CBSE Class 8 Mathematics Factorisation Worksheet Set 03

Factorisation of Algebraic Expressions

Question 1. Find the highest factors of the monomials
(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 \(27abc^3 = 3 \times 3 \times 3 \times a \times b \times c \times c \times c\)
Common numerical factor = 3
Common variables = \(a \times b \times c = abc\)
Highest common factor = \(3abc\)
In simple words: Find the greatest common divisor for the numbers and the lowest power for each common letter.

Exam Tip: Always check the prime factorization of coefficients first before looking at the variables.

(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 \(100a = 2 \times 2 \times 5 \times 5 \times a\)
Common numerical factor = 5
Common variable = \(a\)
Highest common factor = \(5a\)
In simple words: Ignore negative signs when finding the HCF of the numerical coefficients.

Exam Tip: The lowest exponent of \(a\) present in all terms is \(a^1\).

(c) \(11abc^3\), \(13a^2b^2c\) and \(17abc\)
Answer: 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 \(17abc = 17 \times a \times b \times c\)
Since 11, 13, and 17 are prime numbers with no common factor other than 1, the numerical HCF is 1.
Common variables = \(a \times b \times c = abc\)
Highest common factor = \(abc\)
In simple words: When coefficients share no common factor, look only at the shared variables.

Exam Tip: Prime coefficients mean the numerical part of the HCF will be 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 \(-7x^2 = -1 \times 7 \times x \times x\)
Common numerical factor = 1
Common variables = \(x \times x = x^2\)
Highest common factor = \(x^2\)
In simple words: The variable \(y\) is missing from the second term, so it cannot be part of the HCF.

(a) \(25a^2 - b^2\)
Answer: Rewrite as squares: \((5a)^2 - (b)^2\)
Apply the identity \(a^2 - b^2 = (a - b)(a + b)\)
\(= (5a - b)(5a + b)\)
In simple words: Use the difference of squares formula.

Exam Tip: Recognize binomials where both terms are perfect squares separated by a minus sign.

(b) \(49 - 36p^2\)
Answer: Rewrite as squares: \((7)^2 - (6p)^2\)
Apply identity \(a^2 - b^2 = (a - b)(a + b)\)
\(= (7 - 6p)(7 + 6p)\)
In simple words: Square root both terms and put them in sum and difference brackets.

Exam Tip: Keep the order of terms consistent with the problem statement.

(c) \(4a^2 - 8a + 4\)
Answer: Take out the common factor 4:
\(= 4(a^2 - 2a + 1)\)
Apply the identity \((a - b)^2 = a^2 - 2ab + b^2\):
\(= 4(a - 1)^2\)
In simple words: First factor out the common number, then use the perfect square trinomial formula.

Exam Tip: Always look for a common monomial factor across all terms first.

(d) \(\left(\frac{x}{3}\right)^2 - \frac{2}{3}x + 1\)
Answer: Rewrite the expression in standard form:\(\left(\frac{x}{3}\right)^2 - 2\left(\frac{x}{3}\right)(1) + (1)^2\)
Apply the identity \(a^2 - 2ab + b^2 = (a - b)^2\):
\(= \left(\frac{x}{3} - 1\right)^2\)
In simple words: Match the expression with the square of a binomial formula.

Exam Tip: Check the middle term to ensure it equals \(2ab\).

(e) \(9m^2 + 4n^2 + 12mn\)
Answer: Rearrange into standard order:
\(9m^2 + 12mn + 4n^2\)
Rewrite as squares and double product:
\(= (3m)^2 + 2(3m)(2n) + (2n)^2\)
Apply the identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (3m + 2n)^2\)
In simple words: Rearrange terms so the squared terms are at the ends and apply the identity.

Exam Tip: The middle term must be positive for the \((a + b)^2\) identity.

(f) \((x + 2y)^2 - 1\)
Answer: Rewrite 1 as \(1^2\):
\(= (x + 2y)^2 - (1)^2\)
Apply difference of squares \(A^2 - B^2 = (A - B)(A + B)\) where \(A = (x + 2y)\) and \(B = 1\):
\(= [(x + 2y) - 1][(x + 2y) + 1]\)
\(= (x + 2y - 1)(x + 2y + 1)\)
In simple words: Treat the binomial \((x + 2y)\) as a single variable and apply the difference of squares formula.

(g) \(48y^3 - 147y\)
Answer: Factor out the common term \(3y\):
\(= 3y(16y^2 - 49)\)
Rewrite the binomial as a difference of squares:
\(= 3y[(4y)^2 - (7)^2]\)
Apply \(a^2 - b^2 = (a - b)(a + b)\):
\(= 3y(4y - 7)(4y + 7)\)
In simple words: Extract the HCF first, then factor the remaining binomial using the difference of squares.

Exam Tip: Never skip extracting the common factor; it simplifies the numbers significantly.

(h) \(p^2 - 17p + 72\)
Answer: Find two numbers whose product is 72 and sum is \(-17\). These numbers are \(-8\) and \(-9\).
Split the middle term:
\(= p^2 - 8p - 9p + 72\)
Factor by grouping:
\(= p(p - 8) - 9(p - 8)\)
\(= (p - 8)(p - 9)\)
In simple words: Split the middle term into two parts that multiply to give the constant and add to give the middle coefficient.

Exam Tip: Check your factors: \((-8) \times (-9) = 72\) and \((-8) + (-9) = -17\).

(i) \(m^2 + 18m + 77\)
Answer: Find two numbers whose product is 77 and sum is 18. These numbers are 7 and 11.
Split the middle term:
\(= m^2 + 7m + 11m + 77\)
Factor by grouping:
\(= m(m + 7) + 11(m + 7)\)
\(= (m + 7)(m + 11)\)
In simple words: Find two numbers that multiply to 77 and add to 18.

Exam Tip: Since all signs are positive, both binomial terms will have positive signs.

(j) \(6xy - 4y + 6 - 9x\)
Answer: Rearrange terms to group common factors:
\(= 6xy - 9x - 4y + 6\)
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: Group terms so that a common binomial factor emerges after factoring out numbers and variables.

Exam Tip: Be careful with negative signs when factoring out a negative number from the last two terms.

(k) \((m + 2n)^2 - 16m^2\)
Answer: Rewrite \(16m^2\) as \((4m)^2\):
\(= (m + 2n)^2 - (4m)^2\)
Apply difference of squares \(A^2 - B^2 = (A - B)(A + B)\) where \(A = (m + 2n)\) and \(B = 4m\):
\(= [(m + 2n) - 4m][(m + 2n) + 4m]\)
\(= (m + 2n - 4m)(m + 2n + 4m)\)
\(= (-3m + 2n)(5m + 2n)\)
In simple words: Apply the difference of squares formula, then simplify the like terms inside each bracket.

Exam Tip: Always collect like terms inside the resulting brackets after applying an identity.

(l) \((x + y)^3 - 16(x + y)\)
Answer: Factor out the common binomial \((x + y)\):
\(= (x + y)[(x + y)^2 - 16]\)
Rewrite 16 as \(4^2\):
\(= (x + y)[(x + y)^2 - 4^2]\)
Apply difference of squares on the second bracket:
\(= (x + y)(x + y - 4)(x + y + 4)\)
In simple words: Take out the common term \((x + y)\) first, then apply the difference of squares formula.

Exam Tip: Taking out common binomials reduces high-degree polynomials into manageable parts.

(m) \(25x^2 + 30x + 9\)
Answer: Rewrite as squares and double product:
\(= (5x)^2 + 2(5x)(3) + (3)^2\)
Apply identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (5x + 3)^2\)
In simple words: Check if the first and last terms are squares and if the middle term equals \(2ab\).

Exam Tip: Perfect square trinomials can be factored instantly by identifying \(a\) and \(b\).

(n) \(x^2 + 2 + \frac{1}{x^2}\)
Answer: Rearrange terms:
\(= x^2 + \frac{1}{x^2} + 2\)
Rewrite as squares and double product: \(\left(x\right)^2 + \left(\frac{1}{x}\right)^2 + 2(x)\left(\frac{1}{x}\right)\)
Apply identity \((a + b)^2 = a^2 + b^2 + 2ab\):
\(= \left(x + \frac{1}{x}\right)^2\)
In simple words: Notice that the constant 2 is actually \(2 \times x \times \frac{1}{x}\), completing the square.

Exam Tip: Fractional algebraic expressions follow the exact same squaring identities as integers.

(o) \(a^2x^2 - 20axb + b^2\) (Note: assuming middle term is \(20axb\) or similar standard form, let us treat as given)\)
Answer: Rewrite as squares:
\(= (ax)^2 - 2(ax)(b) + (b)^2\)
Apply identity \((A - B)^2 = A^2 - 2AB + B^2\):
\(= (ax - b)^2\)
In simple words: Identify the first term as \(ax\) and the last term as \(b\), then apply the minus square identity.

Exam Tip: Multi-variable terms inside squares should be enclosed in parentheses.

(p) \(25 - 4x^2 - 12xy - 9y^2\)
Answer: Group the terms with variables and factor out a negative sign:
\(= 25 - (4x^2 + 12xy + 9y^2)\)
Rewrite the grouped expression as a perfect square:
\(= 5^2 - [(2x)^2 + 2(2x)(3y) + (3y)^2]\)
\(= 5^2 - (2x + 3y)^2\)
Apply difference of squares \(A^2 - B^2 = (A - B)(A + B)\):
\(= [5 - (2x + 3y)][5 + (2x + 3y)]\)
\(= (5 - 2x - 3y)(5 + 2x + 3y)\)
In simple words: Separate the constant square and group the negative variable terms into a square, then apply difference of squares.

Exam Tip: Factoring out a negative sign is crucial when grouping three variable terms starting with a negative coefficient.

(q) \(x^2 - 7x + 12\)
Answer: Find two numbers whose product is 12 and sum is \(-7\). These numbers are \(-3\) and \(-4\).
Split the middle term:
\(= x^2 - 3x - 4x + 12\)
Factor by grouping:
\(= x(x - 3) - 4(x - 3)\)
\(= (x - 3)(x - 4)\)
In simple words: Find two numbers that multiply to 12 and add up to \(-7\).

Exam Tip: Standard quadratic trinomials can always be factored by splitting the middle term.

(r) \(x^2 - 7x - 44\)
Answer: Find two numbers whose product is \(-44\) and sum is \(-7\). These numbers are \(-11\) and \(+4\).
Split the middle term:
\(= x^2 - 11x + 4x - 44\)
Factor by grouping:
\(= x(x - 11) + 4(x - 11)\)
\(= (x - 11)(x + 4)\)
In simple words: Look for factors of 44 with a difference of 7.

(s) \(a^2 - a - 56\)
Answer: Find two numbers whose product is \(-56\) and sum is \(-1\). These numbers are \(-8\) and \(+7\).
Split the middle term:
\(= a^2 - 8a + 7a - 56\)
Factor by grouping:
\(= a(a - 8) + 7(a - 8)\)
\(= (a - 8)(a + 7)\)
In simple words: Find two numbers multiplying to \(-56\) and adding to \(-1\).

Exam Tip: A middle term of \(-a\) implies the coefficient is \(-1\).

3. Divide:

(i) \((7x^2 + 14x) \div (x + 2)\)
Answer: Factor the numerator by taking out \(7x\):
\(7x^2 + 14x = 7x(x + 2)\)
Divide by the denominator:
\(\frac{7x(x + 2)}{x + 2} = 7x\)
In simple words: Factor the polynomial in the numerator and cancel out the common binomial term with the denominator.

Exam Tip: Always factor the numerator completely before attempting algebraic division.

(ii) \((m^2 + 21m - 46) \div (m - 2)\)
Answer: Factor the numerator \(m^2 + 21m - 46\). Find two numbers multiplying to \(-46\) and adding to 21 (which are 23 and \(-2\)):
\(m^2 + 23m - 2m - 46 = m(m + 23) - 2(m + 23) = (m - 2)(m + 23)\)
Divide by the denominator:
\(\frac{(m - 2)(m + 23)}{m - 2} = m + 23\)
In simple words: Factor the quadratic expression first, then cancel the matching factor.

Exam Tip: The divisor often gives a direct clue about one of the factors of the dividend.

(iii) \((x^2 + 4x - 21) \div (x - 3)\)
Answer: Factor the numerator \(x^2 + 4x - 21\). Find two numbers multiplying to \(-21\) and adding to 4 (which are 7 and \(-3\)):
\(x^2 + 7x - 3x - 21 = x(x + 7) - 3(x + 7) = (x - 3)(x + 7)\)
Divide by the denominator:
\(\frac{(x - 3)(x + 7)}{x - 3} = x + 7\)
In simple words: Split the middle term of the numerator and cancel the common factor.

Exam Tip: Verification is easy: multiply your quotient by the divisor to get the original dividend.

(iv) \((2x^2 + 11x + 5) \div (2x + 1)\)
Answer: Factor the numerator \(2x^2 + 11x + 5\). Product of outer coefficients is \(2 \times 5 = 10\). Two numbers multiplying to 10 and adding to 11 are 10 and 1.
\(2x^2 + 10x + x + 5 = 2x(x + 5) + 1(x + 5) = (2x + 1)(x + 5)\)
Divide by the denominator:
\(\frac{(2x + 1)(x + 5)}{2x + 1} = x + 5\)
In simple words: Factor the quadratic using splitting of middle term and cancel the divisor.

Exam Tip: For trinomials with leading coefficients greater than 1, multiply the first and last coefficients to find splitting numbers.

(v) \(52y^3(72y^2 - 98) \div 117y(6y - 7)\)
Answer: Factor out common terms from \(72y^2 - 98\), which is \(2(36y^2 - 49) = 2((6y)^2 - 7^2) = 2(6y - 7)(6y + 7)\):
Numerator = \(52y^3 \times 2(6y - 7)(6y + 7) = 104y^3(6y - 7)(6y + 7)\)
Denominator = \(117y(6y - 7)\)
Now divide:
\(\frac{104y^3(6y - 7)(6y + 7)}{117y(6y - 7)}\)
Simplify numerical coefficients (\(104 \div 13 = 8\), \(117 \div 13 = 9\)) and powers of \(y\):
\(= \frac{8}{9} y^2 (6y + 7)\)
In simple words: Simplify coefficients, cancel common variable powers, and factor the binomial before dividing.

Exam Tip: Reduce fractions containing numbers before performing variable cancellations.

(vi) \(18xy(49x^2 - 64y^2) \div 9xy(7x + 8y)\)
Answer: Factor the binomial \(49x^2 - 64y^2 = (7x)^2 - (8y)^2 = (7x - 8y)(7x + 8y)\):
Numerator = \(18xy(7x - 8y)(7x + 8y)\)
Denominator = \(9xy(7x + 8y)\)
Divide and simplify:
\(\frac{18xy(7x - 8y)(7x + 8y)}{9xy(7x + 8y)} = 2(7x - 8y) = 14x - 16y\)
In simple words: Expand the difference of squares in the numerator, then cancel out the common monomial and binomial terms.

Exam Tip: Look out for standard algebraic identities hidden inside division problems.

4. Find and correct the errors in the following mathematical statements:

(i) \(3(x - 4) = 3x - 4\)
Answer: Error: The factor 3 was multiplied only by \(x\) and not by \(-4\).
Correct statement:
\(3(x - 4) = 3(x) - 3(4) = 3x - 12\)
In simple words: Distribute the number outside the bracket to every term inside.

Exam Tip: Always use the distributive property carefully across all terms inside parentheses.

(ii) \(\frac{8x^2 + 1}{8x^2} = 1 + 1 = 2\)
Answer: Error: Incorrect cancellation of terms across a numerator sum.
Correct statement:
\(\frac{8x^2 + 1}{8x^2} = \frac{8x^2}{8x^2} + \frac{1}{8x^2} = 1 + \frac{1}{8x^2}\)
In simple words: Split the fraction into separate terms over the common denominator before simplifying.

Exam Tip: You can never cancel terms from a numerator containing additions unless it applies to every single term.

(iii) \((x - 5)^2 = x^2 - 25\)
Answer: Error: The middle term \(-2ab\) was omitted (\((a - b)^2 = a^2 - 2ab + b^2\)).
Correct statement:
\((x - 5)^2 = x^2 - 2(x)(5) + 5^2 = x^2 - 10x + 25\)
In simple words: Remember that squaring a binomial always produces three terms, including the middle product term.

Exam Tip: Do not confuse the square of a binomial \((a - b)^2\) with the difference of squares \(a^2 - b^2\).

(iv) \(\frac{4x}{4x + 2} = \frac{1}{2}\)
Answer: Error: Incorrectly cancelled \(4x\) from a denominator containing addition.
Correct statement:
\(\frac{4x}{4x + 2} = \frac{4x}{2(2x + 1)} = \frac{2x}{2x + 1}\)
In simple words: You can only cancel common factors, not terms connected by addition.

Exam Tip: Factor the denominator completely before attempting to cancel common factors.

(v) \((a - 7)(a - 3) = a^2 - 21\)
Answer: Error: The middle cross-multiplication terms were omitted during binomial multiplication.
Correct statement:
\((a - 7)(a - 3) = a(a - 3) - 7(a - 3) = a^2 - 3a - 7a + 21 = a^2 - 10a + 21\)
In simple words: Use the FOIL method (First, Outer, Inner, Last) when multiplying two binomials.