)
In simple words: Write the expression as \( (4x^2)^2 - (9y^2)^2 \) and factorise it. Then factorise the resulting minus term, \( 4x^2 - 9y^2 \), a second time.
Exam Tip: Be careful not to try to factorise the sum of squares, \( 4x^2 + 9y^2 \), as it cannot be factored further using real numbers.

Question 8. Factorise completely : \( 625 - x^4 \)
Answer:
We rewrite the numbers as squares:
\( 625 - x^4 = (25)^2 - (x^2)^2 \)
Applying the difference of squares identity:
\( = (25 + x^2)(25 - x^2) \)
The second binomial is also a difference of squares:
\( = (25 + x^2)[5^2 - x^2] \)
Factoring this second term gives:
\( = (25 + x^2)(5 + x)(5 - x) \)
In simple words: Break \( 625 - x^4 \) down into \( (25 + x^2)(25 - x^2) \). Then, factorise the \( 25 - x^2 \) part once more.
Exam Tip: Recognizing powers of 5, such as \( 5^4 = 625 \), helps in quickly identifying perfect squares.

Question 9. Factorise completely : \( x^2 - y^2 - 3x - 3y \)
Answer:
We can group the terms in pairs:
\( x^2 - y^2 - 3x - 3y = (x^2 - y^2) - 3(x + y) \)
Use the difference of squares identity on the first group:
\( = (x + y)(x - y) - 3(x + y) \)
Now, factor out the common binomial \( (x + y) \):
\( = (x + y)(x - y - 3) \)
In simple words: Group the first two terms as a difference of squares and factorise them. Group the last two terms by pulling out \( -3 \). Finally, take out the common bracket \( (x + y) \).
Exam Tip: Pay attention to the sign when grouping \( -3x - 3y \) as \( -3(x + y) \); the sign inside the parentheses changes to positive.

Question 10. Factorise completely : \( x^2 - y^2 - 2x + 2y \)
Answer:
Group the terms into two separate parts:
\( x^2 - y^2 - 2x + 2y = (x^2 - y^2) - 2(x - y) \)
Applying the difference of squares formula to the first group:
\( = (x + y)(x - y) - 2(x - y) \)
Factor out the common binomial \( (x - y) \):
\( = (x - y)(x + y - 2) \)
In simple words: Group the first two terms and the last two terms. Factorise the first group as a difference of squares, and factor out \( -2 \) from the second. Then pull out the common bracket \( (x - y) \).
Exam Tip: Factoring out a negative number from a positive term changes that term's sign inside the parentheses, like how \( +2y \) became \( -y \).

Question 11. Factorise completely : \( 3x^2 + 15x - 72 \)
Answer:
First, we factor out the common term \( 3 \) from the entire expression:
\( 3x^2 + 15x - 72 = 3(x^2 + 5x - 24) \)
Now, split the middle term inside the parentheses. We look for two numbers that multiply to \( -24 \) and add to \( 5 \). These are \( 8 \) and \( -3 \):
\( = 3[x^2 + 8x - 3x - 24] \)
Factor by grouping:
\( = 3[x(x + 8) - 3(x + 8)] \)
Factor out the common binomial \( (x + 8) \):
\( = 3(x + 8)(x - 3) \)
In simple words: Start by pulling out the common factor 3. Then, factorise the quadratic trinomial inside by finding two numbers that multiply to \( -24 \) and add to 5, which are 8 and -3.
Exam Tip: Never ignore the numerical coefficient at the beginning of the problem; always try to factor it out first to work with smaller numbers.

Question 12. Factorise completely : \( 2a^2 - 8a - 64 \)
Answer:
We begin by factoring out the common factor of \( 2 \):
\( 2a^2 - 8a - 64 = 2(a^2 - 4a - 32) \)
Now we split the middle term of the trinomial inside. We need two numbers whose product is \( -32 \) and sum is \( -4 \). These are \( -8 \) and \( 4 \):
\( = 2[a^2 - 8a + 4a - 32] \)
Factor by grouping:
\( = 2[a(a - 8) + 4(a - 8)] \)
Factor out the binomial \( (a - 8) \):
\( = 2(a - 8)(a + 4) \)
In simple words: Take out 2 as a common factor first. Then, split the middle term inside the bracket into \( -8a + 4a \), and factorise by grouping.
Exam Tip: Remember to carry the common factor \( 2 \) through all the steps of your working.

Question 13. Factorise completely : \( 5b^2 + 45b + 90 \)
Answer:
First, we pull out the common factor \( 5 \):
\( 5b^2 + 45b + 90 = 5(b^2 + 9b + 18) \)
Now, split the middle term inside the parentheses. We look for two numbers that multiply to \( 18 \) and add to \( 9 \). These are \( 6 \) and \( 3 \):
\( = 5[b^2 + 6b + 3b + 18] \)
Factoring by grouping:
\( = 5[b(b + 6) + 3(b + 6)] \)
Factor out the common bracket \( (b + 6) \):
\( = 5(b + 6)(b + 3) \)
In simple words: Factor out 5 from the expression, then split \( 9b \) into \( 6b + 3b \) because 6 and 3 multiply to 18. Finally, group and factorise.
Exam Tip: Check your final binomial terms by multiplying them to ensure they simplify back to the trinomial.

Question 14. Factorise completely : \( 3x^2y + 11xy + 6y \)
Answer:
First, we factor out the common variable \( y \) from the expression:
\( 3x^2y + 11xy + 6y = y(3x^2 + 11x + 6) \)
For the trinomial inside, we need two numbers that multiply to \( 3 \times 6 = 18 \) and add to \( 11 \). These are \( 9 \) and \( 2 \):
\( = y[3x^2 + 9x + 2x + 6] \)
Factor by grouping:
\( = y[3x(x + 3) + 2(x + 3)] \)
Factor out the binomial \( (x + 3) \):
\( = y(x + 3)(3x + 2) \)
In simple words: Pull out the common factor \( y \). Then, split \( 11x \) into \( 9x + 2x \). Finally, group and factorise the expression.
Exam Tip: Be sure not to lose the common variable factor \( y \) as you work through the grouping inside the brackets.

Question 15. Factorise completely : \( 5ap^2 + 11ap + 2a \)
Answer:
We begin by factoring out the common factor \( a \):
\( 5ap^2 + 11ap + 2a = a(5p^2 + 11p + 2) \)
To factorise the trinomial inside, we need two numbers that multiply to \( 5 \times 2 = 10 \) and add up to \( 11 \). These are \( 10 \) and \( 1 \):
\( = a[5p^2 + 10p + p + 2] \)
Factoring by grouping:
\( = a[5p(p + 2) + 1(p + 2)] \)
Taking out the common factor \( (p + 2) \):
\( = a(p + 2)(5p + 1) \)
In simple words: Extract the common factor \( a \) first. Then split \( 11p \) into \( 10p + 1p \), and factorise by grouping.

Exam Tip: If no visible number is common in the second group, remember that \( 1 \) is always a common factor to write down.

Question 16. Factorise completely : \( a^2 + 2ab + b^2 - c^2 \)
Answer:
Group the first three terms to form a perfect square trinomial:
\( a^2 + 2ab + b^2 - c^2 = (a^2 + 2ab + b^2) - c^2 \)
Since the terms inside the parentheses form the expansion of \( (a + b)^2 \), we can write:
\( = (a + b)^2 - c^2 \)
Now, use the difference of squares identity \( x^2 - y^2 = (x + y)(x - y) \), letting \( x = a + b \) and \( y = c \):
\( = (a + b + c)(a + b - c) \)
In simple words: First, combine the first three terms into a single squared term, \( (a + b)^2 \). Then, use the difference of squares formula to factorise the entire expression into two brackets.

Exam Tip: Recognising perfect square trinomials like \( a^2 + 2ab + b^2 \) immediately helps in grouping terms to apply the difference of squares identity.

Question 17. Factorise completely : \( x^2 + 6xy + 9y^2 + x + 3y \)
Answer:
Group the terms to make factorisation easier:
\( x^2 + 6xy + 9y^2 + x + 3y = [x^2 + 2 \cdot x \cdot 3y + (3y)^2] + (x + 3y) \)
The grouped terms inside the square brackets form the perfect square \( (x + 3y)^2 \):
\( = [x + 3y]^2 + (x + 3y) \)
We can write this by showing the common binomial factor:
\( = (x + 3y)(x + 3y) + (x + 3y) \)
Factoring out \( (x + 3y) \) from both parts gives:
\( = (x + 3y)(x + 3y + 1) \)
In simple words: The first three terms make a perfect square, \( (x + 3y)^2 \). We then pull out the common bracket \( (x + 3y) \) from the whole expression to get the final answer.

Exam Tip: When you have terms left over after grouping a perfect square, look for how they relate to the base of that perfect square to find a common factor.

Question 18. Factorise completely : \( 4a^2 - 12ab + 9b^2 + 4a - 6b \)
Answer:
Group the expression into two sections:
\( 4a^2 - 12ab + 9b^2 + 4a - 6b = [4a^2 - 12ab + 9b^2] + (4a - 6b) \)
Express the first group as a perfect square and extract 2 from the second group:
\( = [(2a)^2 - 2 \cdot 2a \cdot 3b + (3b)^2] + 2(2a - 3b) \)
This simplifies to:
\( = (2a - 3b)^2 + 2(2a - 3b) \)
Take out the common binomial factor \( (2a - 3b) \):
\( = (2a - 3b)(2a - 3b + 2) \)
In simple words: Combine the first three terms into the perfect square \( (2a - 3b)^2 \) and factor out 2 from the remaining terms. Finally, pull out the common bracket \( (2a - 3b) \).

Exam Tip: Be careful with signs when factoring out coefficients from the linear terms; always double-check by expanding the brackets mentally.

Question 19. Factorise completely : \( 2a^2b^2 - 98b^4 \)
Answer:
First, factor out the greatest common term, which is \( 2b^2 \):
\( 2a^2b^2 - 98b^4 = 2b^2(a^2 - 49b^2) \)
Express the terms inside the parentheses as a difference of two squares:
\( = 2b^2[(a)^2 - (7b)^2] \)
Apply the identity \( x^2 - y^2 = (x + y)(x - y) \):
\( = 2b^2(a + 7b)(a - 7b) \)
In simple words: Take out the common factor \( 2b^2 \) first. Then, use the difference of squares formula on the remaining part inside the bracket.

Exam Tip: Always look for the highest common factor (HCF) first before applying any algebraic identities. This simplifies the expression immediately.

Question 20. Factorise completely : \( a^2 - 16b^2 - 2a - 8b \)
Answer:
Group the terms into two parts to factorise them separately:
\( a^2 - 16b^2 - 2a - 8b = (a^2 - 16b^2) - (2a + 8b) \)
Write the first group as a difference of squares and factorise out 2 from the second group:
\( = [(a)^2 - (4b)^2] - 2(a + 4b) \)
Factorise the difference of squares:
\( = (a + 4b)(a - 4b) - 2(a + 4b) \)
Take out the common binomial factor \( (a + 4b) \):
\( = (a + 4b)(a - 4b - 2) \)
In simple words: Split the expression. Use the difference of squares on \( a^2 - 16b^2 \) and factor out 2 from the other part, then pull out the common bracket \( (a + 4b) \).

Exam Tip: Be careful with the minus sign when grouping the last two terms: \( -2a - 8b \) becomes \( -(2a + 8b) \), changing the sign inside the brackets.

Exercise 13(F)

Question 1. Factorise :
(i) \( 6x^3 - 8x^2 \)
(ii) \( 35a^3b^2c + 42ab^2c^2 \)
(iii) \( 36x^2y^2 - 30x^3y^3 + 48x^3y^2 \)
(iv) \( 8(2a + 3b)^3 - 12(2a + 3b)^2 \)
(v) \( 9a(x - 2y)^4 - 12a(x - 2y)^3 \)
Answer:
(i) Factor out the common term \( 2x^2 \):
\( 6x^3 - 8x^2 = 2x^2(3x - 4) \)

(ii) Factor out the common term \( 7ab^2c \):
\( 35a^3b^2c + 42ab^2c^2 = 7ab^2c(5a^2 + 6c) \)

(iii) Factor out the common term \( 6x^2y^2 \):
\( 36x^2y^2 - 30x^3y^3 + 48x^3y^2 = 6x^2y^2(6 - 5xy + 8x) \)

(iv) Factor out the common term \( 4(2a + 3b)^2 \):
\( 8(2a + 3b)^3 - 12(2a + 3b)^2 = 4(2a + 3b)^2 [2(2a + 3b) - 3] \)
\( = 4(2a + 3b)^2 [4a + 6b - 3] \)

(v) Factor out the common term \( 3a(x - 2y)^3 \):
\( 9a(x - 2y)^4 - 12a(x - 2y)^3 = 3a(x - 2y)^3 [3(x - 2y) - 4] \)
\( = 3a(x - 2y)^3 (3x - 6y - 4) \)
In simple words: Find the biggest common number and variables for each expression, pull them outside the brackets, and simplify what remains inside.

Exam Tip: When dealing with powers of binomial expressions, always factor out the binomial raised to the lowest power present in the terms.

Question 2. Factorise :
(i) \( a^2 - ab - 3a + 3b \)
(ii) \( x^2y - xy^2 + 5x - 5y \)
(iii) \( a^2 - ab(1 - b) - b^3 \)
(iv) \( xy^2 + (x - 1)y - 1 \)
(v) \( (ax + by)^2 + (bx - ay)^2 \)
(vi) \( ab(x^2 + y^2) - xy(a^2 + b^2) \)
(vii) \( m - 1 - (m - 1)^2 + am - a \)
Answer:
(i) Group the terms in pairs and factor out common factors:
\( a^2 - ab - 3a + 3b = a(a - b) - 3(a - b) \)
\( = (a - b)(a - 3) \)

(ii) Group the first two terms and the last two terms:
\( x^2y - xy^2 + 5x - 5y = xy(x - y) + 5(x - y) \)
\( = (x - y)(xy + 5) \)

(iii) First, expand the middle term:
\( a^2 - ab(1 - b) - b^3 = a^2 - ab + ab^2 - b^3 \)
Group in pairs:
\( = a(a - b) + b^2(a - b) \)
\( = (a - b)(a + b^2) \)

(iv) Expand the term containing \( y \):
\( xy^2 + (x - 1)y - 1 = xy^2 + xy - y - 1 \)
Group and factorise:
\( = xy(y + 1) - 1(y + 1) \)
\( = (xy - 1)(y + 1) \)

(v) Expand both squared terms using algebraic identities:
\( (ax + by)^2 + (bx - ay)^2 = a^2x^2 + b^2y^2 + 2abxy + b^2x^2 + a^2y^2 - 2abxy \)
The \( 2abxy \) terms cancel out:
\( = a^2x^2 + b^2y^2 + b^2x^2 + a^2y^2 \)
Rearrange and group by \( x^2 \) and \( y^2 \) terms:
\( = x^2(a^2 + b^2) + y^2(a^2 + b^2) \)
\( = (x^2 + y^2)(a^2 + b^2) \)

(vi) Expand first, then regroup:
\( ab(x^2 + y^2) - xy(a^2 + b^2) = abx^2 + aby^2 - a^2xy - b^2xy \)
Rearrange the terms:
\( = abx^2 - a^2xy - b^2xy + aby^2 \)
Factorise each pair:
\( = ax(bx - ay) - by(bx - ay) \)
\( = (bx - ay)(ax - by) \)

(vii) Group the terms to find common factors of \( (m - 1) \):
\( m - 1 - (m - 1)^2 + am - a = (m - 1) - (m - 1)^2 + a(m - 1) \)
Take out the common binomial factor \( (m - 1) \):
\( = (m - 1)[1 - (m - 1) + a] \)
\( = (m - 1)(1 - m + 1 + a) \)
\( = (m - 1)(2 - m + a) \)
In simple words: Expand any bracketed terms first when needed, then group the terms into pairs so that you can factor out a common bracket.

Exam Tip: When expanding expressions like \( (ax + by)^2 \), don't forget the middle \( 2abxy \) term. These middle terms often cancel out with other terms in the expression, leaving a simple factorisable result.

Question 3. Factorise :
(i) \( a^2 - (b - c)^2 \)
(ii) \( 25(2x - y)^2 - 16(x - 2y)^2 \)
(iii) \( 16(5x + 4)^2 - 9(3x - 2)^2 \)
(iv) \( 9x^2 - \frac{1}{16} \)
(v) \( 25(x - 2y)^2 - 4 \)
Answer:
(i) Use the difference of squares identity \( X^2 - Y^2 = (X - Y)(X + Y) \):
\( a^2 - (b - c)^2 = (a - (b - c))(a + (b - c)) \)
\( = (a - b + c)(a + b - c) \)

(ii) Express as the difference of two square terms:
\( 25(2x - y)^2 - 16(x - 2y)^2 = [5(2x - y)]^2 - [4(x - 2y)]^2 \)
Apply \( X^2 - Y^2 = (X - Y)(X + Y) \):
\( = [5(2x - y) - 4(x - 2y)] [5(2x - y) + 4(x - 2y)] \)
Expand and simplify the brackets:
\( = [10x - 5y - 4x + 8y] [10x - 5y + 4x - 8y] \)
\( = (6x + 3y)(14x - 13y) \)
Factor out the common number 3 from the first bracket:
\( = 3(2x + y)(14x - 13y) \)

(iii) Express as a difference of squares:
\( 16(5x + 4)^2 - 9(3x - 2)^2 = [4(5x + 4)]^2 - [3(3x - 2)]^2 \)
Apply \( X^2 - Y^2 = (X - Y)(X + Y) \):
\( = [4(5x + 4) - 3(3x - 2)] [4(5x + 4) + 3(3x - 2)] \)
Expand the terms inside the brackets:
\( = [20x + 16 - 9x + 6] [20x + 16 + 9x - 6] \)
\( = (11x + 22)(29x + 10) \)
Factor out 11 from the first bracket:
\( = 11(x + 2)(29x + 10) \)

(iv) Convert both terms to perfect squares:
\( 9x^2 - \frac{1}{16} = (3x)^2 - \left(\frac{1}{4}\right)^2 \)
Apply \( X^2 - Y^2 = (X - Y)(X + Y) \):
\( = \left(3x - \frac{1}{4}\right)\left(3x + \frac{1}{4}\right) \)

(v) Convert to difference of squares:
\( 25(x - 2y)^2 - 4 = [5(x - 2y)]^2 - 2^2 \)
Apply \( X^2 - Y^2 = (X - Y)(X + Y) \):
\( = [5(x - 2y) - 2] [5(x - 2y) + 2] \)
\( = (5x - 10y - 2)(5x - 10y + 2) \)
In simple words: Change each expression into the format of one square minus another square, then use the formula \( X^2 - Y^2 = (X - Y)(X + Y) \) to get the final brackets.

Exam Tip: Don't forget to look for common constant factors (like 3 in sub-part ii, and 11 in sub-part iii) at the final step to make sure the factorisation is fully complete.

Question 4. Factorise :
(i) \( a^2 - 23a + 42 \)
(ii) \( a^2 - 23a - 108 \)
(iii) \( 1 - 18x - 63x^2 \)
(iv) \( 5x^2 - 4xy - 12y^2 \)
(v) \( x(3x + 14) + 8 \)
(vi) \( 5 - 4x(1 + 3x) \)
(vii) \( x^2y^2 - 3xy - 40 \)
(viii) \( (3x - 2y)^2 - 5(3x - 2y) - 24 \)
(ix) \( 12(a + b)^2 - (a + b) - 35 \)
Answer:
(i) Split the middle term \( -23a \) into \( -21a \) and \( -2a \):
\( a^2 - 23a + 42 = a^2 - 21a - 2a + 42 \)
\( = a(a - 21) - 2(a - 21) \)
\( = (a - 21)(a - 2) \)

(ii) Split the middle term \( -23a \) into \( -27a \) and \( 4a \):
\( a^2 - 23a - 108 = a^2 - 27a + 4a - 108 \)
\( = a(a - 27) + 4(a - 27) \)
\( = (a - 27)(a + 4) \)

(iii) Split the middle term \( -18x \) into \( -21x \) and \( 3x \):
\( 1 - 18x - 63x^2 = 1 - 21x + 3x - 63x^2 \)
\( = 1(1 - 21x) + 3x(1 - 21x) \)
\( = (1 - 21x)(1 + 3x) \)

(iv) Split the middle term \( -4xy \) into \( -10xy \) and \( 6xy \):
\( 5x^2 - 4xy - 12y^2 = 5x^2 - 10xy + 6xy - 12y^2 \)
\( = 5x(x - 2y) + 6y(x - 2y) \)
\( = (x - 2y)(5x + 6y) \)

(v) Expand first, then split the middle term \( 14x \) into \( 12x \) and \( 2x \):
\( x(3x + 14) + 8 = 3x^2 + 14x + 8 \)
\( = 3x^2 + 12x + 2x + 8 \)
\( = 3x(x + 4) + 2(x + 4) \)
\( = (x + 4)(3x + 2) \)

(vi) Expand first, then split the middle term \( -4x \) into \( -10x \) and \( 6x \):
\( 5 - 4x(1 + 3x) = 5 - 4x - 12x^2 \)
\( = 5 - 10x + 6x - 12x^2 \)
\( = 5(1 - 2x) + 6x(1 - 2x) \)
\( = (1 - 2x)(5 + 6x) \)

(vii) Split the middle term \( -3xy \) into \( -8xy \) and \( 5xy \):
\( x^2y^2 - 3xy - 40 = x^2y^2 - 8xy + 5xy - 40 \)
\( = xy(xy - 8) + 5(xy - 8) \)
\( = (xy - 8)(xy + 5) \)

(viii) Treating \( (3x - 2y) \) as a single unit, split the middle term into \( -8(3x - 2y) \) and \( 3(3x - 2y) \):
\( (3x - 2y)^2 - 5(3x - 2y) - 24 = (3x - 2y)^2 - 8(3x - 2y) + 3(3x - 2y) - 24 \)
\( = (3x - 2y)[(3x - 2y) - 8] + 3[(3x - 2y) - 8] \)
\( = (3x - 2y - 8)(3x - 2y + 3) \)

(ix) Split the middle term \( -(a + b) \) into \( -21(a + b) \) and \( 20(a + b) \):
\( 12(a + b)^2 - (a + b) - 35 = 12(a + b)^2 - 21(a + b) + 20(a + b) - 35 \)
\( = 3(a + b)[4(a + b) - 7] + 5[4(a + b) - 7] \)
\( = [4(a + b) - 7][3(a + b) + 5] \)
\( = (4a + 4b - 7)(3a + 3b + 5) \)
In simple words: This question uses the splitting-the-middle-term method. Find two numbers that multiply to the product of the first and last coefficients and add up to the middle coefficient.

Exam Tip: When factorising trinomials with compound terms like \( 3x - 2y \) or \( a + b \), you can temporarily replace them with a single variable (like \( t \)) to make the middle term splitting easier to visualise.

Question 5. Factorise :
(i) \( 15(5x - 4)^3 - 10(5x - 4) \)
(ii) \( 3a^2x - bx + 3a^2 - b \)
(iii) \( b(c - d)^2 + a(d - c) + 3(c - d) \)
(iv) \( ax^2 + b^2y - ab^2 - x^2y \)
(v) \( 1 - 3x - 3y - 4(x + y)^2 \)
Answer:
(i) Factor out the common factor \( 5(5x - 4) \):
\( 15(5x - 4)^3 - 10(5x - 4) = 5(5x - 4) [3(5x - 4)^2 - 2] \)
Expand the term inside the square brackets:
\( = 5(5x - 4) [3(25x^2 - 40x + 16) - 2] \)
\( = 5(5x - 4) [75x^2 - 120x + 48 - 2] \)
\( = 5(5x - 4)(75x^2 - 120x + 46) \)

(ii) Group the terms to find a common factor:
\( 3a^2x - bx + 3a^2 - b = x(3a^2 - b) + 1(3a^2 - b) \)
\( = (x + 1)(3a^2 - b) \)

(iii) Standardize the variables inside the parentheses by using \( (d - c) = -(c - d) \):
\( b(c - d)^2 + a(d - c) + 3(c - d) = b(c - d)^2 - a(c - d) + 3(c - d) \)
Now, pull out the common binomial factor \( (c - d) \):
\( = (c - d)[b(c - d) - a + 3] \)
\( = (c - d)(bc - bd - a + 3) \)

(iv) Rearrange the terms to group them:
\( ax^2 + b^2y - ab^2 - x^2y = ax^2 - ab^2 + b^2y - x^2y \)
\( = a(x^2 - b^2) + y(b^2 - x^2) \)
Rewrite \( (b^2 - x^2) \) as \( -(x^2 - b^2) \):
\( = a(x^2 - b^2) - y(x^2 - b^2) \)
\( = (x^2 - b^2)(a - y) \)
Further factorise the difference of squares:
\( = (x - b)(x + b)(a - y) \)

(v) Group the terms to see a pattern in terms of \( (x + y) \):
\( 1 - 3x - 3y - 4(x + y)^2 = 1 - 3(x + y) - 4(x + y)^2 \)
Factorise this quadratic form by splitting the middle term:
\( = 1 - 4(x + y) + (x + y) - 4(x + y)^2 \)
\( = 1[1 - 4(x + y)] + (x + y)[1 - 4(x + y)] \)
\( = [1 - 4(x + y)][1 + (x + y)] \)
\( = (1 - 4x - 4y)(1 + x + y) \)
In simple words: Look for common binomial terms like \( c - d \) or \( x + y \). You may need to change signs (like turning \( d - c \) into \( -(c - d) \)) to make them match.

Exam Tip: Be mindful when factoring out negative signs inside brackets: remember that \( d - c = -(c - d) \). This is a very common trick used to create matching brackets.

Question 6. Factorise :
(i) \( 2a^3 - 50a \)
(ii) \( 54a^2b^2 - 6 \)
(iii) \( 64a^2b - 144b^3 \)
(iv) \( (2x - y)^3 - (2x - y) \)
(v) \( x^2 - 2xy + y^2 - z^2 \)
(vi) \( x^2 - y^2 - 2yz - z^2 \)
(vii) \( 7a^5 - 567a \)
(viii) \( 5x^2 - \frac{20x^4}{9} \)
Answer:
(i) Take out the common term \( 2a \) and apply the difference of squares identity:
\( 2a^3 - 50a = 2a(a^2 - 25) \)
\( = 2a(a^2 - 5^2) \)
\( = 2a(a - 5)(a + 5) \)

(ii) Factor out the common number 6:
\( 54a^2b^2 - 6 = 6(9a^2b^2 - 1) \)
\( = 6[(3ab)^2 - (1)^2] \)
\( = 6(3ab - 1)(3ab + 1) \)

(iii) Factor out the common term \( 16b \):
\( 64a^2b - 144b^3 = 16b(4a^2 - 9b^2) \)
\( = 16b[(2a)^2 - (3b)^2] \)
\( = 16b(2a - 3b)(2a + 3b) \)

(iv) Pull out the common binomial term \( (2x - y) \):
\( (2x - y)^3 - (2x - y) = (2x - y)[(2x - y)^2 - 1] \)
Apply the difference of squares inside the square brackets:
\( = (2x - y)(2x - y - 1)(2x - y + 1) \)

(v) Group the perfect square trinomial:
\( x^2 - 2xy + y^2 - z^2 = (x - y)^2 - z^2 \)
Factorise using the difference of squares:
\( = (x - y - z)(x - y + z) \)

(vi) Group the last three terms by factoring out a negative sign to get a perfect square trinomial:
\( x^2 - y^2 - 2yz - z^2 = x^2 - (y^2 + 2yz + z^2) \)
\( = x^2 - (y + z)^2 \)
Use the difference of squares:
\( = [x - (y + z)][x + (y + z)] \)
\( = (x - y - z)(x + y + z) \)

(vii) Take out the common factor \( 7a \):
\( 7a^5 - 567a = 7a(a^4 - 81) \)
\( = 7a[(a^2)^2 - 9^2] \)
\( = 7a(a^2 - 9)(a^2 + 9) \)
Factorise \( (a^2 - 9) \) further:
\( = 7a(a - 3)(a + 3)(a^2 + 9) \)

(viii) Factor out the common term \( 5x^2 \):
\( 5x^2 - \frac{20x^4}{9} = 5x^2\left[1 - \frac{4x^2}{9}\right] \)
Express the terms inside the bracket as a difference of squares:
\( = 5x^2\left[1^2 - \left(\frac{2x}{3}\right)^2\right] \)
\( = 5x^2\left[1 - \frac{2x}{3}\right]\left[1 + \frac{2x}{3}\right] \)
In simple words: First extract any common factors. Then, look for ways to write the remaining expression as \( A^2 - B^2 \) so that you can factorise it into \( (A - B)(A + B) \).

Exam Tip: Always check if any bracket in your final answer can be factorised further. For example, in sub-part vii, \( a^2 - 9 \) must be broken down into \( (a - 3)(a + 3) \) to get full marks.

Question 7. Factorise \( xy^2 - xz^2 \), Hence, find the value of:
(i) \( 9 \times 8^2 - 9 \times 2^2 \)
(ii) \( 40 \times 5.5^2 - 40 \times 4.5^2 \)
Answer:
First, factorise the given expression by taking out the common factor \( x \):
\( xy^2 - xz^2 = x(y^2 - z^2) \)
Using the difference of squares identity, we get:
\( = x(y - z)(y + z) \)

(i) Compare \( 9 \times 8^2 - 9 \times 2^2 \) with \( xy^2 - xz^2 \). Here, \( x = 9 \), \( y = 8 \), and \( z = 2 \):
Substitute these values into the factorised form \( x(y - z)(y + z) \):
\( = 9(8 - 2)(8 + 2) \)
\( = 9(6)(10) \)
\( = 540 \)

(ii) Compare \( 40 \times 5.5^2 - 40 \times 4.5^2 \) with \( xy^2 - xz^2 \). Here, \( x = 40 \), \( y = 5.5 \), and \( z = 4.5 \):
Substitute these values into the factorised form:
\( = 40(5.5 - 4.5)(5.5 + 4.5) \)
\( = 40(1)(10) \)
\( = 400 \)
In simple words: First factorise \( xy^2 - xz^2 \) to get \( x(y - z)(y + z) \). Then substitute the numbers from each sub-part into this formula to quickly calculate the answer without squaring large numbers.

Exam Tip: The word "Hence" means you must use the factorised formula to find the values. Direct evaluation by calculating squares like \( 5.5^2 \) may lose marks for method.

Question 8. Factorise :
(i) \( (a - 3b)^2 - 36b^2 \)
(ii) \( 25(a - 5b)^2 - 4(a - 3b)^2 \)
(iii) \( a^2 - 0.36b^2 \)
(iv) \( a^4 - 625 \)
(v) \( x^4 - 5x^2 - 36 \)
(vi) \( 15(2x - y)^2 - 16(2x - y) - 15 \)
Answer:
(i) Express both terms as squares:
\( (a - 3b)^2 - 36b^2 = (a - 3b)^2 - (6b)^2 \)
Apply \( X^2 - Y^2 = (X - Y)(X + Y) \):
\( = (a - 3b - 6b)(a - 3b + 6b) \)
\( = (a - 9b)(a + 3b) \)

(ii) Rewrite the expression as a difference of squares:
\( 25(a - 5b)^2 - 4(a - 3b)^2 = [5(a - 5b)]^2 - [2(a - 3b)]^2 \)
Factorise using the identity \( X^2 - Y^2 = (X - Y)(X + Y) \):
\( = [5(a - 5b) - 2(a - 3b)] [5(a - 5b) + 2(a - 3b)] \)
Expand and combine like terms inside the brackets:
\( = [5a - 25b - 2a + 6b] [5a - 25b + 2a - 6b] \)
\( = (3a - 19b)(7a - 31b) \)

(iii) Convert the decimal term to a perfect square:
\( a^2 - 0.36b^2 = a^2 - (0.6b)^2 \)
Apply the difference of squares:
\( = (a - 0.6b)(a + 0.6b) \)

(iv) Write as a difference of squares of squares:
\( a^4 - 625 = (a^2)^2 - 25^2 \)
\( = (a^2 - 25)(a^2 + 25) \)
Factorise the first bracket further:
\( = (a - 5)(a + 5)(a^2 + 25) \)

(v) Solve by splitting the middle term for \( x^2 \):
\( x^4 - 5x^2 - 36 = (x^2)^2 - 9x^2 + 4x^2 - 36 \)
Group and factorise:
\( = x^2(x^2 - 9) + 4(x^2 - 9) \)
\( = (x^2 - 9)(x^2 + 4) \)
Further factorise the difference of squares:
\( = (x - 3)(x + 3)(x^2 + 4) \)

(vi) Split the middle term \( -16(2x - y) \) into \( -25(2x - y) \) and \( 9(2x - y) \):
\( 15(2x - y)^2 - 16(2x - y) - 15 = 15(2x - y)^2 - 25(2x - y) + 9(2x - y) - 15 \)
Factorise by grouping:
\( = 5(2x - y)[3(2x - y) - 5] + 3[3(2x - y) - 5] \)
\( = [3(2x - y) - 5][5(2x - y) + 3] \)
Expand inside the brackets:
\( = (6x - 3y - 5)(10x - 5y + 3) \)
In simple words: Look for the basic structure of each expression. Some are simple differences of squares, some are quadratic in form (where you split the middle term), and some require a second level of factorisation.

Exam Tip: In sub-part v, always remember that while \( x^2 - 9 \) can be factorised further into \( (x - 3)(x + 3) \), the term \( x^2 + 4 \) is a sum of squares and cannot be factorised any further over real numbers.

Question 9. Factorise \( a^2b - b^3 \) Using this result, find the value of \( 101^2 \times 100 - 100^3 \).
Answer:
First, factorise the expression by taking out the common factor \( b \):
\( a^2b - b^3 = b(a^2 - b^2) \)
Using the difference of squares identity, we get:
\( = b(a - b)(a + b) \)

Now, compare \( 101^2 \times 100 - 100^3 \) with the expression \( a^2b - b^3 \). Here, \( a = 101 \) and \( b = 100 \):
Substitute these values into our factorised form \( b(a - b)(a + b) \):
\( = 100(101 - 100)(101 + 100) \)
\( = 100(1)(201) \)
\( = 20100 \)
In simple words: First factorise \( a^2b - b^3 \) into \( b(a - b)(a + b) \). By matching the numbers \( a = 101 \) and \( b = 100 \), we can find the answer easily without doing heavy multiplication.

Exam Tip: When evaluating large arithmetic expressions, look for patterns that match standard algebraic factorisation identities to save time and reduce errors.

Question 10. Evaluate (using factors): \( 301^2 \times 300 - 300^3 \)
Answer:
Express the given calculation in the algebraic form \( a^2b - b^3 \) by letting \( a = 301 \) and \( b = 300 \):
\( 301^2 \times 300 - 300^3 = 300(301^2 - 300^2) \)
Apply the difference of squares identity \( X^2 - Y^2 = (X - Y)(X + Y) \) to the term inside the parentheses:
\( = 300(301 - 300)(301 + 300) \)
Simplify the expressions within the brackets:
\( = 300(1)(601) \)
Multiply the final values:
\( = 180300 \)
In simple words: Pull out 300 as a common factor, leaving \( 301^2 - 300^2 \). Use the formula \( A^2 - B^2 = (A - B)(A + B) \) to calculate the bracket easily and multiply the result by 300.

Exam Tip: Showing the step \( 301 - 300 = 1 \) is important to demonstrate that you used algebraic factors to solve the problem rather than direct calculations.