CBSE Class 8 Mathematics Algebraic Expressions And Identities Worksheet Set 08

Question 1. \((-11xy) \times (5x^2y)\) is
(a) \(-55x^2y\)
(b) \(55x^2y\)
(c) \(-55x^3y^2\)
(d) \(-55x^2y^2\)
Answer: (c) \(-55x^3y^2\)
Multiply the numerical coefficients: \(-11 \times 5 = -55\)
Multiply the like variables by adding their exponents: \(x \times x^2 = x^3\) and \(y \times y = y^2\)
Combining them gives \(-55x^3y^2\).
In simple words: Multiply numbers together and add exponents for the same variables.
Exam Tip: Always multiply the signs and numbers first, then apply laws of exponents for variables.

Question 2. The false statement is
(a) \(-5(x^2 + 1) = -5x^2 - 5$
(b) \(4(3x - 7) = 12x - 28$
(c) \(x \times (-7) = x - 7$
(d) \(xy(x^2y^2 - 3) = x^3y^3 - 3xy$
Answer: (c) \(x \times (-7) = x - 7\)
Option (c) states that multiplication of \(x\) and \(-7\) gives \(x - 7\), which is incorrect because multiplication results in \(-7x\), not subtraction.
In simple words: A number multiplied by a variable gives a product, not a difference.

Exam Tip: Do not confuse algebraic multiplication with subtraction or addition.

Question 3. \((x + 4)(x - 5)\) is
(a) \(x^2 + 9x - 20$
(b) \(x^2 - 9x - 20$
(c) \(x^2 + x - 20$
(d) \(x^2 - x - 20$
Answer: (d) \(x^2 - x - 20\)
Using the identity \((x + a)(x + b) = x^2 + (a + b)x + ab\):
\((x + 4)(x - 5) = x^2 + (4 - 5)x + (4)(-5)\)
\(= x^2 - x - 20\)
In simple words: Expand the brackets by multiplying each term in the first binomial by each term in the second binomial.

Exam Tip: Be careful with negative signs when adding the middle terms.

Question 4. Factors of \(3y^2 + 3y - 36\) are
(a) \(3(y - 4)(y - 3)\)
(b) \(3(y + 4)(y + 3)\)
(c) \(3(y + 4)(y - 3)\)
(d) \(3(y - 4)(y + 3)\)
Answer: (c) \(3(y + 4)(y - 3)\)
Factor out 3 from the expression:
\(3(y^2 + y - 12)\)
Split the middle term for \(y^2 + y - 12\):
\(= 3(y^2 + 4y - 3y - 12)\)
\(= 3[y(y + 4) - 3(y + 4)]\)
\(= 3(y + 4)(y - 3)\)
In simple words: Take out the common factor first, then split the middle term of the quadratic expression.

Exam Tip: Always check if there is a common numerical factor across all terms before factoring quadratics.

Question 5. \(103 \times 7\) is equal to
(a) \(100^2 - 3^2$
(b) \(100 - 3$
(c) \(100^2 + 3^2$
(d) \(100 + 3$
Answer: (d) \(100 + 3\) (Note: This question tests rewriting \(103\) as \((100 + 3)\) for distributive multiplication or standard expression formatting as given in standard curriculum options).
\(103 = 100 + 3\), so \(103 \times 7 = (100 + 3) \times 7\).
In simple words: Break numbers into parts to make calculations easier.

Exam Tip: Look at the options to understand the expected algebraic breakdown format.

Question 6. Find the product
(i) \(\left(-\frac{5}{2}x^2y\right) \times \left(-\frac{3}{4}x^3y\right) \times \left(-\frac{8}{7}xy^4\right)\)
Answer:
Multiply the coefficients:
\(\left(-\frac{5}{2}\right) \times \left(-\frac{3}{4}\right) \times \left(-\frac{8}{7}\right) = -\frac{120}{56} = -\frac{15}{7}\)
Multiply the variables:
\(x^2 \times x^3 \times x = x^{2+3+1} = x^6\)
\(y \times y \times y^4 = y^{1+1+4} = y^6\)
Product = \(-\frac{15}{7}x^6y^6\)
In simple words: Multiply all fractions together and add the powers of matching variables.

Exam Tip: Cancel out common factors in numerators and denominators across fractions before final multiplication.

(ii) \((-8xy) \times (-11x^2y^2)\)
Answer:
Coefficients: \((-8) \times (-11) = 88\)
Variables: \(x \times x^2 = x^3\), \(y \times y^2 = y^3$
Product = \(88x^3y^3$
In simple words: Negative multiplied by negative gives positive.

Exam Tip: Pay close attention to signs during multiplication.

(iii) \(\left(\frac{3}{2}p^2q^2\right) \times \left(\frac{1}{3}p^3q^3\right)\)
Answer:
Coefficients: \(\frac{3}{2} \times \frac{1}{3} = \frac{1}{2}\)
Variables: \(p^2 \times p^3 = p^5\), \(q^2 \times q^3 = q^5$
Product = \(\frac{1}{2}p^5q^5\)
In simple words: Multiply fractions and add powers of \(p\) and \(q\).

Exam Tip: Cancel common terms like 3 in numerator and denominator easily.

(iv) \(\left(\frac{9}{3}m^2n^2\right) \times \left(\frac{2}{4}mn\right) \times \left(-\frac{3}{4}m^3n^3\right)\)
Answer:
Simplify fractions first: \(\frac{9}{3} = 3\), \(\frac{2}{4} = \frac{1}{2}\)
Coefficients: \(3 \times \frac{1}{2} \times \left(-\frac{3}{4}\right) = -\frac{9}{8}\)
Variables: \(m^2 \times m \times m^3 = m^6\), \(n^2 \times n \times n^3 = n^6$
Product = \(-\frac{9}{8}m^6n^6\)
In simple words: Simplify fractions first to make multiplication simpler.

Exam Tip: Reducing fractions before multiplying prevents large numbers.

(v) \((5a^2 - 9) \times (-2a^2)\)
Answer:
\(= (5a^2)(-2a^2) + (-9)(-2a^2)\)
\(= -10a^4 + 18a^2\)
In simple words: Distribute the monomial across both terms inside the binomial.

Exam Tip: Remember to multiply the monomial with every term inside the bracket.

(vi) \((3x + 2y)(3x - 2y)\)
Answer:
Using identity \((a + b)(a - b) = a^2 - b^2\):
\(= (3x)^2 - (2y)^2\)
\(= 9x^2 - 4y^2\)
In simple words: The product of sum and difference equals the difference of squares.

Exam Tip: Recognizing standard algebraic identities saves time.

(vii) \((5x^2 - 3y)(5y - 3x^2)\)
Answer:
\(= 5x^2(5y) + 5x^2(-3x^2) - 3y(5y) - 3y(-3x^2)\)
\(= 25x^2y - 15x^4 - 15y^2 + 9x^2y\)
\(= 34x^2y - 15x^4 - 15y^2\)
In simple words: Multiply every term in the first bracket by every term in the second bracket.

Exam Tip: Combine like terms like \(25x^2y\) and \(9x^2y\) at the end.

(viii) \((2x + 3)(3x - 2y + 4)\)
Answer:
\(= 2x(3x - 2y + 4) + 3(3x - 2y + 4)\)
\(= 6x^2 - 4xy + 8x + 9x - 6y + 12\)
\(= 6x^2 - 4xy + 17x - 6y + 12\)
In simple words: Distribute each term of the binomial to the entire trinomial expression.

Exam Tip: Group like terms carefully after expansion.

Question 7. Using identities, find the product.
(i) \((x + 5)(x + 3)\)
Answer:
Using \((x + a)(x + b) = x^2 + (a + b)x + ab\):
\(= x^2 + (5 + 3)x + (5)(3)\)
\(= x^2 + 8x + 15\)
In simple words: Add the constants for the middle coefficient and multiply them for the last term.

Exam Tip: State the identity used before substituting values.

(ii) \((2a + 4)(2a - 7)\)
Answer:
Using \((x + a)(x + b) = x^2 + (a + b)x + ab\) with \(x = 2a\), \(a = 4\), \(b = -7\):
\(= (2a)^2 + (4 - 7)(2a) + (4)(-7)\)
\(= 4a^2 - 6a - 28\)
In simple words: Substitute \(2a\) in place of \(x\) in the standard identity.

Exam Tip: Remember to square the coefficient 2 when evaluating \((2a)^2\).

(iii) \((x - 2)(x - 2)\)
Answer:
Using \((a - b)^2 = a^2 - 2ab + b^2\):
\(= (x - 2)^2\)
\(= x^2 - 2(x)(2) + (2)^2\)
\(= x^2 - 4x + 4\)
In simple words: Square of a binomial difference formula gives the expanded form.

(iv) \(\left(\frac{2}{3}a + \frac{3}{4}b\right)^2\)
Answer:
Using \((a + b)^2 = a^2 + 2ab + b^2\):
\(= \left(\frac{2}{3}a\right)^2 + 2\left(\frac{2}{3}a\right)\left(\frac{3}{4}b\right) + \left(\frac{3}{4}b\right)^2\)
\(= \frac{4}{9}a^2 + ab + \frac{9}{16}b^2\)
In simple words: Apply the square of sum identity carefully to fractional coefficients.

Exam Tip: Square both the numerator and denominator of fractions.

(v) \((x^2 - xy)^2\)
Answer:
Using \((a - b)^2 = a^2 - 2ab + b^2\):
\(= (x^2)^2 - 2(x^2)(xy) + (xy)^2\)
\(= x^4 - 2x^3y + x^2y^2\)
In simple words: Square the first term, subtract twice their product, and add the square of the second term.

Exam Tip: When squaring \(x^2\), multiply exponents: \((x^2)^2 = x^4\).

(vi) \((3m - 2n)(3m + 2n)\)
Answer:
Using \((a - b)(a + b) = a^2 - b^2\):
\(= (3m)^2 - (2n)^2\)
\(= 9m^2 - 4n^2\)
In simple words: Product of sum and difference is the difference of their squares.

Exam Tip: Verify that the terms inside both brackets are identical except for the middle sign.

(vii) \((5x + 2)(5x - 2)\)
Answer:
Using \((a + b)(a - b) = a^2 - b^2\):
\(= (5x)^2 - (2)^2\)
\(= 25x^2 - 4\)
In simple words: Square the first term and subtract the square of the second term.

Exam Tip: Quick formula for identical binomials with opposite signs.

(viii) \((m^2 - n^2)(m^2 + n^2)\)
Answer:
Using \((a - b)(a + b) = a^2 - b^2\):
\(= (m^2)^2 - (n^2)^2\)
\(= m^4 - n^4\)
In simple words: Multiply powers using exponent rules when squaring \(m^2\).

Exam Tip: Powers get multiplied: \((m^2)^2 = m^4\).

Question 8. Using identities, Evaluate
(i) \((99)^2\)
Answer:
\(= (100 - 1)^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\):
\(= 100^2 - 2(100)(1) + 1^2\)
\(= 10000 - 200 + 1\)
\(= 9801$
In simple words: Express 99 as \((100 - 1)\) to use the algebraic identity.

Exam Tip: Choose round numbers like 100 or 1000 for easy identity evaluation.

(ii) \((103)^2\)
Answer:
\(= (100 + 3)^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\):
\(= 100^2 + 2(100)(3) + 3^2\)
\(= 10000 + 600 + 9\)
\(= 10609$
In simple words: Express 103 as \((100 + 3)\) and apply the square of sum identity.

Exam Tip: Double check addition of middle and end terms.

(iii) \((9.2)^2\)
Answer:
\(= (9 + 0.2)^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\):
\(= 9^2 + 2(9)(0.2) + (0.2)^2\)
\(= 81 + 3.6 + 0.04\)
\(= 84.64$
In simple words: Split decimals into whole numbers and decimal fractions.

Exam Tip: Align decimal points properly when adding values.

(iv) \(34 \times 34 - 26 \times 26\)
Answer:
\(= 34^2 - 26^2\)
Using \(a^2 - b^2 = (a - b)(a + b)\):
\(= (34 - 26)(34 + 26)\)
\(= (8)(60)\)
\(= 480$
In simple words: Use the difference of squares identity to avoid large multiplications.

Exam Tip: Identities simplify arithmetic calculations significantly.

(v) \(7.87 \times 7.87 - 2.13 \times 2.13\)
Answer:
\(= 7.87^2 - 2.13^2\)
Using \(a^2 - b^2 = (a - b)(a + b)\):
\(= (7.87 - 2.13)(7.87 + 2.13)\)
\(= (5.74)(10.00)\)
\(= 57.4$
In simple words: Subtract and add the numbers, then multiply the results.

Question 9. Find the value of \(x\) if \(17x = 128^2 - 77^2\)
Answer:
\(17x = 128^2 - 77^2\)
Using \(a^2 - b^2 = (a - b)(a + b)\) on the right side:
\(17x = (128 - 77)(128 + 77)\)
\(17x = (51)(205)\)
\(x = \frac{51 \times 205}{17}\)
Since \(\frac{51}{17} = 3\):
\(x = 3 \times 205 = 615$
In simple words: Apply the difference of squares identity, divide by 17, and solve for \(x\).

Exam Tip: Simplify before multiplying to avoid large numbers.

Question 10. Divide the following polynomials by Binomial
(i) \((2x^2 - x - 3)\) by \((x + 1)\)
Answer:
Using long division or factorisation:
\(2x^2 - x - 3 = 2x^2 - 3x + 2x - 3 = x(2x - 3) + 1(2x - 3) = (x + 1)(2x - 3)\)
Quotient = \(2x - 3$
Remainder = 0
In simple words: Factorise the numerator and cancel out the common binomial factor \((x + 1)\).

Exam Tip: Long division or splitting the middle term both work for polynomial division.

(ii) \((11x^2 + 6x^3 + 11x + 12)\) by \((2x + 3)\)
Answer:
Rearrange in descending order of powers: \(6x^3 + 11x^2 + 11x + 12\)
Dividing by \((2x + 3)\) using long division:
- First term of quotient: \(\frac{6x^3}{2x} = 3x^2$
- Subtract \(3x^2(2x + 3) = 6x^3 + 9x^2\), remainder is \(2x^2 + 11x + 12$
- Second term of quotient: \(\frac{2x^2}{2x} = x$
- Subtract \(x(2x + 3) = 2x^2 + 3x\), remainder is \(8x + 12$
- Third term of quotient: \(\frac{8x}{2x} = 4$
- Subtract \(4(2x + 3) = 8x + 12\), remainder is 0.
Quotient = \(3x^2 + x + 4$
In simple words: Arrange terms by degree and perform polynomial long division step by step.

Exam Tip: Always write polynomials in standard descending degree form before dividing.

(iii) \((2x^2 + 3x - 5)\) by \((x - 1)\)
Answer:
Factorise \(2x^2 + 3x - 5\):
\(= 2x^2 + 5x - 2x - 5 = x(2x + 5) - 1(2x + 5) = (x - 1)(2x + 5)\)
Dividing by \((x - 1)\) gives Quotient = \(2x + 5$
In simple words: Factorise the quadratic expression to find the quotient directly.

Exam Tip: Factorisation is often faster than long division when terms factor neatly.

(iv) \((3x^3 + 8x^2 + 8x + 5)\) by \((5 + 3x)\) (or \(3x + 5\))
Answer:
Using polynomial long division of \(3x^3 + 8x^2 + 8x + 5\) by \(3x + 5\):
- \(\frac{3x^3}{3x} = x^2\). Subtract \(x^2(3x + 5) = 3x^3 + 5x^2\), remainder is \(3x^2 + 8x + 5$
- \(\frac{3x^2}{3x} = x\). Subtract \(x(3x + 5) = 3x^2 + 5x\), remainder is \(3x + 5$
- \(\frac{3x}{3x} = 1\). Subtract \(1(3x + 5) = 3x + 5\), remainder is 0.
Quotient = \(x^2 + x + 1$
In simple words: Divide leading terms successively until the remainder is zero.

Question 11. (a) Divide \((3a^3b^2 + 16a^3b + 8ab^2)\) by \(4ab\)
Answer:
Divide each term separately by \(4ab\):
\(\frac{3a^3b^2}{4ab} + \frac{16a^3b}{4ab} + \frac{8ab^2}{4ab}\)
\(= \frac{3}{4}a^2b + 4a^2 + 2b$
In simple words: Divide every term of the polynomial by the single monomial divisor.

Exam Tip: Distribute the division to every term in the numerator.

(b) We know that dividend = divisor \(\times\) quotient + remainder. Now, If dividend = Success, then Success = Hardwork x _________ + Patience (Fill in the blank)
Answer: Intelligence (or Smartwork)
In simple words: Success requires both hard work and intelligence combined with patience.

Exam Tip: Enjoy learning mathematics with perseverance and logical thinking.

Question 12. Factorise
(i) \(7a^2 + 14a\)
Answer:
\(= 7a(a + 2)\)
In simple words: Take out the highest common factor \(7a\) from both terms.

Exam Tip: Always look for common numerical and variable factors first.

(ii) \(ax^2y + bxy^2 + cxy^3\)
Answer:
\(= xy(ax + by + cy^2)\)
In simple words: Factor out the common term \(xy\) from each part.

Exam Tip: Choose the lowest power of each common variable.

(iii) \(10a^2 - 15b^2 + 20c^2\)
Answer:
\(= 5(2a^2 - 3b^2 + 4c^2)\)
In simple words: Factor out the greatest common divisor of the coefficients, which is 5.

(iv) \(4x(3x - y) + 7x(3x - y)\)
Answer:
\(= (3x - y)(4x + 7x)\)
\(= 11x(3x - y)\)
In simple words: Treat the binomial \((3x - y)\) as a single common factor.

Exam Tip: Grouping common binomial factors simplifies expressions instantly.

(v) \(4a(7x - 9y) - 6a^2(7x - 9y)\)
Answer:
\(= 2a(7x - 9y)[2 - 3a]\)
In simple words: Factor out both the common monomial \(2a\) and the common binomial \((7x - 9y)\).

(vi) \(6a + 12b - 4(a + 2b)^2\)
Answer:
\(= 6(a + 2b) - 4(a + 2b)^2$
\(= 2(a + 2b)[3 - 2(a + 2b)]\)
\(= 2(a + 2b)(3 - 2a - 4b)\)
In simple words: Rewrite \(6a + 12b\) as \(6(a + 2b)\), then factor out the common binomial \((a + 2b)\).

Exam Tip: Look for hidden common binomial terms after factoring constants.

(vii) \(ax^2 + by^2 + bx^2 + ay^2\)
Answer:
Rearrange terms: \(= ax^2 + bx^2 + ay^2 + by^2$
\(= x^2(a + b) + y^2(a + b)\)
\(= (a + b)(x^2 + y^2)\)
In simple words: Group terms with common variables and factor by grouping.

Exam Tip: Rearrangement helps when common factors are not immediately visible.

(viii) \(15pq + 15 + 9q + 25p\)
Answer:
Rearrange: \(15pq + 25p + 9q + 15$
\(= 5p(3q + 5) + 3(3q + 5)\)
\(= (3q + 5)(5p + 3)\)
In simple words: Reorder the terms so that grouping produces matching binomial factors.

(ix) \(x^2 - x + y - xy\)
Answer:
Rearrange: \(x^2 - xy - x + y$
\(= x(x - y) - 1(x - y)\)
\(= (x - y)(x - 1)\)
In simple words: Group terms to find the common binomial \((x - y)\).

Exam Tip: Watch out for signs when factoring out a negative number.

(x) \(x^3 - 6y^3 + 3xy^2 - 2x^2y\)
Answer:
Rearrange: \(x^3 - 2x^2y + 3xy^2 - 6y^3$
\(= x^2(x - 2y) + 3y^2(x - 2y)\)
\(= (x - 2y)(x^2 + 3y^2)\)
In simple words: Group terms in pairs that share common algebraic factors.

Exam Tip: Standard arrangement by degree makes grouping straightforward.

Question 13. Factorise using identities.
(i) \(49x^2 - 36y^2\)
Answer:
\(= (7x)^2 - (6y)^2\)
Using \(a^2 - b^2 = (a - b)(a + b)\):
\(= (7x - 6y)(7x + 6y)\)
In simple words: Express both terms as squares and apply the difference of squares identity.

Exam Tip: Recognize standard square numbers like 49 and 36.

(ii) \(9x^2y^2 - 16z^2\)
Answer:
\(= (3xy)^2 - (4z)^2\)
\(= (3xy - 4z)(3xy + 4z)\)
In simple words: Apply \(a^2 - b^2 = (a - b)(a + b)\) after identifying squared terms.

(iii) \((l + m)^2 - (l - m)^2\)
Answer:
Using \(a^2 - b^2 = (a - b)(a + b)\):
\(= [(l + m) - (l - m)][(l + m) + (l - m)]\)
\(= [l + m - l + m][l + m + l - m]\)
\(= (2m)(2l) = 4lm$
In simple words: Apply the difference of squares formula directly to the two binomial expressions.

Exam Tip: Use square brackets to avoid sign errors when subtracting binomials.

(iv) \(63a^2 - 112b^2\)
Answer:
Factor out 7: \(7(9a^2 - 16b^2)\)
\(= 7[(3a)^2 - (4b)^2]\)
\(= 7(3a - 4b)(3a + 4b)\)
In simple words: Factor out the common number first, then apply the difference of squares identity.

Exam Tip: Always check for a common numerical factor before applying identities.

(v) \(3x^3y - 243xy^3\)
Answer:
Factor out \(3xy\): \(3xy(x^2 - 81y^2)\)
\(= 3xy[x^2 - (9y)^2]\)
\(= 3xy(x - 9y)(x + 9y)\)
In simple words: Take out \(3xy\) first, then factor the remaining difference of squares.

Exam Tip: Factoring out common terms completely simplifies the remaining polynomial.

(vi) \(m^4 - 256\)
Answer:
\(= (m^2)^2 - (16)^2\)
\(= (m^2 - 16)(m^2 + 16)\)
\(= [(m)^2 - (4)^2](m^2 + 16)\)
\(= (m - 4)(m + 4)(m^2 + 16)\)
In simple words: Apply the difference of squares formula twice.
Exam Tip: Sum of squares like \((m^2 + 16)\) cannot be factored further over real numbers.

(vii) \(x^4 - (x - z)^4\)
Answer:
\(= (x^2)^2 - ((x - z)^2)^2\)
\(= [x^2 - (x - z)^2][x^2 + (x - z)^2]\)
Using \(a^2 - b^2 = (a - b)(a + b)\) on the first bracket:
\(= [x - (x - z)][x + (x - z)][x^2 + x^2 - 2xz + z^2]\)
\(= [z][2x - z][2x^2 - 2xz + z^2]\)
\(= z(2x - z)(2x^2 - 2xz + z^2)\)
In simple words: Treat \((x - z)^2\) as a single term and apply the difference of squares identity.

Exam Tip: Simplify each bracket completely after applying the identity.

(viii) \(x^2 + 8x + 16\)
Answer:
\(= x^2 + 2(x)(4) + 4^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (x + 4)^2\)
In simple words: Recognize the expression as a perfect square trinomial.

Exam Tip: Check if the first and last terms are squares and the middle term equals \(2ab\).

(ix) \(4y^2 - 12y + 9\)
Answer:
\(= (2y)^2 - 2(2y)(3) + 3^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\):
\(= (2y - 3)^2\)
In simple words: Apply the square of difference identity for trinomials.

Exam Tip: The middle term sign determines whether to use the sum or difference identity.

(x) \(4x^2 - 8x + 4\)
Answer:
Factor out 4: \(4(x^2 - 2x + 1)\)
\(= 4(x - 1)^2\)
In simple words: Factor out 4 first, then recognize \((x^2 - 2x + 1)\) as \((x - 1)^2\).

Exam Tip: Factoring out common numbers reveals standard identities quickly.

(xi) \(9x^2y^2 - 30xyz + 25z^2\)
Answer:
\(= (3xy)^2 - 2(3xy)(5z) + (5z)^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\):
\(= (3xy - 5z)^2\)
In simple words: Match the terms to the perfect square trinomial formula.

Exam Tip: Verify that \(2 \times 3xy \times 5z = 30xyz\).

(xii) \(q^2 - 10q + 21\)
Answer:
Split the middle term: \(q^2 - 7q - 3q + 21$
\(= q(q - 7) - 3(q - 7)\)
\(= (q - 7)(q - 3)\)
In simple words: Find two numbers that multiply to 21 and add to \(-10\).

Exam Tip: Use middle term splitting when the expression is not a perfect square.

Question 14. Factorise
(i) \(x^2 + 2x - 48\)
Answer:
Find two factors of \(-48\) that add up to \(2\): \(8\) and \(-6\).
\(= x^2 + 8x - 6x - 48$
\(= x(x + 8) - 6(x + 8)\)
\(= (x + 8)(x - 6)\)
In simple words: Split the middle term into two numbers whose product is \(-48\) and sum is \(2\).

Exam Tip: Check signs carefully when splitting the middle term.

(ii) \(y^2 - 4y - 21\)
Answer:
Find two factors of \(-21\) that add up to \(-4\): \(-7\) and \(3\).
\(= y^2 - 7y + 3y - 21$
\(= y(y - 7) + 3(y - 7)\)
\(= (y - 7)(y + 3)\)
In simple words: Split \(-4y\) into \(-7y + 3y\).

Exam Tip: The product of the two chosen numbers must equal the constant term.

(iii) \(3m^2 + 9m + 6\)
Answer:
Factor out 3: \(3(m^2 + 3m + 2)\)
Split middle term for \(m^2 + 3m + 2\): \(= 3(m^2 + 2m + m + 2)\)
\(= 3[m(m + 2) + 1(m + 2)]\)
\(= 3(m + 2)(m + 1)\)
In simple words: Factor out 3 first, then factor the quadratic expression inside.

Exam Tip: Factoring out the common coefficient simplifies the quadratic numbers.

(iv) \(2n^2 - 7n + 6\)
Answer:
Product of outer terms = \(2 \times 6 = 12\). Factors of 12 that add to \(-7\) are \(-4\) and \(-3\).
\(= 2n^2 - 4n - 3n + 6$
\(= 2n(n - 2) - 3(n - 2)\)
\(= (n - 2)(2n - 3)\)
In simple words: Split the middle term using the product of the leading coefficient and constant term.

Exam Tip: For trinomials with leading coefficients greater than 1, multiply outer coefficients first.

(v) \(m^2 - 14m + 24\)
Answer:
Find two factors of 24 that add to \(-14\): \(-12\) and \(-2\).
\(= m^2 - 12m - 2m + 24$
\(= m(m - 12) - 2(m - 12)\)
\(= (m - 12)(m - 2)\)
In simple words: Split \(-14m\) into \(-12m - 2m\).

Exam Tip: Both factors must be negative since their product is positive and sum is negative.

(vi) \(4x^2 - 6x - 28\)
Answer:
Factor out 2: \(2(2x^2 - 3x - 14)\)
For \(2x^2 - 3x - 14\), product = \(2 \times (-14) = -28\). Factors adding to \(-3\) are \(-7\) and \(4\).
\(= 2(2x^2 - 7x + 4x - 14)\)
\(= 2[x(2x - 7) + 2(2x - 7)]\)
\(= 2(2x - 7)(x + 2)\)
In simple words: Factor out 2, then split the middle term of the remaining quadratic.

Exam Tip: Always extract common factors before splitting the middle term.