Factorisation

Q1: Factorising completely (Multiple Choice)

i. \(x^3 – 4x\) is equal to:

Step 1: Take out the common factor \(x\):
\(x^3 – 4x = x(x^2 – 4)\)
Step 2: Factor \(x^2 – 4\) as difference of squares:
\(x^2 – 4 = (x+2)(x-2)\)
Answer: b. \(x(x+2)(x-2)\)

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

Step 1: Group terms:
\((x^4 – y^4) + (x^2 – y^2)\)
Step 2: Factor each group:
\(x^4 – y^4 = (x^2 – y^2)(x^2 + y^2)\), \(x^2 – y^2 = (x – y)(x + y)\)
Step 3: Combine common factors:
\((x^2 – y^2) + 1?\) → Check factorisation carefully → Correct factorisation:
\(x^4 – y^4 + x^2 – y^2 = (x-y)(x+y)(x^2+y^2+1)\)
Answer: c. \((x+y)(x-y)(x^2+y^2+1)\)

iii. \(x^3 – x^2 + ax + x – a – 1\) is equal to:

Step 1: Group terms:
\((x^3 – x^2) + (ax + x) + (-a – 1)\)
Step 2: Factor each group:
\(x^2(x-1) + x(a+1) -1(a+1) = x^2(x-1) + (a+1)(x-1)\)
Step 3: Factor out common \((x-1)\):
\((x-1)(x^2 + a + 1)\)
Answer: b. \((x-1)(x^2 + a + 1)\)

iv. \(8x^3 – 18x\) is equal to:

Step 1: Take out common factor \(2x\):
\(8x^3 – 18x = 2x(4x^2 – 9)\)
Step 2: Factor \(4x^2 – 9\) as difference of squares:
\(4x^2 – 9 = (2x+3)(2x-3)\)
Answer: c. \(2x(2x+3)(2x-3)\)

v. \(x^2 – (a-b)x – ab\) is equal to:

Step 1: Factor as quadratic:
\(x^2 – (a-b)x – ab\)
Step 2: Find two numbers whose product = -ab and sum = -(a-b) → -a and b
Step 3: Split middle term:
\(x^2 – ax + bx – ab\)
Step 4: Group:
\((x^2 – ax) + (bx – ab) = x(x-a) + b(x-a)\)
Step 5: Factor out \((x-a)\):
\((x-a)(x+b)\)
Answer: c. \((x-a)(x+b)\)

Q2: \(8x^2y – 18y^3\)

Step 1: Take out the common factor from both terms:
The common factor of \(8x^2y\) and \(18y^3\) is \(2y\)
\(8x^2y – 18y^3 = 2y(4x^2 – 9y^2)\)
Step 2: Factor \(4x^2 – 9y^2\) as a difference of squares:
\(4x^2 – 9y^2 = (2x + 3y)(2x – 3y)\)
Answer: \(2y(2x + 3y)(2x – 3y)\)

Q3: \(25x^3 – x\)

Step 1: Take out the common factor from both terms:
The common factor of \(25x^3\) and \(-x\) is \(x\)
\(25x^3 – x = x(25x^2 – 1)\)
Step 2: Factor \(25x^2 – 1\) as a difference of squares:
\(25x^2 – 1 = (5x + 1)(5x – 1)\)
Answer: \(x(5x + 1)(5x – 1)\)

Q4: \(16x^4 – 81y^4\)

Step 1: Recognize the expression as a difference of squares:
\(16x^4 – 81y^4 = (4x^2)^2 – (9y^2)^2\)
Step 2: Apply the difference of squares formula \(a^2 – b^2 = (a+b)(a-b)\):
\((4x^2 + 9y^2)(4x^2 – 9y^2)\)
Step 3: Factor the second term further as a difference of squares:
\(4x^2 – 9y^2 = (2x + 3y)(2x – 3y)\)
Answer: \((4x^2 + 9y^2)(2x + 3y)(2x – 3y)\)

Q5: \(x^2 – y^2 – 3x – 3y\)

Step 1: Group terms to factor easily:
\((x^2 – y^2) – (3x + 3y)\)
Step 2: Factor each group:
\( (x+y)(x-y) – 3(x+y) \)
Step 3: Take \((x+y)\) as common factor:
\((x+y)\left[(x-y)/ (x+y)?\right] \) Actually:
\((x+y)(x-y – 3)\)
Answer: \((x+y)(x-y-3)\)

Q6: \(x^2 – y^2 – 2x + 2y\)

Step 1: Group terms for easier factorisation:
\((x^2 – y^2) – (2x – 2y)\)
Step 2: Factor each group:
\((x+y)(x-y) – 2(x-y)\)
Step 3: Take \((x-y)\) as common factor:
\((x-y)\left[(x+y) – 2\right]\)
Step 4: Simplify inside the bracket:
\((x-y)(x+y-2)\)
Answer: \((x-y)(x+y-2)\)

Q7: \(3x^2 + 15x – 72\)

Step 1: Take out the greatest common factor (GCF):
\(3x^2 + 15x – 72 = 3(x^2 + 5x – 24)\)
Step 2: Factor the trinomial \(x^2 + 5x – 24\):
We need two numbers whose product = -24 and sum = 5.
These numbers are 8 and -3.
Step 3: Write as factors:
\(x^2 + 5x – 24 = (x + 8)(x – 3)\)
Step 4: Combine with the GCF:
\(3(x + 8)(x – 3)\)
Answer: \(3(x+8)(x-3)\)

Q8: \(2a^2 – 8a – 64\)

Step 1: Take out the greatest common factor (GCF):
\(2a^2 – 8a – 64 = 2(a^2 – 4a – 32)\)
Step 2: Factor the trinomial \(a^2 – 4a – 32\):
We need two numbers whose product = -32 and sum = -4.
These numbers are -8 and 4.
Step 3: Write as factors:
\(a^2 – 4a – 32 = (a – 8)(a + 4)\)
Step 4: Combine with the GCF:
\(2(a – 8)(a + 4)\)
Answer: \(2(a-8)(a+4)\)

Q9: \(3x^2y + 11xy + 6y\)

Step 1: Take out the common factor \(y\):
\(3x^2y + 11xy + 6y = y(3x^2 + 11x + 6)\)
Step 2: Factor the trinomial \(3x^2 + 11x + 6\):
We need two numbers whose product = \(3 \times 6 = 18\) and sum = 11.
These numbers are 9 and 2.
Step 3: Split the middle term:
\(3x^2 + 9x + 2x + 6\)
Step 4: Group terms:
\((3x^2 + 9x) + (2x + 6)\)
Step 5: Factor each group:
\(3x(x + 3) + 2(x + 3)\)
Step 6: Factor out the common binomial:
\((3x + 2)(x + 3)\)
Step 7: Combine with the GCF \(y\):
\(y(x + 3)(3x + 2)\)
Answer: \(y(x+3)(3x+2)\)

Q10: \(5ap^2 + 11ap + 2a\)

Step 1: Take out the common factor \(a\):
\(5ap^2 + 11ap + 2a = a(5p^2 + 11p + 2)\)
Step 2: Factor the trinomial \(5p^2 + 11p + 2\):
We need two numbers whose product = \(5 \times 2 = 10\) and sum = 11.
These numbers are 10 and 1.
Step 3: Split the middle term:
\(5p^2 + 10p + p + 2\)
Step 4: Group terms:
\((5p^2 + 10p) + (p + 2)\)
Step 5: Factor each group:
\(5p(p + 2) + 1(p + 2)\)
Step 6: Factor out the common binomial:
\((5p + 1)(p + 2)\)
Step 7: Combine with the GCF \(a\):
\(a(p + 2)(5p + 1)\)
Answer: \(a(p+2)(5p+1)\)

Q11: \(a^2 + 2ab + b^2 – c^2\)

Step 1: Recognize a perfect square trinomial:
\(a^2 + 2ab + b^2 = (a + b)^2\)
Step 2: Rewrite the expression:
\((a + b)^2 – c^2\)
Step 3: Apply the difference of squares formula:
\((a + b)^2 – c^2 = (a + b + c)(a + b – c)\)
Answer: \((a+b+c)(a+b-c)\)

Q12: \(x^2 + 6xy + 9y^2 + x + 3y\)

Step 1: Group terms:
\((x^2 + 6xy + 9y^2) + (x + 3y)\)
Step 2: Factorize each group:
\((x + 3y)^2 + 1(x + 3y)\)
Step 3: Take out the common factor \((x + 3y)\):
\((x + 3y)((x + 3y) + 1)\)
Answer: \((x+3y)(x+3y+1)\)

Q13: \(4a^2 – 12ab + 9b^2 + 4a – 6b\)

Step 1: Group terms:
\((4a^2 – 12ab + 9b^2) + (4a – 6b)\)
Step 2: Factorize each group:
\((2a – 3b)^2 + 2(2a – 3b)\)
Step 3: Take out the common factor \((2a – 3b)\):
\((2a – 3b)((2a – 3b) + 2)\)
Answer: \((2a-3b)(2a-3b+2)\)

Q14: \(2a^2b^2 – 98b^4\)

Step 1: Take out the common factor \(2b^2\):
\(2b^2(a^2 – 49b^2)\)
Step 2: Factorize the difference of squares \(a^2 – 49b^2\):
\(a^2 – 49b^2 = (a+7b)(a-7b)\)
Step 3: Combine the factors:
\(2b^2(a+7b)(a-7b)\)
Answer: \(2b^2(a+7b)(a-7b)\)

Q15: \(a^2-16b^2-2a-8b\)

Step 1: Group terms conveniently:
\((a^2 – 16b^2) – (2a + 8b)\)
Step 2: Factor out common factors from each group:
\((a^2 – 16b^2) – 2(a + 4b)\)
Step 3: Factorize \(a^2 – 16b^2\) as difference of squares:
\((a+4b)(a-4b) – 2(a+4b)\)
Step 4: Take out the common factor \((a+4b)\):
\((a+4b)(a-4b-2)\)
Answer: \((a+4b)(a-4b-2)\)