Factorisation

Q1: Multiple Choice Type Factorisation

i. \((a+b)^2-4ab\) is equal to:

Step 1: \[ (a+b)^2 – 4ab = a^2 + 2ab + b^2 – 4ab \\ = a^2 – 2ab + b^2 = (a-b)^2 \]Answer: d. (a-b)(a-b)

ii. \(a^4+4a^2-32\) is equal to:

Step 1: Factor out 1 (no common factor here) and write as quadratic in \(a^2\):
\(a^4+4a^2-32 = (a^2)^2 + 4(a^2) – 32\)
Step 2: Factor quadratic \(a^4+4a^2-32\) as \((a^2+8)(a^2-4)\)
Step 3: Factor \(a^2-4\) as \((a-2)(a+2)\)
Answer: d. \((a^2+8)(a-2)(a+2)\)

iii. \(36-60y+25y^2\) is equal to:

\[ 36-60y+25y^2 = (6-5y)^2 \]None of the given options exactly match \((6-5y)(6-5y)\).

Answer:d. None of these

iv. \((x-2y)^2 – 3x + 6y\) is equal to:

Step 1: Group terms: \((x-2y)^2 – 3(x-2y)\)
Step 2: Factor out common \((x-2y)\):
\((x-2y)((x-2y)-3) = (x-2y)(x-2y-3)\)
Answer: d. \((x-2y)(x-2y-3)\)

v. \(a(x-y)^2 – by + bx\) is equal to:

Step 1: Group terms: \(a(x-y)^2 + b(x-y)\)
Step 2: Factor out common \((x-y)\):
\((x-y)(a(x-y)+b) = (x-y)(ax – ay + b)\)
Answer: d. \((x-y)(ax-ay+b)\)

vi. Statement 1: The product of two binomials is a trinomial, conversely if we factorise a trinomial we always obtain two binomial factors.
Statement 2: The square of the difference of two terms = The sum of the same two terms × their difference.
Which of the following options is correct?

Step 1: Statement 1: Product of two binomials is a trinomial → False.
beacuse Product of \((x+2)(x-2)\) is binomial.
Step 2: Statement 2: Square of difference = sum × difference → False
because \((a-b)^2 = a^2 – 2ab + b^2\) (this is LHS)
\((a+b)(a-b) = a^2-b^2\) (this is RHS)
As LHS ≠ RHS, so statement is also false.
Answer: d. Both statements are false

vii. Assertion (A): \(25x^2-5x+1\) is a perfect square trinomial.
Reason (R): Any trinomial which can be expressed as \(x^2+y^2+2xy\) or \(x^2+y^2-2xy\) is a perfect square trinomial.

Step 1: Check whether \(25x^2-5x+1\) is a perfect square.
\[ (5x-1)^2 = 25x^2-10x+1 \neq 25x^2-5x+1 \] So, Assertion is false.
Step 2: Reason gives the correct definition of a perfect square trinomial.
So, Reason is true.
Answer: d. A is false, but R is true.

viii. Assertion (A): \(x^2+7x+12=x^2+(4+3)x+4×3=x^2+4x+3x+4×3=(x+4)(x+3)\).
Reason (R): To factorise a given trinomial, product of the first and the last term of the trinomial is always the sum of two parts when we split the middle term.

Step 1: \[ x^2+7x+12=x^2+4x+3x+12=(x+4)(x+3) \] Assertion is true.
Step 2: The reason correctly explains the splitting method.
Reason is true and explains Assertion.
Answer: a. Both A and R are correct, and R is the correct explanation for A.

ix. Assertion (A): The value of k so that the factors of \((x^2-kx+121/16)\) the same is \(11/2\).
Reasons (R): \((x+a)(x+b)=a^2+(a+b)x+ab\)

Step 1: For equal factors: \[ \left(x-\frac{11}{4}\right)^2=x^2-\frac{11}{2}x+\frac{121}{16} \] So, \(k=\frac{11}{2}\). Assertion is true.
Step 2:The identity given in Reason is correct and used here.
Reason is true and explains Assertion.
Answer: a. Both A and R are correct, and R is the correct explanation for A.

x. Assertion (A): There are two values of b so that \(x^2+by-24\) is factorisable.
Reason (R): Two values have: Product = -24 and sum = 2.

Step 1: Multiply to get −24 (the constant term)
Add to b (the coefficient of the middle term)
Values of b such that their sum is 2 and product is -24 are 6 and -4.
So Assertion is True.
Step 2: Reason:
\[ ⇒ x^2 + bx – 24 \\ ⇒ x^2 + 6x – 4x – 24 \\ ⇒ x(x + 6) – 4(x + 6) \\ ⇒ (x – 4)(x + 6) \] Reason is True.
Answer: a. Both A and R are correct, and R is the correct explanation for A.

Q2: Factorise:

i. \(6x^3-8x^2\)

Step 1: Take the common factor: \(2x^2\)
Step 2: Factorise: \(6x^3-8x^2 = 2x^2(3x-4)\)
Answer: \(2x^2(3x-4)\)

ii. \(36x^2y^2-30x^3y^3+48x^3y^2\)

Step 1: Take the common factor: \(6x^2y^2\)
Step 2: Factorise the remaining terms: \(36x^2y^2-30x^3y^3+48x^3y^2 = 6x^2y^2(6-5xy+8x)\)
Answer: \(6x^2y^2(6-5xy+8x)\)

iii. \(8(2a+3b)^3-12(2a+3b)^2\)

Step 1: Take the common factor: \(4(2a+3b)^2\)
Step 2: Factorise remaining: \(8(2a+3b)^3-12(2a+3b)^2 = 4(2a+3b)^2(2(2a+3b)-3)\)
Step 3: Simplify inside bracket: \(2(2a+3b)-3 = 4a+6b-3\)
Answer: \(4(2a+3b)^2(4a+6b-3)\)

iv. \(9a(x-2y)^4-12a(x-2y)^3\)

Step 1: Take the common factor: \(3a(x-2y)^3\)
Step 2: Factorise remaining: \(9a(x-2y)^4-12a(x-2y)^3 = 3a(x-2y)^3(3(x-2y)-4)\)
Step 3: Simplify inside bracket: \(3(x-2y)-4 = 3x-6y-4\)
Answer: \(3a(x-2y)^3(3x-6y-4)\)

Q3: Factorise:

i. \(a^2 – ab(1-b) – b^3\)

Step 1: Expand the middle term: \(-ab(1-b) = -ab + ab^2\)
Step 2: Write the expression: \(a^2 – ab + ab^2 – b^3\)
Step 3: Group terms: \((a^2 – ab) + (ab^2 – b^3)\)
Step 4: Factor each group: \(a(a-b) + b^2(a-b)\)
Step 5: Factor common: \((a-b)(a+b^2)\)
Answer: \((a-b)(a+b^2)\)

ii. \(xy^2 + (x-1)y – 1\)

Step 1: Group terms: \((xy^2 + xy) – (y + 1)\)
Step 2: Factor each group: \(xy(y+1) – 1(y+1)\)
Step 3: Factor common: \((y+1)(xy-1)\)
Answer: \((y+1)(xy-1)\)

iii. \((ax+by)^2 + (bx-ay)^2\)

Step 1: Expand both squares:
\((ax+by)^2 = a^2x^2 + 2abxy + b^2y^2\)
\((bx-ay)^2 = b^2x^2 – 2abxy + a^2y^2\)
Step 2: Add them: \(a^2x^2 + b^2y^2 + b^2x^2 + a^2y^2 = a^2(x^2+y^2) + b^2(x^2+y^2)\)
Step 3: Factor common: \((a^2+b^2)(x^2+y^2)\)
Answer: \((a^2+b^2)(x^2+y^2)\)

iv. \(ab(x^2+y^2)-xy(a^2+b^2)\)

Step 1: Rearrange: \(ab(x^2+y^2) – a^2xy – b^2xy = abx^2 + aby^2 – a^2xy – b^2xy\)
Step 2: Group terms: \((abx^2 – a^2xy) + (aby^2 – b^2xy)\)
Step 3: Factor each group: \(x(ab – a^2y) + y(ab – b^2x)\)
Step 4: Factor common: \((bx-ay)(ax-by)\)
Answer: \((bx-ay)(ax-by)\)

v. \(m-1-(m-1)^2 + am – a\)

Step 1: Expand the square: \(-(m-1)^2 = -m^2 + 2m – 1\)
Step 2: Combine all terms: \(m-1 – m^2 + 2m -1 + am – a = -m^2 + (3+a)m – (2+a)\)
Step 3: Factor quadratic: \((m-1)(2-m+a)\)
Answer: \((m-1)(2-m+a)\)

Q4: Factorise:

i. \(25(2x-y)^2 – 16(x-2y)^2\)

Step 1: Recognise difference of squares: \(25(2x-y)^2 – 16(x-2y)^2 = [5(2x-y)]^2 – [4(x-2y)]^2\)
Step 2: Apply the formula \(A^2-B^2 = (A+B)(A-B)\)
Step 3: Expand each factor:
\(5(2x-y)+4(x-2y) = 10x-5y+4x-8y = 14x-13y\)
\(5(2x-y)-4(x-2y) = 10x-5y-4x+8y = 6x+3y = 3(2x+y)\)
Answer: \(3(14x-13y)(2x+y)\)

ii. \(16(5x+4)^2 – 9(3x-2)^2\)

Step 1: Recognise difference of squares: \(16(5x+4)^2 – 9(3x-2)^2 = [4(5x+4)]^2 – [3(3x-2)]^2\)
Step 2: Apply \(A^2-B^2 = (A+B)(A-B)\)
Step 3: Expand each factor:
\(4(5x+4)+3(3x-2) = 20x+16+9x-6 = 29x+10\)
\(4(5x+4)-3(3x-2) = 20x+16-9x+6 = 11x+22 = 11(x+2)\)
Answer: \(11(29x+10)(x+2)\)

iii. \(25(x-2y)^2 – 4\)

Step 1: Recognise difference of squares: \(25(x-2y)^2 – 4 = [5(x-2y)]^2 – 2^2\)
Step 2: Apply \(A^2-B^2 = (A+B)(A-B)\)
Step 3: Expand each factor:
\(5(x-2y)+2 = 5x-10y+2\)
\(5(x-2y)-2 = 5x-10y-2\)
Answer: \((5x-10y+2)(5x-10y-2)\)

Q5: Factorise:

i. \(a^2-23a+42\)

Step 1: Find two numbers whose sum = -23 and product = 42 → -21 and -2
Step 2: Split middle term: \(a^2-21a-2a+42\)
Step 3: Factor by grouping: \(a(a-21)-2(a-21) = (a-21)(a-2)\)
Answer: \((a-21)(a-2)\)

ii. \(a^2-23a-108\)

Step 1: Find two numbers whose sum = -23 and product = -108 → -27 and 4
Step 2: Split middle term: \(a^2-27a+4a-108\)
Step 3: Factor by grouping: \(a(a-27)+4(a-27) = (a-27)(a+4)\)
Answer: \((a-27)(a+4)\)

iii. \(1-18x-63x^2\)

Step 1: Rearrange: \(-63x^2-18x+1\)
Step 2: Find two numbers whose sum=-18 and product=-63 → -21 and 3
Step 3: Split middle term: \(-63x^2-21x+3x+1\)
Step 4: Factor by grouping: \(-3x(21x+7)+1(3x+1) = (1-21x)(1+3x)\)
Answer: \((1-21x)(1+3x)\)

iv. \(5x^2-4xy-12y^2\)

Step 1: Multiply first and last: 5*-12=-60, find numbers sum=-4 → 6 and -10
Step 2: Split middle: \(5x^2+6xy-10xy-12y^2\)
Step 3: Factor by grouping: \(x(5x+6y)-2y(5x+6y) = (x-2y)(5x+6y)\)
Answer: \((x-2y)(5x+6y)\)

v. \(x(3x+14)+8\)

Step 1: Expand: \(3x^2+14x+8\)
Step 2: Multiply first and last: 3*8=24, find two numbers sum=14 → 12 and 2
Step 3: Split middle: \(3x^2+12x+2x+8\)
Step 4: Factor by grouping: \(3x(x+4)+2(x+4)=(3x+2)(x+4)\)
Answer: \((x+4)(3x+2)\)

vi. \(5-4x(1+3x)\)

Step 1: Expand: \(5-4x-12x^2\)
Step 2: Rearrange: \(-12x^2-4x+5\)
Step 3: Multiply first and last: -12*5=-60, sum=-4 → -10 and 6
Step 4: Split middle: \(-12x^2-10x+6x+5\)
Step 5: Factor by grouping: \(-2x(6x+5)+1(6x+5) = (1-2x)(5+6x)\)
Answer: \((1-2x)(5+6x)\)

vii. \(x^2y^2-3xy-40\)

Step 1: Treat xy terms: \(x^2y^2-3xy-40\)
Step 2: Multiply first and last: 1*-40=-40, sum=-3 → -8 and 5
Step 3: Split middle: \(x^2y^2-8xy+5xy-40\)
Step 4: Factor by grouping: \(xy(xy-8)+5(xy-8)=(xy-8)(xy+5)\)
Answer: \((xy-8)(xy+5)\)

viii. \((3x-2y)^2-5(3x-2y)-24\)

Step 1: Expand: \((3x-2y)^2-5(3x-2y)-24 = 9x^2-12xy+4y^2 -15x+10y -24\)
Step 2: Factor by inspection: \((3x-2y-8)(3x-2y+3)\)
Answer: \((3x-2y-8)(3x-2y+3)\)

ix. \(12(a+b)^2-(a+b)-35\)

Step 1: Expand: \(12a^2+24ab+12b^2 – a – b -35\)
Step 2: Factor by grouping: \((4a+4b-7)(3a+3b+5)\)
Answer: \((4a+4b-7)(3a+3b+5)\)

Q6: Factorise:

i. \(15(5x-4)^2 – 10(5x-4)\)

Step 1: Take common factor: \(5(5x-4)(3(5x-4)-2)\)
Step 2: Simplify: \(5(5x-4)(15x-12-2) = 5(5x-4)(15x-14)\)
Answer: \(5(5x-4)(15x-14)\)

ii. \(3a^2x – bx + 3a^2 – b\)

Step 1: Factor by grouping: \((3a^2x + 3a^2) – (bx + b)\)
Step 2: Take common factors: \(3a^2(x+1) – b(x+1)\)
Step 3: Factor out \((x+1)\): \((3a^2-b)(x+1)\)
Answer: \((3a^2-b)(x+1)\)

iii. \(b(c-d)^2 + a(d-c) + 3(c-d)\)

Step 1: Note that \(a(d-c) = -a(c-d)\)
Step 2: Combine terms: \(b(c-d)^2 – a(c-d) + 3(c-d)\)
Step 3: Take common factor \((c-d)\): \((c-d)(b(c-d) – a + 3) = (c-d)(bc-bd-a+3)\)
Answer: \((c-d)(bc-bd-a+3)\)

iv. \(ax^2 + b^2y – ab^2 – x^2y\)

Step 1: Group terms: \((ax^2 – x^2y) + (b^2y – ab^2)\)
Step 2: Factor each group: \(x^2(a-y) + b^2(y-a)\)
Step 3: Rewrite second term: \(b^2(y-a) = -b^2(a-y)\)
Step 4: Factor out \((a-y)\): \((a-y)(x^2 – b^2) = (a-y)(x+b)(x-b)\)
Answer: \((a-y)(x+b)(x-b)\)

v. \(1 – 3x – 3y – 4(x+y)^2\)

Step 1: Common: \(1-3(x+y)-4(x+y)^2\)
Step 2: Put \(x+y=z\):
\(1-3z-4z^2\)
Step 3: Solve this quadratic equation such that product is -4 and sum is -3.
\[ 1-4z+z-4z^2 \\ (1-4z)+z(1-4z) \]Step 4: Common:
\[ (1-4z)(1+z) \\ \] Step 5: Now put \( z = (x+y)\):
\[ [1-4(x+y)](1+x+y) \\ (1-4x-4y)(1+x+y) \]Answer: \((1-4x-4y)(1+x+y)\)

Q7: Factorise:

i. \(2a^3 – 50a\)

Step 1: Take common factor: \(2a(a^2 – 25)\)
Step 2: Factorise difference of squares: \(2a(a+5)(a-5)\)
Answer: \(2a(a+5)(a-5)\)

ii. \(54a^2b^2 – 6\)

Step 1: Take common factor: \(6(9a^2b^2 – 1)\)
Step 2: Factorise difference of squares: \(6(3ab+1)(3ab-1)\)
Answer: \(6(3ab+1)(3ab-1)\)

iii. \(64a^2b – 144b^3\)

Step 1: Take common factor: \(16b(4a^2 – 9b^2)\)
Step 2: Factorise difference of squares: \(16b(2a+3b)(2a-3b)\)
Answer: \(16b(2a+3b)(2a-3b)\)

iv. \((2x-y)^3 – (2x-y)\)

Step 1: Take common factor: \((2x-y)((2x-y)^2 – 1)\)
Step 2: Factorise difference of squares: \((2x-y)((2x-y)+1)((2x-y)-1)\)
Answer: \((2x-y)(2x-y+1)(2x-y-1)\)

v. \(x^2 – 2xy + y^2 – z^2\)

Step 1: Combine squares: \((x-y)^2 – z^2\)
Step 2: Factorise difference of squares: \((x-y+z)(x-y-z)\)
Answer: \((x-y+z)(x-y-z)\)

vi. \(x^2 – y^2 – 2yz – z^2\)

Step 1: Combine squares: \(x^2 – (y+z)^2\)
Step 2: Factorise difference of squares: \((x+y+z)(x-y-z)\)
Answer: \((x+y+z)(x-y-z)\)

vii. \(7a^5 – 567a\)

Step 1: Take common factor: \(7a(a^4 – 81)\)
Step 2: Factorise difference of squares: \(7a((a^2)^2 – 9^2) = 7a(a^2+9)(a+3)(a-3)\)
Answer: \(7a(a^2+9)(a+3)(a-3)\)

viii. \(5x^2 – \frac{20x^4}{9}\)

Step 1: Take common factor: \(5x^2(1 – \frac{4x^2}{9})\)
Step 2: Factorise difference of squares: \(5x^2(1+\frac{2x}{3})(1-\frac{2x}{3})\)
Answer: \(5x^2(1+\frac{2x}{3})(1-\frac{2x}{3})\)

Q8: Factorise \(xy^2-xz^2\). Hence, find the value of:

Factorise \(xy^2 – xz^2\)

Step 1: Take common factor \(x\): \(x(y^2 – z^2)\)
Step 2: Factorise difference of squares: \(x(y+z)(y-z)\)
Answer: \(x(y+z)(y-z)\)

i. \(9×8^2-9×2^2\)

Step 1: Take common factor 9: \(9(8^2 – 2^2)\)
Step 2: Factorise difference of squares: \(9(8+2)(8-2)\)
Step 3: Simplify: \(9 \times 10 \times 6 = 540\)
Answer: 540

ii. \(40×〖5.5〗^2-40×〖4.5〗^2\)

Step 1: Take common factor 40: \(40(5.5^2 – 4.5^2)\)
Step 2: Factorise difference of squares: \(40(5.5 + 4.5)(5.5 – 4.5)\)
Step 3: Simplify: \(40 \times 10 \times 1 = 400\)
Answer: 400

Q9: Factorise:

i. \((a-3b)^2 – 36b^2\)

Step 1: Recognise difference of squares: \((a-3b)^2 – (6b)^2\)
Step 2: Factorise: \((a-3b+6b)(a-3b-6b)\)
Step 3: Simplify: \((a+3b)(a-9b)\)
Answer: \((a+3b)(a-9b)\)

ii. \(25(a-5b)^2 – 4(a-3b)^2\)

Step 1: Recognise difference of squares: \([5(a-5b)]^2 – [2(a-3b)]^2\)
Step 2: Factorise: \([5(a-5b)+2(a-3b)][5(a-5b)-2(a-3b)]\)
Step 3: Simplify each bracket:
5(a-5b)+2(a-3b) = 5a-25b + 2a-6b = 7a-31b
5(a-5b)-2(a-3b) = 5a-25b – 2a+6b = 3a-19b
Answer: \((7a-31b)(3a-19b)\)

iii. \(a^2 – 0.36b^2\)

Step 1: Recognise difference of squares: \(a^2 – (0.6b)^2\)
Step 2: Factorise: \((a + 0.6b)(a – 0.6b)\)
Answer: \((a+0.6b)(a-0.6b)\)

iv. \(x^4 – 5x^2 – 36\)

Step 1: Substitute \(y = x^2\), equation becomes \(y^2 – 5y – 36\)
Step 2: Factorise \(y^2 – 5y – 36 = (y+4)(y-9)\)
Step 3: Substitute back \(y = x^2\): \((x^2+4)(x^2-9)\)
Step 4: Factorise \(x^2-9 = (x+3)(x-3)\)
Answer: \((x^2+4)(x+3)(x-3)\)

v. \(15(2x-y)^2 – 16(2x-y) – 15\)

Step 1: Substitute \(z = 2x – y\), equation becomes \(15z^2 – 16z – 15\)
Step 2: Factorise \(15z^2 – 16z – 15 = (3z-5)(5z+3)\)
Step 3: Substitute back \(z = 2x – y\): (3(2x-y)-5)(5(2x-y)+3)
Step 4: Simplify brackets: \((6x-3y-5)(10x-5y+3)\)
Answer: \((6x-3y-5)(10x-5y+3)\)

Q10: Evaluate (using factors): \(301^2 \times 300 – 300^3\)

Step 1: Factor out common term \(300\):
\(301^2 \times 300 – 300^3 = 300 \left(301^2 – 300^2\right)\)
Step 2: Recognise difference of squares:
\(301^2 – 300^2 = (301 – 300)(301 + 300) = 1 \times 601 = 601\)
Step 3: Multiply the factor back:
\(300 \times 601 = 180300\)
Answer: 180300

Q11: Use factor method to evaluate

i. \((5z^2-80)\div(z-4)\)

Step 1: Factor out common term from numerator:
\(5z^2-80 = 5(z^2-16)\)
Step 2: Factorise as difference of squares:
\(z^2-16 = (z-4)(z+4)\)
Step 3: Divide by \((z-4)\):
\(\frac{5(z-4)(z+4)}{(z-4)} = 5(z+4)\)
Answer: \(5(z+4)\)

ii. \(10y(6y+21)\div(2y+7)\)

Step 1: Factorise numerator:
\(6y+21 = 3(2y+7)\)
So numerator = \(10y \times 3 (2y+7) = 30y(2y+7)\)
Step 2: Divide by \((2y+7)\):
\(\frac{30y(2y+7)}{2y+7} = 30y\)
Answer: 30y

iii. \((a^2-14a-32)\div(a+2)\)

Step 1: Factorise numerator:
Find two numbers whose product = -32 and sum = -14 → -16 and 2
So, \(a^2-14a-32 = (a-16)(a+2)\)
Step 2: Divide by \((a+2)\):
\(\frac{(a-16)(a+2)}{(a+2)} = a-16\)
Answer: \(a-16\)

iv. \(39x^3(50x^2-98)\div26x^2(5x+7)\)

Step 1: Factorise 50x^2-98:
\(50x^2-98 = 2(25x^2-49) = 2(5x+7)(5x-7)\)
So numerator = \(39x^3 \cdot 2(5x+7)(5x-7) = 78x^3(5x+7)(5x-7)\)
Denominator = \(26x^2(5x+7)\)
Step 2: Simplify fraction:
\(\frac{78x^3(5x+7)(5x-7)}{26x^2(5x+7)} = 3x(5x-7)\)
Answer: \(3x(5x-7)\)