Factorisation

Q1: Multiple Choice Type

i. \(-7x^2-14y\)

Step 1: Identify the common factor of the terms \(-7x^2\) and \(-14y\).
Both terms have a common factor of \(-7\).
Step 2: Factor out \(-7\). \[ -7x^2 – 14y = -7(x^2 + 2y) \]Answer: d. \(-7(x^2 + 2y)\)

ii. \(a(x+y)-b(x-y)^2\)

Step 1: Factor common if possible. Rewrite \(-(b(x-y)^2)\) as \(-b(x-y)^2\).
Step 2: Factor \((x-y)\) from both terms. \[ a(x+y)-b(x-y)^2 = (x-y)(a – bx + by) \]Answer: c. \((x-y)(a – bx + by)\)

iii. \(a^2 + bc + ab + ac\)

Step 1: Group terms: \[ (a^2 + ab) + (ac + bc) \]Step 2: Factor each group: \[ a(a+b) + c(a+b) \]Step 3: Factor out \((a+b)\): \[ (a+b)(a+c) \]Answer: a. \((a+b)(a+c)\)

iv. \(1 – 2x – 2x^2 + 4x^3\)

Step 1: Group terms: \[ 1 – 2x – 2x^2 + 4x^3 \] Rearrange: \[ = (1 – 2x) – 2x^2(1 – 2x) \]Step 2: Take common factor \((1 – 2x)\): \[ = (1 – 2x)(1 – 2x^2) \]Answer: c. \((1-2x)(1-2x^2)\)

v. \(a(x-y)-b(y-x)^2\)

Step 1: Simplify \((y-x)^2 = -(x-y)^2\).
Step 2: Rewrite expression: \[ a(x-y) + b(x-y)^2 \]Step 3: Common \((x-y)\): \[ (x-y)[a+b(x-y)] \\ (x-y)(a + bx – by) \]Answer: a. \((x-y)(a – bx + by)\)

Q2: \(17a^6b^8-34a^4b^6+51a^2b^4\)

Step 1: Identify the common factors of all terms.
– Coefficients: \(17, 34, 51\) → common factor = 17
– Powers of \(a\): \(a^6, a^4, a^2\) → common factor = \(a^2\)
– Powers of \(b\): \(b^8, b^6, b^4\) → common factor = \(b^4\)
Step 2: Factor out \(17a^2b^4\) from all terms. \[ 17a^6b^8 – 34a^4b^6 + 51a^2b^4 = 17a^2b^4 \big(a^4b^4 – 2a^2b^2 + 3\big) \]Step 3: Check if the remaining polynomial can be further factorised. \[ a^4b^4 – 2a^2b^2 + 3 \] cannot be factorised further using integers.
Answer: \(17a^2b^4(a^4b^4 – 2a^2b^2 + 3)\)

Q3: \(3x^5y-27x^4y^2+12x^3y^3\)

Step 1: Identify the common factors of all terms.
– Coefficients: \(3, 27, 12\) → common factor = 3
– Powers of \(x\): \(x^5, x^4, x^3\) → common factor = \(x^3\)
– Powers of \(y\): \(y, y^2, y^3\) → common factor = \(y\)
Step 2: Factor out \(3x^3y\) from all terms. \[ 3x^5y – 27x^4y^2 + 12x^3y^3 = 3x^3y \big(x^2 – 9xy + 4y^2\big) \]Step 3: Check if the remaining quadratic \(x^2 – 9xy + 4y^2\) can be factorised further.
– The discriminant is \(\Delta = (-9y)^2 – 4 \cdot 1 \cdot 4y^2 = 81y^2 – 16y^2 = 65y^2\) → Not a perfect square.
– So, it cannot be factorised further using integers.
Answer: \(3x^3y(x^2 – 9xy + 4y^2)\)

Q4: \(x^2(a-b)-y^2(a-b)+z^2(a-b)\)

Step 1: Identify the common factor in all terms.
– Each term contains the factor \((a-b)\).
Step 2: Factor out \((a-b)\) from all terms. \[ x^2(a-b) – y^2(a-b) + z^2(a-b) = (a-b)(x^2 – y^2 + z^2) \]Step 3: Check if the remaining polynomial \(x^2 – y^2 + z^2\) can be factorised further.
– \(x^2 – y^2 = (x-y)(x+y)\), so we can write: \[ x^2 – y^2 + z^2 = (x-y)(x+y) + z^2 \] – No further factorisation with integers is possible.
Answer: \((a-b)(x^2 – y^2 + z^2)\)

Q5: \((x+y)(a+b)+(x-y)(a+b)\)

Step 1: Identify the common factor in both terms.
– Both terms contain the factor \((a+b)\).
Step 2: Factor out \((a+b)\). \[ (x+y)(a+b) + (x-y)(a+b) = (a+b)\big((x+y) + (x-y)\big) \]Step 3: Simplify the expression inside the parentheses. \[ (x+y) + (x-y) = x + y + x – y = 2x \]Answer: \(2x(a+b)\)

Q6: \(2b(2a+b)-3c(2a+b)\)

Step 1: Identify the common factor in both terms.
– Both terms contain the factor \((2a+b)\).
Step 2: Factor out \((2a+b)\). \[ 2b(2a+b) – 3c(2a+b) = (2a+b)(2b – 3c) \]Answer: \((2a+b)(2b – 3c)\)

Q7: \(12abc-6a^2b^2c^2+3a^3b^3c^3\)

Step 1: Identify the common factors of all terms.
– Coefficients: \(12, 6, 3\) → common factor = 3
– Powers of \(a\): \(a, a^2, a^3\) → common factor = \(a\)
– Powers of \(b\): \(b, b^2, b^3\) → common factor = \(b\)
– Powers of \(c\): \(c, c^2, c^3\) → common factor = \(c\)
Step 2: Factor out \(3abc\) from all terms. \[ 12abc – 6a^2b^2c^2 + 3a^3b^3c^3 = 3abc \big(4 – 2abc + a^2b^2c^2\big) \]Step 3: Check if the remaining polynomial \(4 – 2abc + a^2b^2c^2\) can be factorised further.
– It cannot be factorised further using integers.
Answer: \(3abc(4 – 2abc + a^2b^2c^2)\)

Q8: \(4x(3x-2y)-2y(3x-2y)\)

Step 1: Identify the common factor in both terms.
– Both terms contain the factor \((3x-2y)\).
Step 2: Factor out \((3x-2y)\). \[ 4x(3x-2y) – 2y(3x-2y) = (3x-2y)(4x – 2y) \]Step 3: Check if further factorisation is possible.
– Factor 2 from \((4x – 2y)\): \[ 4x – 2y = 2(2x – y) \\ \text{So, } (3x-2y)(4x – 2y) = 2(3x-2y)(2x – y) \]Answer: \(2(3x-2y)(2x – y)\)

Q9: \((a+2b)(3a+b)-(a+b)(a+2b)+(a+2b)^2\)

Step 1: Identify the common factor in all terms.
– All three terms contain the factor \((a+2b)\).
Step 2: Factor out \((a+2b)\) from all terms. \[ (a+2b)(3a+b) – (a+b)(a+2b) + (a+2b)^2 = (a+2b)\big((3a+b) – (a+b) + (a+2b)\big) \]Step 3: Simplify the expression inside the parentheses. \[ (3a+b) – (a+b) + (a+2b) = 3a + b – a – b + a + 2b = 3a + 2b \]Answer: \((a+2b)(3a+2b)\)

Q10: \(6xy(a^2+b^2)+8yz(a^2+b^2)-10xz(a^2+b^2)\)

Step 1: Identify the common factor in all terms.
– Each term contains the factor \((a^2+b^2)\).
Step 2: Factor out \((a^2+b^2)\). \[ 6xy(a^2+b^2) + 8yz(a^2+b^2) – 10xz(a^2+b^2) = (a^2+b^2)\big(6xy + 8yz – 10xz\big) \]Step 3: Look for common factors in the expression \(6xy + 8yz – 10xz\).
– Group terms: \((6xy – 10xz) + 8yz\)
– Factor each group: \(2(3xy – 5xz + 8yz)\)
Answer: \(2(a^2+b^2)(3xy + 4yz – 5xz)\)

Q11: \(xy – ay – ax + a^2 + bx – ab\)

Step 1: Group the terms for factorisation. \[ (xy – ax) + (-ay + a^2) + (bx – ab) \]Step 2: Factor each group. \[ x(y – a) + a(-y + a) + b(x – a) \]Step 3: Simplify each group. \[ x(y – a) – a(y – a) + b(x – a) = (x – a)(y – a) + b(x – a) \]Step 4: Factor out \((x – a)\) from the remaining expression. \[ (x – a)(y – a) + b(x – a) = (x – a)\big((y – a) + b\big) = (x – a)(y + b – a) \]Answer: \((x – a)(y + b – a)\)

Q12: \(3x^5 – 6x^4 – 2x^3 + 4x^2 + x – 2\)

Step 1: Group terms for factorisation. \[ (3x^5 – 6x^4) + (-2x^3 + 4x^2) + (x – 2) \]Step 2: Factor each group. \[ 3x^4(x – 2) – 2x^2(x – 2) + 1(x – 2) \]Step 3: Factor out the common binomial \((x – 2)\). \[ 3x^4(x – 2) – 2x^2(x – 2) + 1(x – 2) = (x – 2)(3x^4 – 2x^2 + 1) \]Answer: \((x – 2)(3x^4 – 2x^2 + 1)\)

Q13: \(-x^2y – x + 3xy + 3\)

Step 1: Group terms for factorisation. \[ (-x^2y + 3xy) + (-x + 3) \]Step 2: Factor out common factors from each group. \[ xy(-x + 3) + (-1)(x – 3) \]Step 3: Adjust signs to make common binomial factor identical. \[ xy(-x + 3) – 1(x – 3) = xy(3 – x) – 1(x – 3) \\ 3 – x = -(x – 3) \Rightarrow -1(3 – x) = + 1(3 – x) \\ \text{So, } xy(3 – x) + 1(3 – x) = (3 – x)(xy + 1) \]Answer: \((3 – x)(xy + 1)\)

Q14: \(6a^2 – 3a^2b – bc^2 + 2c^2\)

Step 1: Group terms for factorisation. \[ (6a^2 – 3a^2b) + (-bc^2 + 2c^2) \]Step 2: Factor out common factors from each group. \[ 3a^2(2 – b) + c^2(-b + 2) \]Step 3: Adjust signs to make the binomial identical. \[ -b + 2 = 2 – b \Rightarrow 3a^2(2 – b) + c^2(2 – b) \]Step 4: Factor out the common binomial \((2 – b)\). \[ (2 – b)(3a^2 + c^2) \]Answer: \((2 – b)(3a^2 + c^2)\)

Q15: \(3a^2b – 12a^2 – 9b + 36\)

Step 1: Group terms for factorisation. \[ (3a^2b – 12a^2) + (-9b + 36) \]Step 2: Factor out the common factors from each group. \[ 3a^2(b – 4) – 9(b – 4) \]Step 3: Factor out the common binomial \((b – 4)\). \[ 3a^2(b – 4) – 9(b – 4) = (b – 4)(3a^2 – 9) \]Step 4: Factor the remaining quadratic \(3a^2 – 9\). \[ 3a^2 – 9 = 3(a^2 – 3) \]Answer: \((b – 4) \cdot 3(a^2 – 3) = 3(b – 4)(a^2 – 3)\)

Q16: \(x^2 – (a-3)x – 3a\)

Step 1: Identify the quadratic in the form \(x^2 + px + q\). \[ x^2 – (a-3)x – 3a = x^2 + (-a+3)x – 3a \]Step 2: Find two numbers whose product is \(-3a\) (coefficient of constant term) and sum is \(-a+3\) (coefficient of \(x\)).
– Numbers: \(3\) and \(-a\), because \(3 \cdot (-a) = -3a\) and \(3 + (-a) = -a + 3\).
Step 3: Split the middle term using these numbers. \[ x^2 + 3x – ax – 3a \]Step 4: Group terms and factor each group. \[ (x^2 + 3x) + (-ax – 3a) = x(x + 3) – a(x + 3) \]Step 5: Factor out the common binomial \((x+3)\). \[ x(x+3) – a(x+3) = (x – a)(x + 3) \]Answer: \((x – a)(x + 3)\)

Q17: \(ab^2 – (a-c)b – c\)

Step 1: Rewrite the expression clearly. \[ ab^2 – (a-c)b – c = ab^2 – ab + cb – c \]Step 2: Group terms for factorisation. \[ (ab^2 – ab) + (cb – c) \]Step 3: Factor out common factors from each group. \[ a b (b – 1) + c (b – 1) \]Step 4: Factor out the common binomial \((b – 1)\). \[ a b (b – 1) + c (b – 1) = (b – 1)(ab + c) \]Answer: \((b – 1)(ab + c)\)

Q18: \((a^2-b^2)c + (b^2-c^2)a\)

Step 1: Rewrite the expression clearly. \[ (a^2 – b^2)c + (b^2 – c^2)a = a^2c – b^2c + ab^2 – ac^2 \]Step 2: Group terms for factorisation. \[ (a^2c + ab^2) – (b^2c + ac^2) \]Step 3: Factor out common factors from each group. \[ a(ac + b^2) – c(b^2 + ac) \]Step 4: Factor out the common binomial \((ac + b^2)\). \[ a(ac + b^2) – c(ac + b^2) = (a – c) (b^2 + ac) \]Answer: \((a – c) (b^2 + ac)\)

Q19: \(a^3 – a^2 – ab + a + b + 1\)

Step 1: Group terms strategically. \[ (a^3 – a^2) + (-ab + b) + (a + 1) \]Step 2: Factor out common factors from each group. \[ a^2(a – 1) – b(a – 1) + 1(a – 1) \]Step 3: Factor out the common binomial \((a – 1)\). \[ a^2(a – 1) – b(a – 1) + 1(a – 1) = (a – 1)(a^2 – b + 1) \]Answer: \((a – 1)(a^2 – b + 1)\)

Q20: \(ab(c^2 + d^2) – a^2cd – b^2cd\)

Step 1: Rewrite the expression strategically. \[ ab(c^2 + d^2) – a^2cd – b^2cd = abc^2 + abd^2 – a^2cd – b^2cd \]Step 2: Group terms for factorisation. \[ (abc^2 – a^2cd) + (abd^2 – b^2cd) \]Step 3: Factor common terms from each group. \[ ac(bc – ad) + bd(ab – bc) \]Step 4: Rearrange and factor to get the final product. \[ (ac – bd)(bc – ad) \]Answer: \((ac – bd)(bc – ad)\)

Q21: \(2ab^2 – aby + 2cby – cy^2\)

Step 1: Group terms for factorisation. \[ (2ab^2 – aby) + (2cby – cy^2) \]Step 2: Factor out common factors from each group. \[ ab(2b – y) + cy(2b – y) \]Step 3: Factor out the common binomial \((2b – y)\). \[ ab(2b – y) + cy(2b – y) = (2b – y)(ab + cy) \]Answer: \((2b – y)(ab + cy)\)

Q22: \(ax + 2bx + 3cx – 3a – 6b – 9c\)

Step 1: Group terms for factorisation. \[ (ax + 2bx + 3cx) – (3a + 6b + 9c) \]Step 2: Factor out common factors from each group. \[ x(a + 2b + 3c) – 3(a + 2b + 3c) \]Step 3: Factor out the common binomial \((a + 2b + 3c)\). \[ x(a + 2b + 3c) – 3(a + 2b + 3c) = (a + 2b + 3c)(x – 3) \]Answer: \((a + 2b + 3c)(x – 3)\)

Q23: \(2ab^2c – 2a + 3b^3c – 3b – 4b^2c^2 + 4c\)

Step 1: Group terms for factorisation. \[ (2ab^2c – 2a) + (3b^3c – 3b) + (-4b^2c^2 + 4c) \]Step 2: Factor out common factors from each group. \[ 2a(b^2c – 1) + 3b(b^2c – 1) – 4c(b^2c – 1) \]Step 3: Factor out the common binomial \((b^2c – 1)\). \[ 2a(b^2c – 1) + 3b(b^2c – 1) – 4c(b^2c – 1) = (b^2c – 1)(2a + 3b – 4c) \]Answer: \((b^2c – 1)(2a + 3b – 4c)\)