Factorisation

Q1: Multiple Choice Type

i. \(x^2 – 9x – 10\) is equal to:

Step 1: Product = \(1\times(-10)=-10\), Sum = \(-9\).
Step 2: Two numbers: \(-10\) and \(+1\).
Step 3: Factorise: \[ x^2 – 9x – 10 = (x-10)(x+1) \]Answer: a. \((x-10)(x+1)\)

ii. \(x^2 – 23x + 42\) is equal to:

Step 1: Product = \(1\times42=42\), Sum = \(-23\).
Step 2: Two numbers: \(-21\) and \(-2\).
Step 3: Factorise: \[ x^2 – 23x + 42 = (x-21)(x-2) \]Answer: b. \((x-21)(x-2)\)

iii. \((4x^2 – 4x + 1)÷(2x – 1)\) is equal to:

Step 1: Observe numerator: \[ 4x^2 – 4x + 1 = (2x – 1)^2 \]Step 2: Divide: \[ \frac{(2x – 1)^2}{2x – 1} = 2x – 1 \]Answer: b. \(2x – 1\)

iv. \((x+y)^2 – 3(x+y) – 4\) is equal to:

Step 1: Put \(t = x+y\). Then expression becomes \(t^2 – 3t – 4\).
Step 2: Factorise \(t^2 – 3t – 4 = (t-4)(t+1)\).
Step 3: Substitute back \(t = x+y\): \[ (x+y-4)(x+y+1) \]Answer: c. \((x+y-4)(x+y+1)\)

v. \(60 + 11x – x^2\) is equal to:

Step 1: Rearrange: \[ 60 + 11x – x^2 = -(x^2 – 11x – 60) \]Step 2: Factor inside: find numbers with product \(-60\) and sum \(-11\): \(-15\) and \(+4\). \[ x^2 – 11x – 60 = (x-15)(x+4) \]Step 3: Apply negative sign: \[ -(x-15)(x+4) = (15-x)(x+4) = (x+4)(15-x) \]Answer: a. \((x+4)(15-x)\)

Q2: \(a^2 + 5a + 6\)

Step 1: Product = \(1 \times 6 = 6\), Sum = \(5\).
Step 2: Numbers = \(2\) and \(3\).
Step 3: Split the middle term: \[ a^2 + 5a + 6 = a^2 + 2a + 3a + 6 \]Step 4: Group terms: \[ = (a^2 + 2a) + (3a + 6) \\ = a(a+2) + 3(a+2) \]Step 5: Factor common binomial: \[ = (a+2)(a+3) \]Answer: \((a+2)(a+3)\)

Q3: \(a^2 – 5a + 6\)

Step 1: Product = \(1 \times 6 = 6\), Sum = \(-5\).
Step 2: Numbers = \(-2\) and \(-3\).
Step 3: Split the middle term: \[ a^2 – 5a + 6 = a^2 – 2a – 3a + 6 \]Step 4: Group terms: \[ = (a^2 – 2a) – (3a – 6) \\ = a(a-2) – 3(a-2) \]Step 5: Factor common binomial: \[ = (a-2)(a-3) \]Answer: \((a-2)(a-3)\)

Q4: \(a^2 + 5a – 6\)

Step 1: Product = \(1 \times -6 = -6\), Sum = \(+5\).
Step 2: Numbers = \(+6\) and \(-1\).
Step 3: Split the middle term: \[ a^2 + 5a – 6 = a^2 + 6a – a – 6 \]Step 4: Group terms: \[ = (a^2 + 6a) – (a + 6) \\ = a(a+6) – 1(a+6) \]Step 5: Factor common binomial: \[ = (a+6)(a-1) \]Answer: \((a+6)(a-1)\)

Q5: \(x^2 + 5xy + 4y^2\)

Step 1: Product = \(1 \times 4 = 4\), Sum = \(+5\).
Step 2: Numbers = \(+4\) and \(+1\).
Step 3: Split the middle term: \[ x^2 + 5xy + 4y^2 = x^2 + 4xy + xy + 4y^2 \]Step 4: Group terms: \[ = (x^2 + 4xy) + (xy + 4y^2) \\ = x(x+4y) + y(x+4y) \]Step 5: Factor common binomial: \[ = (x+y)(x+4y) \]Answer: \((x+y)(x+4y)\)

Q6: \(a^2 – 3a – 40\)

Step 1: Product = \(1 \times -40 = -40\), Sum = \(-3\).
Step 2: Numbers = \(-8\) and \(+5\).
Step 3: Split the middle term: \[ a^2 – 3a – 40 = a^2 – 8a + 5a – 40 \]Step 4: Group terms: \[ = (a^2 – 8a) + (5a – 40) \\ = a(a – 8) + 5(a – 8) \]Step 5: Factor common binomial: \[ = (a+5)(a-8) \]Answer: \((a+5)(a-8)\)

Q7: \(x^2 – x – 72\)

Step 1: Product = \(1 \times -72 = -72\), Sum = \(-1\).
Step 2: Numbers = \(-9\) and \(+8\).
Step 3: Split the middle term: \[ x^2 – x – 72 = x^2 – 9x + 8x – 72 \]Step 4: Group terms: \[ = (x^2 – 9x) + (8x – 72) \\ = x(x – 9) + 8(x – 9) \]Step 5: Factor common binomial: \[ = (x+8)(x-9) \]Answer: \((x+8)(x-9)\)

Q8: \(3a^2 – 5a + 2\)

Step 1: Product = \(3 \times 2 = 6\), Sum = \(-5\).
Step 2: Numbers = \(-3\) and \(-2\).
Step 3: Split the middle term: \[ 3a^2 – 5a + 2 = 3a^2 – 3a – 2a + 2 \]Step 4: Group terms: \[ = (3a^2 – 3a) – (2a – 2) \\ = 3a(a – 1) – 2(a – 1) \]Step 5: Factor common binomial: \[ = (3a – 2)(a – 1) \]Answer: \((3a – 2)(a – 1)\)

Q9: \(2a^2 – 17ab + 26b^2\)

Step 1: Multiply \(2 \times 26 = 52\). We need two numbers whose product = \(52\) and sum = \(-17\).
Numbers are \(-13\) and \(-4\).
Step 2: Split the middle term: \[ 2a^2 – 17ab + 26b^2 = 2a^2 – 13ab – 4ab + 26b^2 \]Step 3: Group terms: \[ = (2a^2 – 13ab) – (4ab – 26b^2) \]Step 4: Factorise each group: \[ = a(2a – 13b) – 2b(2a – 13b) \]Step 5: Take out common binomial: \[ = (a – 2b)(2a – 13b) \]Answer: \((a – 2b)(2a – 13b)\)

Q10: \(2x^2 + xy – 6y^2\)

Step 1: Multiply \(2 \times -6 = -12\). We need two numbers whose product = \(-12\) and sum = \(1\) (coefficient of \(xy\)).
Numbers are \(4\) and \(-3\).
Step 2: Split the middle term: \[ 2x^2 + xy – 6y^2 = 2x^2 + 4xy – 3xy – 6y^2 \]Step 3: Group terms: \[ = (2x^2 + 4xy) – (3xy + 6y^2) \]Step 4: Factorise each group: \[ = 2x(x+2y) – 3y(x+2y) \]Step 5: Take out common binomial: \[ = (2x – 3y)(x + 2y) \]Answer: \((2x – 3y)(x + 2y)\)

Q11: \(4c^2 + 3c – 10\)

Step 1: Multiply coefficient of \(c^2\) and constant: \[ 4 \times (-10) = -40 \]Step 2: Find two numbers whose product = \(-40\) and sum = \(+3\).
Numbers: \(8\) and \(-5\).
Step 3: Split the middle term using those numbers: \[ 4c^2 + 3c – 10 = 4c^2 + 8c – 5c – 10 \]Step 4: Group terms: \[ (4c^2 + 8c) – (5c + 10) \]Step 5: Factor each group: \[ 4c(c+2) – 5(c+2) \]Step 6: Factor out the common binomial \((c+2)\): \[ (4c – 5)(c + 2) \]Answer: \((4c – 5)(c + 2)\)

Q12: \(14x^2 + x – 3\)

Step 1: Multiply coefficient of \(x^2\) and constant: \[ 14 \times (-3) = -42 \]Step 2: Find two numbers whose product = \(-42\) and sum = \(+1\).
Numbers: \(7\) and \(-6\).
Step 3: Split the middle term using those numbers: \[ 14x^2 + 7x – 6x – 3 \]Step 4: Group terms: \[ (14x^2 + 7x) – (6x + 3) \]Step 5: Factor each group: \[ 7x(2x+1) – 3(2x+1) \]Step 6: Factor out the common binomial \((2x+1)\): \[ (7x – 3)(2x + 1) \]Answer: \((7x – 3)(2x + 1)\)

Q13: \(6 + 7b – 3b^2\)

Step 1: Write the trinomial in standard quadratic form. \[-3b^2 + 7b + 6\]Step 2: Take out \(-1\) common from the expression. \[-(3b^2 – 7b – 6)\]Step 3: Multiply the coefficient of \(b^2\) with the constant term: \(3 \times -6 = -18\).
Now, find two numbers whose product = \(-18\) and sum = \(-7\).
These numbers are \(-9\) and \(2\).
Step 4: Split the middle term using \(-9\) and \(2\). \[-(3b^2 – 9b + 2b – 6)\]Step 5: Group the terms. \[-\big[(3b^2 – 9b) + (2b – 6)\big]\]Step 6: Take out common factors from each group. \[-\big[3b(b – 3) + 2(b – 3)\big]\]Step 7: Take out \((b – 3)\) as common. \[-(3b + 2)(b – 3)\]Step 8: Simplify the negative sign. \[(3b + 2)(3 – b)\]Answer: (3b + 2)(3 – b)

Q14: \(5 + 7x – 6x^2\)

Step 1: Write the trinomial in standard quadratic form. \[-6x^2 + 7x + 5\]Step 2: Take out \(-1\) common from the expression. \[-(6x^2 – 7x – 5)\]Step 3: Multiply the coefficient of \(x^2\) with the constant term: \(6 \times -5 = -30\).
Now, find two numbers whose product = \(-30\) and sum = \(-7\).
These numbers are \(-10\) and \(3\).
Step 4: Split the middle term using \(-10\) and \(3\). \[-(6x^2 – 10x + 3x – 5)\]Step 5: Group the terms. \[-\big[(6x^2 – 10x) + (3x – 5)\big]\]Step 6: Take out common factors from each group. \[-\big[2x(3x – 5) + 1(3x – 5)\big]\]Step 7: Take out \((3x – 5)\) as common. \[-(2x + 1)(3x – 5)\]Step 8: Simplify the negative sign. \[(2x + 1)(5 – 3x)\]Answer: (2x + 1)(5 – 3x)

Q15: \(4 + y – 14y^2\)

Step 1: Write the trinomial in standard quadratic form. \[-14y^2 + y + 4\]Step 2: Take out \(-1\) common to make the coefficient of \(y^2\) positive. \[-(14y^2 – y – 4)\]Step 3: Multiply the coefficient of \(y^2\) with the constant term: \(14 \times -4 = -56\).
Now, find two numbers whose product = \(-56\) and sum = \(-1\).
These numbers are \(-8\) and \(7\).
Step 4: Split the middle term using \(-8\) and \(7\). \[-(14y^2 – 8y + 7y – 4)\]Step 5: Group the terms. \[-\big[(14y^2 – 8y) + (7y – 4)\big]\]Step 6: Take out common factors from each group. \[-\big[2y(7y – 4) + 1(7y – 4)\big]\]Step 7: Take out \((7y – 4)\) as common. \[-(2y + 1)(7y – 4)\]Step 8: Simplify the negative sign. \[(2y + 1)(4 – 7y)\]Answer: (2y + 1)(4 – 7y)

Q16: \(5 + 3a – 14a^2\)

Step 1: Write the trinomial in standard quadratic form. \[-14a^2 + 3a + 5\]Step 2: Take out \(-1\) common to make the coefficient of \(a^2\) positive. \[-(14a^2 – 3a – 5)\]Step 3: Multiply the coefficient of \(a^2\) with the constant term: \(14 \times -5 = -70\).
Now, find two numbers whose product = \(-70\) and sum = \(-3\).
These numbers are \(-10\) and \(7\).
Step 4: Split the middle term using \(-10\) and \(7\). \[-(14a^2 – 10a + 7a – 5)\]Step 5: Group the terms. \[-\big[(14a^2 – 10a) + (7a – 5)\big]\]Step 6: Take out common factors from each group. \[-\big[2a(7a – 5) + 1(7a – 5)\big]\]Step 7: Take out \((7a – 5)\) as common. \[-(2a + 1)(7a – 5)\]Step 8: Simplify the negative sign. \[(2a + 1)(5 – 7a)\]Answer: (2a + 1)(5 – 7a)

Q17: \((2a+b)^2 + 5(2a+b) + 6\)

Step 1: Write the trinomial clearly:
\((2a+b)^2 + 5(2a+b) + 6\)
Step 2: Factorise the quadratic in \((2a+b)\) by splitting the middle term:
Find two numbers whose product = 6 (coefficient of constant term) and sum = 5 (coefficient of \((2a+b)\)) → 2 and 3
Step 3: Split the middle term using 2 and 3:
\((2a+b)^2 + 2(2a+b) + 3(2a+b) + 6\)
Step 4: Group terms and factorise each group:
\(((2a+b)^2 + 2(2a+b)) + (3(2a+b) + 6) = (2a+b)((2a+b)+2) + 3((2a+b)+2)\)
Step 5: Factor out the common factor \(((2a+b)+2)\):
\(((2a+b)+2)((2a+b)+3) = (2a+b+2)(2a+b+3)\)
Answer: (2a+b+2)(2a+b+3)

Q18: \(1 – (2x+3y) – 6(2x+3y)^2\)

Step 1: Rewrite the trinomial in standard quadratic form:
\(-6(2x+3y)^2 – (2x+3y) + 1\)
Step 2: Factor out \(-1\) to make the leading coefficient positive:
\(-1 \cdot \left(6(2x+3y)^2 + (2x+3y) – 1\right)\)
Step 3: Factorise the quadratic in \((2x+3y)\) by splitting the middle term:
\(6(2x+3y)^2 + 3(2x+3y) – 2(2x+3y) – 1 = (6(2x+3y)^2 + 3(2x+3y)) – (2(2x+3y) + 1) = 3(2x+3y)(2(2x+3y)+1) – 1(2(2x+3y)+1) = (3(2x+3y)-1)(2(2x+3y)+1)\)
Step 4: Include the \(-1\) factor:
\(-1 \cdot (3(2x+3y)-1)(2(2x+3y)+1) = (-3(2x+3y)+1)(2(2x+3y)+1) = (1-6x-9y)(4x+6y+1)\)
Answer: \((1-6x-9y)(4x+6y+1)\)

Q19: \((x-2y)^2 – 12(x-2y) + 32\)

Step 1: Write the trinomial clearly:
\((x-2y)^2 – 12(x-2y) + 32\)
Step 2: Factor the quadratic by splitting the middle term:
Find two numbers whose product = 32 (constant term) and sum = -12 (coefficient of \((x-2y)\)) → -8 and -4
Step 3: Split the middle term using -8 and -4:
\((x-2y)^2 – 8(x-2y) – 4(x-2y) + 32\)
Step 4: Group terms and factorise each group:
\(((x-2y)^2 – 8(x-2y)) – (4(x-2y) – 32) = (x-2y)((x-2y)-8) – 4((x-2y)-8)\)
Step 5: Factor out the common factor \(((x-2y)-8)\):
\(((x-2y)-8)((x-2y)-4) = (x-2y-8)(x-2y-4)\)
Answer: (x-2y-8)(x-2y-4)

Q20: \(8 + 6(a+b) – 5(a+b)^2\)

Step 1: Rewrite the trinomial in standard quadratic form:
\(-5(a+b)^2 + 6(a+b) + 8\)
Step 2: Factor out \(-1\) to make the leading coefficient positive:
\(-1 \cdot \left(5(a+b)^2 – 6(a+b) – 8\right)\)
Step 3: Factor the quadratic in \((a+b)\) by splitting the middle term:
Find two numbers whose product = 5×(-8) = -40 and sum = -6 → 4 and -10
\(\Rightarrow 5(a+b)^2 + 4(a+b) – 10(a+b) – 8\)
Step 4: Group terms and factorise:
\([5(a+b)^2 + 4(a+b)] – [10(a+b) + 8] = (a+b)(5(a+b)+4) – 2(5(a+b)+4)\)
Step 5: Factor out the common factor \((5(a+b)+4)\):
\((5(a+b)+4)((a+b)-2) = (4 + 5a + 5b)(2 – a – b)\)
Answer: (4+5a+5b)(2-a-b)

Q21: \(2(x+2y)^2 – 5(x+2y) + 2\)

Step 1: Treat \((x+2y)\) as a single variable for factoring:
\(2(x+2y)^2 – 5(x+2y) + 2\)
Step 2: Multiply the leading coefficient and constant term: 2 × 2 = 4.
We need two numbers whose product = 4 and sum = -5 → -4 and -1
Step 3: Split the middle term using -4 and -1:
\(2(x+2y)^2 – 4(x+2y) – (x+2y) + 2\)
Step 4: Group terms and factor each group:
\([2(x+2y)^2 – 4(x+2y)] – [(x+2y) – 2] = 2(x+2y)((x+2y)-2) -1((x+2y)-2)\)
Step 5: Factor out the common factor \(((x+2y)-2)\):
\(((x+2y)-2)(2(x+2y)-1) = (x+2y-2)(2x+4y-1)\)
Answer: (x+2y-2)(2x+4y-1)

Q22: In each case, find whether the trinomial is a perfect square or not:

i. \(x^2 + 14x + 49\)

Step 1: Check if first term and last term are perfect squares:
\(x^2 = (x)^2\), \(49 = 7^2\)
Step 2: Check if middle term = \(2 \times x \times 7 = 14x\)
Answer: Yes, perfect square: \((x+7)^2\)

ii. \(a^2 – 10a + 25\)

Step 1: \(a^2 = (a)^2\), \(25 = 5^2\)
Step 2: Check middle term: \(2 \cdot a \cdot 5 = 10a\), sign = negative → \(-10a\)
Answer: Yes, perfect square: \((a-5)^2\)

iii. \(4x^2 + 4x + 1\)

Step 1: \(4x^2 = (2x)^2\), \(1 = 1^2\)
Step 2: Check middle term: \(2 \cdot 2x \cdot 1 = 4x\)
Answer: Yes, perfect square: \((2x+1)^2\)

iv. \(9b^2 + 12b + 16\)

Step 1: \(9b^2 = (3b)^2\), \(16 = 4^2\)
Step 2: Check middle term: \(2 \cdot 3b \cdot 4 = 24b\), actual = 12b → does not match
Answer: No, not a perfect square

v. \(16x^2 – 16xy + y^2\)

Step 1: \(16x^2 = (4x)^2\), \(y^2 = (y)^2\)
Step 2: Check middle term: \(2 \cdot 4x \cdot y = 8xy\), actual = -16xy → does not match
Answer: No, not a perfect square

vi. \(x^2 – 4x + 16\)

Step 1: \(x^2 = (x)^2\), \(16 = 4^2\)
Step 2: Check middle term: \(2 \cdot x \cdot 4 = 8x\), actual = -4x → does not match
Answer: No, not a perfect square