Factorisation

Q1: Multiple Choice Type

i. \(\left(2x+y\right)^2-\left(2y+x\right)^2\) is equal to:

Step 1: Recall the formula for difference of squares: \(p^2 – q^2 = (p-q)(p+q)\) \[ (2x+y)^2 – (2y+x)^2 \]Step 2: Identify \(p = 2x+y\) and \(q = 2y+x\), then apply the formula. \[ (2x+y-(2y+x))((2x+y)+(2y+x)) = (x-y)(3x+3y) \]Step 3: Factor out 3 from the second bracket. \[ (x-y)3(x+y) = 3(x+y)(x-y) \]Answer: a. 3(x+y)(x-y)

ii. \(49-\left(x+5\right)^2\) is equal to:

Step 1: Recognise \(49 – (x+5)^2\) as \(7^2 – (x+5)^2\).
Step 2: Apply difference of squares formula. \[ 7^2 – (x+5)^2 = (7 – (x+5))(7 + (x+5)) \]Step 3: Simplify each bracket. \[ (2-x)(12+x) \]Answer: b. (2-x)(12+x)

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

Step 1: Group the quadratic terms: \(a^2 – 2ab + b^2 + a – b = (a-b)^2 + (a-b)\)
Step 2: Factor out \((a-b)\) from the expression. \[ (a-b)((a-b)+1) = (a-b)(a-b+1) \]Answer: d. (a-b)(a-b+1)

iv. \(x^2+y^2-2xy-1\) is equal to:

Step 1: Rewrite \(x^2 + y^2 – 2xy – 1 = (x-y)^2 – 1\)
Step 2: Apply difference of squares formula: \(p^2 – q^2 = (p-q)(p+q)\) \[ (x-y)^2 – 1 = ((x-y)-1)((x-y)+1) \]Step 3: Simplify brackets. \[ (x-y-1)(x-y+1) \]Answer: d. (x – y + 1)(x – y – 1)

v. \(a^2+2a+1-b^2-x^2+2bx\) is equal to:

Step 1: Rearrange and group squares: \(a^2+2a+1 – (b^2 – 2bx + x^2) = (a+1)^2 – (b-x)^2\)
Step 2: Apply difference of squares formula. \[ (a+1 – (b-x))(a+1 + (b-x)) = (a+1 – b + x)(a+1 + b – x) \]Answer: b. (a + 1 + b – x)(a + 1 – b + x)

Q2: \((a+2b)^2 – a^2\)

Step 1: Recognise the pattern \(p^2 – q^2 = (p-q)(p+q)\) with \(p=a+2b,\; q=a\). \[ (a+2b)^2 – a^2 = \big((a+2b)-a\big)\big((a+2b)+a\big) \]Step 2: Simplify each factor. \[ \big((a+2b)-a\big) = 2b,\qquad \big((a+2b)+a\big) = 2a+2b = 2(a+b) \]Step 3: Multiply the simplified factors. \[ (2b)\cdot 2(a+b) = 4b(a+b) \]Answer: \(4b(a+b)\)

Q3: \((5a-3b)^2 – 16b^2\)

Step 1: Recognise the formula of difference of squares: \[ p^2 – q^2 = (p-q)(p+q) \] Here, \(p = (5a – 3b)\) and \(q = 4b\).
Step 2: Apply the formula. \[ (5a – 3b)^2 – (4b)^2 = \big((5a – 3b) – 4b\big)\big((5a – 3b) + 4b\big) \]Step 3: Simplify each factor. \[ (5a – 3b – 4b)(5a – 3b + 4b) = (5a – 7b)(5a + b) \]Answer: \((5a – 7b)(5a + b)\)

Q4: \(a^4 – (a^2 – 3b^2)^2\)

Step 1: Recognise the difference of squares pattern \(p^2 – q^2 = (p-q)(p+q)\). Here take \(p = a^2\) and \(q = (a^2 – 3b^2)\).
Step 2: Apply the formula. \[ a^4 – (a^2 – 3b^2)^2 = \big(a^2 – (a^2 – 3b^2)\big)\big(a^2 + (a^2 – 3b^2)\big) \]Step 3: Simplify each bracket. \[ a^2 – (a^2 – 3b^2) = 3b^2,\qquad a^2 + (a^2 – 3b^2) = 2a^2 – 3b^2 \]Step 4: Multiply the simplified factors (or factor out common numerical factor if desired). \[ a^4 – (a^2 – 3b^2)^2 = 3b^2(2a^2 – 3b^2) \]Answer: \(3b^2(2a^2 – 3b^2)\)

Q5: \((5a-2b)^2 – (2a-b)^2\)

Step 1: Identify the difference of two squares formula: \[ p^2 – q^2 = (p-q)(p+q) \] Here, \(p = (5a – 2b)\) and \(q = (2a – b)\).
Step 2: Apply the formula. \[ (5a-2b)^2 – (2a-b)^2 = \big((5a-2b) – (2a-b)\big)\big((5a-2b) + (2a-b)\big) \]Step 3: Simplify each bracket. \[ (5a – 2b – 2a + b) = 3a – b \] \[ (5a – 2b + 2a – b) = 7a – 3b \]Step 4: Write the factorised form. \[ (5a-2b)^2 – (2a-b)^2 = (3a – b)(7a – 3b) \]Answer: \((3a – b)(7a – 3b)\)

Q6: \(1 – 25(a+b)^2\)

Step 1: Recognise the difference of two squares identity: \[ p^2 – q^2 = (p-q)(p+q) \] Here, \(p = 1\) and \(q = 5(a+b)\), because \(25(a+b)^2 = \big(5(a+b)\big)^2\).
Step 2: Apply the formula. \[ 1 – 25(a+b)^2 = \big(1 – 5(a+b)\big)\big(1 + 5(a+b)\big) \]Step 3: Write the factors in simplified form. \[ = (1 – 5a – 5b)(1 + 5a + 5b) \]Answer: \((1 – 5a – 5b)(1 + 5a + 5b)\)

Q7: \(4(2a+b)^2 – (a-b)^2\)

Step 1: Recognise the difference of two squares identity: \[ p^2 – q^2 = (p-q)(p+q) \] Here, \(p = 2(2a+b) = (4a+2b)\) and \(q = (a-b)\).
Step 2: Apply the formula. \[ 4(2a+b)^2 – (a-b)^2 = \big((4a+2b) – (a-b)\big)\big((4a+2b) + (a-b)\big) \]Step 3: Simplify each bracket. \[ (4a+2b – a + b) = (3a+3b) \] \[ (4a+2b + a – b) = (5a+b) \]Step 4: Write the final factorised form. \[ = (3a+3b)(5a+b) = 3 (a + b)(5a + b) \]Answer: \(3 (a + b)(5a + b)\)

Q8: \(25(2x+y)^2 – 16(x-y)^2\)

Step 1: Recognise that this is a difference of two squares. \[ 25(2x+y)^2 – 16(x-y)^2 = \big(5(2x+y)\big)^2 – \big(4(x-y)\big)^2 \]Step 2: Apply the identity: \[ p^2 – q^2 = (p-q)(p+q) \] Here, \(p = 5(2x+y)\), \(q = 4(x-y)\).
Step 3: Write in factorised form. \[ = \big(5(2x+y) – 4(x-y)\big)\big(5(2x+y) + 4(x-y)\big) \]Step 4: Simplify each bracket.
First bracket: \[ 5(2x+y) – 4(x-y) = (10x+5y) – (4x-4y) = 6x+9y \] Second bracket: \[ 5(2x+y) + 4(x-y) = (10x+5y) + (4x-4y) = 14x+y \]Step 5: Final factorised form. \[ = (6x+9y)(14x+y) = 3(2x+3y)(14x+y) \]Answer: \(3(2x+3y)(14x+y)\)

Q9: \(\left(6\frac{2}{3}\right)^2 – \left(2\frac{1}{3}\right)^2\)

Step 1: Convert the mixed fractions into improper fractions. \[ 6\frac{2}{3} = \frac{20}{3}, \quad 2\frac{1}{3} = \frac{7}{3} \]So the given expression becomes: \[ \left(\frac{20}{3}\right)^2 – \left(\frac{7}{3}\right)^2 \]Step 2: Apply the identity of difference of squares: \[ p^2 – q^2 = (p-q)(p+q) \] Here, \(p = \frac{20}{3}, \, q = \frac{7}{3}\).
Step 3: Substitute values. \[ = \left(\frac{20}{3} – \frac{7}{3}\right)\left(\frac{20}{3} + \frac{7}{3}\right) \]Step 4: Simplify each bracket. \[ \frac{20}{3} – \frac{7}{3} = \frac{13}{3}, \quad \frac{20}{3} + \frac{7}{3} = \frac{27}{3} = 9 \]Step 5: Multiply the factors. \[ = \frac{13}{3} \times 9 = 39 \]Answer: \(39\)

Q10: \((0.7)^2 – (0.3)^2\)

Step 1: Recall the identity of difference of two squares: \[ p^2 – q^2 = (p-q)(p+q) \] Here, \(p = 0.7, \, q = 0.3\).
Step 2: Substitute the values. \[ (0.7)^2 – (0.3)^2 = (0.7 – 0.3)(0.7 + 0.3) \]Step 3: Simplify each bracket. \[ (0.7 – 0.3) = 0.4, \quad (0.7 + 0.3) = 1.0 \]Step 4: Multiply the factors. \[ 0.4 \times 1.0 = 0.4 \]Answer: \(0.4\)

Q11: \(75(x+y)^2 – 48(x-y)^2\)

Step 1: Take out the common factor of \(3\). \[ 75(x+y)^2 – 48(x-y)^2 = 3\big(25(x+y)^2 – 16(x-y)^2\big) \]Step 2: Recognize the difference of two squares. \[ 25(x+y)^2 – 16(x-y)^2 = \big(5(x+y)\big)^2 – \big(4(x-y)\big)^2 \]Step 3: Apply the identity: \[ p^2 – q^2 = (p-q)(p+q) \] Here, \(p = 5(x+y), \, q = 4(x-y)\).
Step 4: Substitute and expand. \[ = \big(5(x+y) – 4(x-y)\big)\big(5(x+y) + 4(x-y)\big) \]Step 5: Simplify each bracket. \[ 5(x+y) – 4(x-y) = 5x+5y -4x +4y = (x+9y) \\ 5(x+y) + 4(x-y) = 5x+5y +4x -4y = (9x+y) \]Step 6: Write the final factorisation with the common factor \(3\). \[ 75(x+y)^2 – 48(x-y)^2 = 3(x+9y)(9x+y) \]Answer:\(3(x+9y)(9x+y)\)

Q12: \(a^2 + 4a + 4 – b^2\)

Step 1: Observe that \(a^2 + 4a + 4\) is a perfect square trinomial. \[ a^2 + 4a + 4 = (a+2)^2 \]Step 2: Rewrite the expression using the square form. \[ a^2 + 4a + 4 – b^2 = (a+2)^2 – b^2 \]Step 3: Apply the identity of difference of squares: \[ p^2 – q^2 = (p-q)(p+q) \] Here, \(p = (a+2), \; q = b\).
Step 4: Substitute and simplify. \[ (a+2)^2 – b^2 = \big((a+2) – b\big)\big((a+2) + b\big) \\ = (a+2-b)(a+2+b) \]Answer:\((a+2-b)(a+2+b)\)

Q13: \(a^2 – b^2 – 2b – 1\)

Step 1: Group the terms involving \(b\). \[ a^2 – (b^2 + 2b + 1) \]Step 2: Recognize that \(b^2 + 2b + 1\) is a perfect square trinomial. \[ b^2 + 2b + 1 = (b+1)^2 \]Step 3: Rewrite the expression. \[ a^2 – (b+1)^2 \]Step 4: Apply the difference of squares identity: \[ p^2 – q^2 = (p-q)(p+q) \] Here, \(p = a, \; q = (b+1)\).
Step 5: Simplify. \[ a^2 – (b+1)^2 = (a-(b+1))(a+(b+1)) \\ = (a-b-1)(a+b+1) \]Answer:\((a-b-1)(a+b+1)\)

Q14: \(x^2 + 6x + 9 – 4y^2\)

Step 1: Group the terms properly. \[ (x^2 + 6x + 9) – 4y^2 \]Step 2: Recognize that \(x^2 + 6x + 9\) is a perfect square trinomial. \[ x^2 + 6x + 9 = (x+3)^2 \]Step 3: Rewrite the expression. \[ (x+3)^2 – (2y)^2 \]Step 4: Apply the difference of squares identity: \[ p^2 – q^2 = (p-q)(p+q) \] Here, \(p = (x+3), \; q = 2y\).
Step 5: Factorise. \[ (x+3)^2 – (2y)^2 = (x+3-2y)(x+3+2y) \]Answer:\((x+3-2y)(x+3+2y)\)