CBSE Class 8 Mathematics Algebraic Expressions And Identities Worksheet Set 05

FACTORISATION OF ALGEBEREIC EXPRESSION

1) Find the highest factors of the monomials

Question 1(a). \(6a^2b^2c\) and \(27abc^3\)
Answer: Factors of \(6a^2b^2c\) = \(2 \times 3 \times a \times a \times b \times b \times c\)
Factors of \(27abc^3\) = \(3 \times 3 \times 3 \times a \times b \times c \times c \times c\)
Highest common factor = \(3 \times a \times b \times c = 3abc\)
In simple words: Find the greatest common number and variables that divide both terms completely.

Exam Tip: List the prime factors of numerical coefficients and take the lowest power for each common variable.

Question 1(b). \(5a^2\), \(-25a^4\) and \(100a\)
Answer: Factors of \(5a^2\) = \(5 \times a \times a\)
Factors of \(-25a^4\) = \(-1 \times 5 \times 5 \times a \times a \times a \times a\)
Factors of \(100a\) = \(2 \times 2 \times 5 \times 5 \times a\)
Highest common factor = \(5 \times a = 5a\)
In simple words: The highest common factor is \(5a\) because 5 divides all coefficients and \(a\) is the lowest power present in all terms.

Exam Tip: Ignore negative signs when finding the highest common factor of magnitudes unless specified.

Question 1(c). \(11abc^3\), \(13a^2b^2c\) and \(17abc\)
Answer: Notice that 11, 13, and 17 are all prime numbers with no common numerical factor other than 1.
Factors of \(11abc^3\) = \(11 \times a \times b \times c \times c \times c\)
Factors of \(13a^2b^2c\) = \(13 \times a \times a \times b \times b \times c\)
Factors of \(17abc\) = \(17 \times a \times b \times c\)
Highest common factor = \(abc\)
In simple words: Since the coefficients share no common factor, only the common variables with lowest powers \(abc\) form the highest factor.

Exam Tip: When coefficients are prime numbers, check if they share any divisors; if not, the numerical HCF is 1.

Question 1(d). \(x^3y^2\) and \(-7x^2\)
Answer: Factors of \(x^3y^2\) = \(x \times x \times x \times y \times y\)
Factors of \(-7x^2\) = \(-1 \times 7 \times x \times x\)
Highest common factor = \(x^2\)
In simple words: The variable \(y\) is missing in the second term, so only \(x^2\) is common.

Question 2(a). \(25a^2 - b^2\)
Answer: Using the identity \(a^2 - b^2 = (a - b)(a + b)\):
\(25a^2 - b^2 = (5a)^2 - (b)^2\)
\(= (5a - b)(5a + b)\)
In simple words: Express both terms as squares and apply the difference of squares formula.

Exam Tip: Always look for the difference of two squares pattern: \(x^2 - y^2 = (x - y)(x + y)\).

Question 2(b). \(49 - 36p^2\)
Answer: Rewrite using squares: \(7^2 - (6p)^2\)
Using \(a^2 - b^2 = (a - b)(a + b)\):
\(= (7 - 6p)(7 + 6p)\)
In simple words: Rewrite 49 as \(7^2\) and \(36p^2\) as \((6p)^2\), then factorise.

Exam Tip: The order matters; keep the terms in the same relative position as in the original expression.

Question 2(c). \(4a^2 - 8a + 4\)
Answer: Take out the common factor 4:
\(= 4(a^2 - 2a + 1)\)
Recognize the perfect square trinomial inside: \(a^2 - 2a + 1 = (a - 1)^2\)
\(= 4(a - 1)^2\)
In simple words: First factor out the number 4 from all terms, then use the square formula.

Exam Tip: Always check for a common monomial factor across all terms before applying identities.

Question 2(d). \((\frac{1}{3}x)^2 - \frac{2}{3}x + 1\)
Answer: Rewrite the expression as: \((\frac{1}{3}x)^2 - 2(\frac{1}{3}x)(1) + (1)^2\)
Using the identity \(a^2 - 2ab + b^2 = (a - b)^2\):
\(= (\frac{1}{3}x - 1)^2\)
In simple words: Match the expression to the square of a difference formula.

Exam Tip: Verify the middle term: \(2 \times \frac{1}{3}x \times 1 = \frac{2}{3}x\).

Question 2(e). \(9m^2 + 4n^2 + 12mn\)
Answer: Rearrange terms in standard form: \(9m^2 + 12mn + 4n^2\)
Rewrite as squares: \((3m)^2 + 2(3m)(2n) + (2n)^2\)
Using the identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (3m + 2n)^2\)
In simple words: Rewrite the first and last terms as squares and check if the middle term is twice their product.

Exam Tip: Rearranging terms into standard descending order makes recognizing identities much easier.

Question 2(f). \((x + 2y)^2 - 1\)
Answer: Treat \((x + 2y)\) as a single variable \(A\):
\(A^2 - 1^2 = (A - 1)(A + 1)\)
Substitute back \(A = x + 2y\):
\(= (x + 2y - 1)(x + 2y + 1)\)
In simple words: Apply the difference of squares formula directly by viewing the bracket as a single term.

Exam Tip: Substitution helps avoid confusion when binomials are squared.

Question 2(g). \(48y^3 - 147y\)
Answer: Factor out the common terms \(3y\):
\(= 3y(16y^2 - 49)\)
Express inside as a difference of squares: \(16y^2 - 49 = (4y)^2 - 7^2\)
\(= 3y(4y - 7)(4y + 7)\)
In simple words: Take out \(3y\) first, then apply the difference of squares formula to the remaining binomial.

Exam Tip: Never stop factoring after taking out the common term; check if the remaining factor can be factored further.

Question 2(h). \(p^2 - 17p + 72\)
Answer: Split the middle term by finding two numbers whose product is 72 and sum is -17. These numbers are -8 and -9.
\(= p^2 - 8p - 9p + 72\)
\(= p(p - 8) - 9(p - 8)\)
\(= (p - 8)(p - 9)\)
In simple words: Find two numbers that multiply to 72 and add up to -17, then factor by grouping.

Exam Tip: When the product is positive and sum is negative, both numbers must be negative.

Question 2(i). \(m^2 + 18m + 77\)
Answer: Find two numbers whose product is 77 and sum is 18. These numbers are 7 and 11.
\(= m^2 + 7m + 11m + 77\)
\(= m(m + 7) + 11(m + 7)\)
\(= (m + 7)(m + 11)\)
In simple words: Split the middle term 18m into 7m and 11m, then factor by grouping.

Exam Tip: Check your factors by multiplying them back: \((m+7)(m+11) = m^2 + 18m + 77\).

Question 2(j). \(6xy - 4y + 6 - 9x\)
Answer: Rearrange terms to group them conveniently:
\(= 6xy - 9x - 4y + 6\)
Factor out \(3x\) from the first two terms and \(-2\) from the last two terms:
\(= 3x(2y - 3) - 2(2y - 3)\)
\(= (2y - 3)(3x - 2)\)
In simple words: Rearrange and group terms so you can factor out common binomials.

Exam Tip: Ensure the binomials in both groups are identical in signs and terms before factoring out.

Question 2(k). \((m + 2n)^2 - 16m^2\)
Answer: Rewrite as a difference of squares:
\(= (m + 2n)^2 - (4m)^2\)
Using \(A^2 - B^2 = (A - B)(A + B)\) where \(A = m + 2n\) and \(B = 4m\):
\(= (m + 2n - 4m)(m + 2n + 4m)\)
\(= (-3m + 2n)(5m + 2n)\)
In simple words: Write \(16m^2\) as \((4m)^2\) and apply the difference of squares formula, then simplify.

Exam Tip: Combine like terms inside each bracket after applying the identity.

Question 2(l). \((x + y)^3 - 16(x + y)\)
Answer: Factor out the common binomial term \((x + y)\):
\(= (x + y) \left[ (x + y)^2 - 16 \right]\)
Apply the difference of squares on the bracket: \((x + y)^2 - 4^2 = (x + y - 4)(x + y + 4)\)
\(= (x + y)(x + y - 4)(x + y + 4)\)
In simple words: Take out \((x + y)\) as a common factor, then use the difference of squares on the remaining terms.

Exam Tip: Never expand cubic expressions if taking out a common factor simplifies the problem directly.

Question 2(m). \(25x^2 + 30x + 9\)
Answer: Rewrite as squares: \((5x)^2 + 2(5x)(3) + (3)^2\)
Using the identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (5x + 3)^2\)
In simple words: Express the first and last terms as squares of \(5x\) and 3, and check the middle term.

Exam Tip: Perfect square trinomials always have the form \((ax + b)^2\).

Question 2(n). \(x^2 + 2 + \frac{1}{x^2}\)
Answer: Rewrite as: \((x)^2 + 2(x)(\frac{1}{x}) + (\frac{1}{x})^2\)
Using the identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (x + \frac{1}{x})^2\)
In simple words: Recognize that 2 can be written as \(2(x)(\frac{1}{x})\), completing the square.

Exam Tip: Reciprocal quadratic expressions follow standard square identities.

Question 2(o). \(a^2x^2 - 20axb + b^2\)
Answer: Rewrite as squares: \((ax)^2 - 2(ax)(b) + (b)^2\)
Using the identity \((a - b)^2 = a^2 - 2ab + b^2\):
\(= (ax - b)^2\)
In simple words: Identify the squared terms as \(ax\) and \(b\), and verify the middle term.

Exam Tip: Pay close attention to multi-variable terms when identifying the base of squares.

Question 2(p). \(25 - 12xy - 9y^2\)
Answer: Let us rewrite with standard signs (Note: typical textbook problem assumes \(25x^2 - 12xy - 9y^2\) or similar, let us factor as given or check standard quadratic form. Assuming the expression is \(25x^2 - 12xy - 9y^2\)):
Let us solve for \(25x^2 - 12xy - 9y^2\) (assuming \(x\) is missing in the constant 25 in OCR, standard form is \(25x^2 - 12xy - 9y^2\)):
Product of outer coefficients = \(25 \times (-9) = -225\), middle coefficient = -12. Split middle term:
\(25x^2 - 15xy + 9xy - 9y^2\) (Wait, \(-15 \times 15 = -225\), \(-15 + 15 = 0\)). Let us use \(-25\) and \(13\) - actually, let us re-verify: \(25x^2 - 30xy + 9y^2 = (5x-3y)^2\). If the question is \(25 - 12xy - 9y^2\), it cannot be factored cleanly unless it is \(4 - 12xy - 9y^2\) or \(25x^2 - 12xy\) etc. Let us provide the standard factorisation for \(25x^2 - 12xy - 9y^2\):
\(= (5x)^2 - 2(5x)(...)\) - Let us factor using splitting of middle term if applicable, or present as: Expression as given does not form a standard square, assuming misprint for \(25x^2 - 30xy + 9y^2\):
\(= (5x - 3y)^2\)
In simple words: Group and apply algebraic identities for trinomials.

Exam Tip: Look for signs of typographical errors in mixed variable quadratic expressions.

Question 2(q). \(x^2 - 7x + 12\)
Answer: Find two numbers whose product is 12 and sum is -7. These numbers are -3 and -4.
\(= x^2 - 3x - 4x + 12\)
\(= x(x - 3) - 4(x - 3)\)
\(= (x - 3)(x - 4)\)
In simple words: Find two numbers that multiply to 12 and add to -7, which are -3 and -4.

Exam Tip: Splitting the middle term is the most reliable method for quadratic trinomials.

Question 2(r). \(x^2 - 7x - 44\)
Answer: Find two numbers whose product is -44 and sum is -7. These numbers are -11 and 4.
\(= x^2 - 11x + 4x - 44\)
\(= x(x - 11) + 4(x - 11)\)
\(= (x - 11)(x + 4)\)
In simple words: Find two numbers multiplying to -44 and adding to -7, namely -11 and 4.

Exam Tip: When the product is negative, one number is positive and the other is negative, with the larger number taking the sign of the middle term.

Question 2(s). \(a^2 - a - 56\)
Answer: Find two numbers whose product is -56 and sum is -1. These numbers are -8 and 7.
\(= a^2 - 8a + 7a - 56\)
\(= a(a - 8) + 7(a - 8)\)
\(= (a - 8)(a + 7)\)
In simple words: Split -a into -8a and +7a because -8 and 7 multiply to -56 and add to -1.

Exam Tip: A coefficient of -a implies -1 as the coefficient of the middle term.

3) Divide

Question 3(i). \((7x^2 + 14x) \div (x + 2)\)
Answer: Factorize the numerator by taking out the common factor \(7x\):
\(7x^2 + 14x = 7x(x + 2)\)
Now divide by \((x + 2)\):
\(\frac{7x(x + 2)}{x + 2} = 7x\)
In simple words: Factor out the common term from the numerator so that the divisor cancels out.

Exam Tip: Always factorize the polynomial completely before performing algebraic division.

Question 3(ii). \((m^2 + 21m - 46) \div (m - 2)\)
Answer: Factorize the quadratic expression \(m^2 + 21m - 46\):
Find two numbers whose product is -46 and sum is 21. These numbers are 23 and -2.
\(m^2 + 23m - 2m - 46 = m(m + 23) - 2(m + 23) = (m - 2)(m + 23)\)
Now divide by \((m - 2)\):
\(\frac{(m - 2)(m + 23)}{m - 2} = m + 23\)
In simple words: Factor the numerator into two binomials, then cancel the matching divisor term.

Exam Tip: The divisor is almost always one of the factors of the numerator in such worksheet problems.

Question 3(iii). \((x^2 + 4x - 21) \div (x - 3)\)
Answer: Factorize the quadratic expression \(x^2 + 4x - 21\):
Find two numbers whose product is -21 and sum is 4. These numbers are 7 and -3.
\(x^2 + 7x - 3x - 21 = x(x + 7) - 3(x + 7) = (x - 3)(x + 7)\)
Now divide by \((x - 3)\):
\(\frac{(x - 3)(x + 7)}{x - 3} = x + 7\)
In simple words: Factor the dividend into \((x - 3)(x + 7)\) and cancel \((x - 3)\).

Exam Tip: Double check your middle term splitting by expanding the factors.

Question 3(iv). \((2x^2 + 11x + 5) \div (2x + 1)\)
Answer: Factorize the quadratic expression \(2x^2 + 11x + 5\):
Product = \(2 \times 5 = 10\), sum = 11. Numbers are 10 and 1.
\(2x^2 + 10x + x + 5 = 2x(x + 5) + 1(x + 5) = (2x + 1)(x + 5)\)
Now divide by \((2x + 1)\):
\(\frac{(2x + 1)(x + 5)}{2x + 1} = x + 5\)
In simple words: Split the middle term to factorize the numerator, then cancel out \((2x + 1)\).

Exam Tip: For polynomials with a leading coefficient greater than 1, multiply the outer coefficients to find the split product.

4) Find the highest common factor of the monomials

Question 4(a). \(3abc\)
Answer: Since it is a single monomial, its highest common factor with itself is \(3abc\).
In simple words: The HCF of a single monomial is the monomial itself.

Exam Tip: When only one expression is given, its HCF refers to its own prime factor decomposition.

Question 4(b). \(5a\)
Answer: The highest common factor of the single monomial \(5a\) is \(5a\).
In simple words: The expression itself is the factor.

Exam Tip: Understand that standalone questions test the basic definition of monomial factors.

Question 4(c). \(1abc\)
Answer: The highest common factor is \(abc\) (or \(1abc\)).
In simple words: Coefficient 1 means the variables alone form the core factor.

Exam Tip: Coefficient 1 is usually omitted in final algebraic writing.

Question 4(d). \(x^2\)
Answer: The highest common factor is \(x^2\).
In simple words: The monomial \(x^2\) has factors 1 and \(x^2\).

Exam Tip: Powers of variables are fundamental building blocks in algebraic expressions.

5) Factorise

Question 5(a). \((5a + b)(5a - b)\)
Answer: This expression is already in factored form as the product of two binomials.
\(= (5a + b)(5a - b)\)
In simple words: The expression is already fully factorised.

Exam Tip: Recognize when an expression is already given in its factored form to avoid redundant work.

Question 5(b). \((5m + 2n)(-3m + 2n)\)
Answer: The expression is already factorised into two binomial factors:
\(= (5m + 2n)(-3m + 2n)\)
In simple words: It is already expressed as a product of factors.

Exam Tip: Check if any negative sign can be factored out of the brackets to clean up the expression.

Question 5(c). \(x + y[x + y + 4][x + y - 4]\)
Answer: The expression is already in its factored form:
\(= (x + y)(x + y + 4)(x + y - 4)\)
In simple words: The given expression is already written as a product of factors.

Exam Tip: Identify bracket structures to confirm if expansion or factorisation is requested.

Question 5(d). \((7 + 6p)(7 - 6p)\)
Answer: Already in factored form (difference of squares expansion):
\(= 49 - 36p^2\) (If expanded) or kept as \((7 + 6p)(7 - 6p)\).
In simple words: The expression is already factorised.

Exam Tip: Product of sum and difference is the factored form of \(a^2 - b^2\).

Question 5(e). \((5x + 3)(5x + 3)\)
Answer: Written as a repeated factor:
\(= (5x + 3)^2\)
In simple words: Combine identical factors using exponents.

Exam Tip: Write repeated identical factors as a squared binomial.

Question 5(f). \((2a - 2)(2a - 2)\)
Answer: Factor out 2 from each binomial:
\(= 2(a - 1) \cdot 2(a - 1)\)
\(= 4(a - 1)^2\)
In simple words: Factor out 2 from each bracket and combine them into \(4(a - 1)^2\).

Exam Tip: Always extract common numerical factors from binomials before squaring.

Question 5(g). \((\frac{1}{3}x - 1)(\frac{1}{3}x - 1)\)
Answer: Express as a squared binomial:
\(= (\frac{1}{3}x - 1)^2\)
In simple words: Combine the two identical fractional binomials into a square.

Exam Tip: Identical factors multiply to give a square.

Question 5(h). \((x + \frac{1}{x})(x + \frac{1}{x})\)
Answer: Express as a squared binomial:
\(= (x + \frac{1}{x})^2\)
In simple words: Write the product of identical terms with an exponent of 2.

Exam Tip: Reciprocal binomial products follow standard exponent rules.

Question 5(i). \((3m + 2n)(3m + 2n)\)
Answer: Express as a squared binomial:
\(= (3m + 2n)^2\)
In simple words: Simplify the product of identical binomials using a square.

Exam Tip: Keep multi-variable terms enclosed properly in parentheses.

Question 5(j). \((ax + b)(ax + b)\)
Answer: Express as a squared binomial:
\(= (ax + b)^2\)
In simple words: Combine the identical brackets into a squared term.

Exam Tip: Algebraic constants and coefficients remain grouped inside the parentheses.

Question 5(k). \(x + 2y - 1\)
Answer: This expression cannot be factored further using integer coefficients.
\(= x + 2y - 1\)
In simple words: The expression is irreducible in its current form.

Exam Tip: Not all polynomials can be factorised; recognize irreducible expressions.

Question 5(l). \(3y((4y + 7)(4y - 7))\)
Answer: The expression is already fully factorised:
\(= 3y(4y + 7)(4y - 7)\)
In simple words: The expression is presented as a product of a monomial and two binomial factors.

Exam Tip: Check that all sub-factors like \(16y^2 - 49\) are also fully broken down.

Question 5(m). \((-3 + x)(-4 + x)\)
Answer: Rewrite terms inside brackets in standard variable-first order:
\(= (x - 3)(x - 4)\)
In simple words: Rearrange terms inside each bracket to place the variable first.

Exam Tip: Writing variables before constants improves readability and standard form.

Question 5(n). \((p + (-8))(p - 9)\)
Answer: Simplify the first bracket:
\(= (p - 8)(p - 9)\)
In simple words: Simplify \(p + (-8)\) to \(p - 8\).

Exam Tip: Clean up double signs like \(+(-)\) before final presentation.

Question 5(o). \((x - 11)(x + 4)\)
Answer: The expression is already in factored form:
\(= (x - 11)(x + 4)\)
In simple words: The expression is already factorised.

Exam Tip: Verify that expansion yields the corresponding quadratic expression.

Question 5(p). \((m + 7)(m + 4)\)
Answer: The expression is already in factored form:
\(= (m + 7)(m + 4)\)
In simple words: The expression is already factorised.

Exam Tip: Product of two linear binomials represents a factorised quadratic.

Question 5(q). \((a - 8)(a + 7)\)
Answer: The expression is already in factored form:
\(= (a - 8)(a + 7)\)
In simple words: The expression is already factorised.

Exam Tip: Confirm that no further grouping or splitting is possible.

Question 5(r). \((2y + 3)(2 - 3x)\)
Answer: The expression is already in factored form:
\(= (2y + 3)(2 - 3x)\)
In simple words: The expression is already factorised with distinct variables.

Exam Tip: Different variables in each binomial mean no further combination of terms can occur.

6) Divide

Question 6(1). \(3x\)
Answer: As a standalone problem heading or monomial expression, its simplest form is \(3x\).
In simple words: Represents a monomial expression.

Exam Tip: Review context for implied division expressions in heading lists.

Question 6(2). \((x + 7)\)
Answer: Represents a linear binomial expression in simplest form.
In simple words: Irreducible linear binomial.

Exam Tip: Linear terms cannot be factored further without additional equations.

Question 6(3). \((m + 23)\)
Answer: Represents a linear binomial expression in simplest form.
In simple words: Irreducible linear binomial.

Exam Tip: Linear factors resulting from polynomial divisions are typically presented in this form.