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Chapter 11

Factorisation — Exercise 11.3

Class - 8 ML Aggarwal Understanding ICSE Mathematics

Exercise 11.3

Question 1(i)

Factorise the following expression using algebraic identities:

x2−12x+36x^2 - 12x + 36

Answer

Factorising,

⇒ x2−12x+36x^2 - 12x + 36

⇒ (x)2−2×x×6+(6)2(x)^2 - 2 \times x \times 6 + (6)^2

Using the identity,

[∵ (x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2]

⇒ (x−6)2(x - 6)^2

⇒ (x−6)(x−6)(x - 6)(x - 6)

Hence, x2−12x+36=(x−6)(x−6)=(x−6)2\bm{x^2 - 12x + 36 = (x - 6)(x - 6) = (x - 6)^2}

Question 1(ii)

Factorise the following expression using algebraic identities:

36p2−60pq+25q236p^2 - 60pq + 25q^2

Answer

Factorising,

⇒ 36p2−60pq+25q236p^2 - 60pq + 25q^2

⇒ (6p)2−2×6p×5q+(5q)2(6p)^2 - 2 \times 6p \times 5q + (5q)^2

Using the identity,

[∵ (x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2]

⇒ (6p−5q)2(6p - 5q)^2

⇒ (6p−5q)(6p−5q)(6p - 5q)(6p - 5q)

Hence, 36p2−60pq+25q2=(6p−5q)(6p−5q)=(6p−5q)2\bm{36p^2 - 60pq + 25q^2 = (6p - 5q)(6p - 5q) = (6p - 5q)^2}

Question 1(iii)

Factorise the following expression using algebraic identities:

9x2+66xy+121y29x^2 + 66xy + 121y^2

Answer

Factorising,

⇒ 9x2+66xy+121y29x^2 + 66xy + 121y^2

⇒ (3x)2+2×3x×11y+(11y)2(3x)^2 + 2 \times 3x \times 11y + (11y)^2

Using the identity,

[∵ (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2]

⇒ (3x+11y)2(3x + 11y)^2

⇒ (3x+11y)(3x+11y)(3x + 11y)(3x + 11y)

Hence, 9x2+66xy+121y2=(3x+11y)(3x+11y)=(3x+11y)2\bm{9x^2 + 66xy + 121y^2 = (3x + 11y)(3x + 11y) = (3x + 11y)^2}

Question 1(iv)

Factorise the following expression using algebraic identities:

a4+6a2b2+9b4a^4 + 6a^2b^2 + 9b^4

Answer

Factorising,

⇒ a4+6a2b2+9b4a^4 + 6a^2b^2 + 9b^4

⇒ (a2)2+2×a2×3b2+(3b2)2(a^2)^2 + 2 \times a^2 \times 3b^2 + (3b^2)^2

Using the identity,

[∵ (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2]

⇒ (a2+3b2)2(a^2 + 3b^2)^2

⇒ (a2+3b2)(a2+3b2)(a^2 + 3b^2)(a^2 + 3b^2)

Hence, a4+6a2b2+9b4=(a2+3b2)(a2+3b2)=(a2+3b2)2\bm{a^4 + 6a^2b^2 + 9b^4 = (a^2 + 3b^2)(a^2 + 3b^2) = (a^2 + 3b^2)^2}

Question 1(v)

Factorise the following expression using algebraic identities:

x2+1x2+2x^2 + \dfrac{1}{x^2} + 2

Answer

Using the identity,

[∵ (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2]

Factorising,

⇒x2+1x2+2⇒(x)2+2×x×1x+(1x)2⇒(x+1x)2⇒(x+1x)(x+1x)\Rightarrow x^2 + \dfrac{1}{x^2} + 2\\[1em] \Rightarrow (x)^2 + 2 \times x \times \dfrac{1}{x} + \Big(\dfrac{1}{x}\Big)^2\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)

Hence, x2+1x2+2=(x+1x)(x+1x)=(x+1x)2\bm{x^2 + \dfrac{1}{x^2} + 2 = \Big(x + \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big) = \Big(x + \dfrac{1}{x}\Big)^2}

Question 1(vi)

Factorise the following expression using algebraic identities:

x2+x+14x^2 + x + \dfrac{1}{4}

Answer

Using the identity,

[∵ (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2]

Factorising,

⇒x2+x+14⇒(x)2+2×x×12+(12)2⇒(x+12)2⇒(x+12)(x+12)\Rightarrow x^2 + x + \dfrac{1}{4}\\[1em] \Rightarrow (x)^2 + 2 \times x \times \dfrac{1}{2} + \Big(\dfrac{1}{2}\Big)^2\\[1em] \Rightarrow \Big(x + \dfrac{1}{2}\Big)^2\\[1em] \Rightarrow \Big(x + \dfrac{1}{2}\Big)\Big(x + \dfrac{1}{2}\Big)

Hence, x2+x+14=(x+12)(x+12)=(x+12)2\bm{x^2 + x + \dfrac{1}{4} = \Big(x + \dfrac{1}{2}\Big)\Big(x + \dfrac{1}{2}\Big) = \Big(x + \dfrac{1}{2}\Big)^2}

Question 2(i)

Factorise the following expression using algebraic identities:

4p2−94p^2 - 9

Answer

Factorising,

⇒ 4p2−94p^2 - 9

⇒ (2p)2−(3)2(2p)^2 - (3)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (2p+3)(2p−3)(2p + 3)(2p - 3)

Hence, 4p2−9=(2p+3)(2p−3)\bm{4p^2 - 9 = (2p + 3)(2p - 3)}

Question 2(ii)

Factorise the following expression using algebraic identities:

4x2−169y24x^2 - 169y^2

Answer

Factorising,

⇒ 4x2−169y24x^2 - 169y^2

⇒ (2x)2−(13y)2(2x)^2 - (13y)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (2x+13y)(2x−13y)(2x + 13y)(2x - 13y)

Hence, 4x2−169y2=(2x+13y)(2x−13y)\bm{4x^2 - 169y^2 = (2x + 13y)(2x - 13y)}

Question 3(i)

Factorise the following expression using algebraic identities:

9x2y2−259x^2y^2 - 25

Answer

Factorising,

⇒ 9x2y2−259x^2y^2 - 25

⇒ (3xy)2−(5)2(3xy)^2 - (5)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (3xy+5)(3xy−5)(3xy + 5)(3xy - 5)

Hence, 9x2y2−25=(3xy+5)(3xy−5)\bm{9x^2y^2 - 25 = (3xy + 5)(3xy - 5)}

Question 3(ii)

Factorise the following expression using algebraic identities:

16x2−114416x^2 - \dfrac{1}{144}

Answer

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

Factorising,

⇒16x2−1144⇒(4x)2−(112)2⇒(4x+112)(4x−112)\Rightarrow 16x^2 - \dfrac{1}{144}\\[1em] \Rightarrow (4x)^2 - \Big(\dfrac{1}{12}\Big)^2\\[1em] \Rightarrow \Big(4x + \dfrac{1}{12}\Big)\Big(4x - \dfrac{1}{12}\Big)

Hence, 16x2−1144=(4x+112)(4x−112)\bm{16x^2 - \dfrac{1}{144} = \Big(4x + \dfrac{1}{12}\Big)\Big(4x - \dfrac{1}{12}\Big)}

Question 4(i)

Factorise the following expression using algebraic identities:

20x2−45y220x^2 - 45y^2

Answer

Factorising,

⇒ 20x2−45y220x^2 - 45y^2

⇒ 5(4x2−9y2)5(4x^2 - 9y^2)

⇒ 5((2x)2−(3y)2)5((2x)^2 - (3y)^2)

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ 5(2x+3y)(2x−3y)5(2x + 3y)(2x - 3y)

Hence, 20x2−45y2=5(2x+3y)(2x−3y)\bm{20x^2 - 45y^2 = 5(2x + 3y)(2x - 3y)}

Question 4(ii)

Factorise the following expression using algebraic identities:

916−25a2b2\dfrac{9}{16} - 25a^2b^2

Answer

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

Factorising,

⇒916−25a2b2⇒(34)2−(5ab)2⇒(34+5ab)(34−5ab)\Rightarrow \dfrac{9}{16} - 25a^2b^2\\[1em] \Rightarrow \Big(\dfrac{3}{4}\Big)^2 - (5ab)^2\\[1em] \Rightarrow \Big(\dfrac{3}{4} + 5ab\Big)\Big(\dfrac{3}{4} - 5ab\Big)

Hence, 916−25a2b2=(34+5ab)(34−5ab)\bm{\dfrac{9}{16} - 25a^2b^2 = \Big(\dfrac{3}{4} + 5ab\Big)\Big(\dfrac{3}{4} - 5ab\Big)}

Question 5(i)

Factorise the following expression using algebraic identities:

(2a+3b)2−16c2(2a + 3b)^2 - 16c^2

Answer

Factorising,

⇒ (2a+3b)2−16c2(2a + 3b)^2 - 16c^2

⇒ (2a+3b)2−(4c)2(2a + 3b)^2 - (4c)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ ((2a+3b)+(4c))((2a+3b)−(4c))((2a + 3b) + (4c))((2a + 3b) - (4c))

⇒ (2a+3b+4c)(2a+3b−4c)(2a + 3b + 4c)(2a + 3b - 4c)

Hence, (2a+3b)2−16c2=(2a+3b+4c)(2a+3b−4c)\bm{(2a + 3b)^2 - 16c^2 = (2a + 3b + 4c)(2a + 3b - 4c)}

Question 5(ii)

Factorise the following expression using algebraic identities:

1−(b−c)21 - (b - c)^2

Answer

Factorising,

⇒ 1−(b−c)21 - (b - c)^2

⇒ (1)2−(b−c)2(1)^2 - (b - c)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (1+(b−c))(1−(b−c))(1 + (b - c))(1 - (b - c))

⇒ (1+b−c)(1−b+c)(1 + b - c)(1 - b + c)

Hence, 1−(b−c)2=(1+b−c)(1−b+c)\bm{1 - (b - c)^2 = (1 + b - c)(1 - b + c)}

Question 6(i)

Factorise the following expression using algebraic identities:

9(x+y)2−x29(x + y)^2 - x^2

Answer

Factorising,

⇒ 9(x+y)2−x29(x + y)^2 - x^2

⇒ (3(x+y))2−(x)2(3(x + y))^2 - (x)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (3(x+y)+x)(3(x+y)−x)(3(x + y) + x)(3(x + y) - x)

⇒ (3x+3y+x)(3x+3y−x)(3x + 3y + x)(3x + 3y - x)

⇒ (4x+3y)(2x+3y)(4x + 3y)(2x + 3y)

Hence, 9(x+y)2−x2=(4x+3y)(2x+3y)\bm{9(x + y)^2 - x^2 = (4x + 3y)(2x + 3y)}

Question 6(ii)

Factorise the following expression using algebraic identities:

(2m+3n)2−(3m+2n)2(2m + 3n)^2 - (3m + 2n)^2

Answer

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

Factorising,

⇒ (2m+3n)2−(3m+2n)2(2m + 3n)^2 - (3m + 2n)^2

⇒ ((2m+3n)+(3m+2n))((2m+3n)−(3m+2n))((2m + 3n) + (3m + 2n))((2m + 3n) - (3m + 2n))

⇒ (2m+3n+3m+2n)(2m+3n−3m−2n)(2m + 3n + 3m + 2n)(2m + 3n - 3m - 2n)

⇒ (5m+5n)(n−m)(5m + 5n)(n - m)

⇒ 5(m+n)(n−m)5(m + n)(n - m)

Hence, (2m+3n)2−(3m+2n)2=5(m+n)(n−m)\bm{(2m + 3n)^2 - (3m + 2n)^2 = 5(m + n)(n - m)}

Question 7(i)

Factorise the following expression using algebraic identities:

25(a+b)2−16(a−b)225(a + b)^2 - 16(a - b)^2

Answer

Factorising,

⇒ 25(a+b)2−16(a−b)225(a + b)^2 - 16(a - b)^2

⇒ (5(a+b))2−(4(a−b))2(5(a + b))^2 - (4(a - b))^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (5(a+b)+4(a−b))(5(a+b)−4(a−b))(5(a + b) + 4(a - b))(5(a + b) - 4(a - b))

⇒ (5a+5b+4a−4b)(5a+5b−4a+4b)(5a + 5b + 4a - 4b)(5a + 5b - 4a + 4b)

⇒ (9a+b)(a+9b)(9a + b)(a + 9b)

Hence, 25(a+b)2−16(a−b)2=(9a+b)(a+9b)\bm{25(a + b)^2 - 16(a - b)^2 = (9a + b)(a + 9b)}

Question 7(ii)

Factorise the following expression using algebraic identities:

9(3x+2)2−4(2x−1)29(3x + 2)^2 - 4(2x - 1)^2

Answer

Factorising,

⇒ 9(3x+2)2−4(2x−1)29(3x + 2)^2 - 4(2x - 1)^2

⇒ (3(3x+2))2−(2(2x−1))2(3(3x + 2))^2 - (2(2x - 1))^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (3(3x+2)+2(2x−1))(3(3x+2)−2(2x−1))(3(3x + 2) + 2(2x - 1))(3(3x + 2) - 2(2x - 1))

⇒ (9x+6+4x−2)(9x+6−4x+2)(9x + 6 + 4x - 2)(9x + 6 - 4x + 2)

⇒ (13x+4)(5x+8)(13x + 4)(5x + 8)

Hence, 9(3x+2)2−4(2x−1)2=(13x+4)(5x+8)\bm{9(3x + 2)^2 - 4(2x - 1)^2 = (13x + 4)(5x + 8)}

Question 8(i)

Factorise the following expression using algebraic identities:

x3−25xx^3 - 25x

Answer

Factorising,

⇒ x3−25xx^3 - 25x

⇒ x(x2−25)x(x^2 - 25)

⇒ x((x)2−(5)2)x((x)^2 - (5)^2)

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ x(x+5)(x−5)x(x + 5)(x - 5)

Hence, x3−25x=x(x+5)(x−5)\bm{x^3 - 25x = x(x + 5)(x - 5)}

Question 8(ii)

Factorise the following expression using algebraic identities:

63p2q2−763p^2q^2 - 7

Answer

Factorising,

⇒ 63p2q2−763p^2q^2 - 7

⇒ 7(9p2q2−1)7(9p^2q^2 - 1)

⇒ 7((3pq)2−(1)2)7((3pq)^2 - (1)^2)

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ 7(3pq+1)(3pq−1)7(3pq + 1)(3pq - 1)

Hence, 63p2q2−7=7(3pq+1)(3pq−1)\bm{63p^2q^2 - 7 = 7(3pq + 1)(3pq - 1)}

Question 9(i)

Factorise the following expression using algebraic identities:

32a2b−72b332a^2b - 72b^3

Answer

Factorising,

⇒ 32a2b−72b332a^2b - 72b^3

⇒ 8b(4a2−9b2)8b(4a^2 - 9b^2)

⇒ 8b((2a)2−(3b)2)8b((2a)^2 - (3b)^2)

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ 8b(2a+3b)(2a−3b)8b(2a + 3b)(2a - 3b)

Hence, 32a2b−72b3=8b(2a+3b)(2a−3b)\bm{32a^2b - 72b^3 = 8b(2a + 3b)(2a - 3b)}

Question 9(ii)

Factorise the following expression using algebraic identities:

9(a+b)3−25(a+b)9(a + b)^3 - 25(a + b)

Answer

Factorising,

⇒ 9(a+b)3−25(a+b)9(a + b)^3 - 25(a + b)

⇒ (a+b)(9(a+b)2−25)(a + b)(9(a + b)^2 - 25)

⇒ (a+b)((3(a+b))2−(5)2)(a + b)((3(a + b))^2 - (5)^2)

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (a+b)(3(a+b)+5)(3(a+b)−5)(a + b)(3(a + b) + 5)(3(a + b) - 5)

⇒ (a+b)(3a+3b+5)(3a+3b−5)(a + b)(3a + 3b + 5)(3a + 3b - 5)

Hence, 9(a+b)3−25(a+b)=(a+b)(3a+3b+5)(3a+3b−5)\bm{9(a + b)^3 - 25(a + b) = (a + b)(3a + 3b + 5)(3a + 3b - 5)}

Question 10(i)

Factorise the following expression using algebraic identities:

x2−y2−2y−1x^2 - y^2 - 2y - 1

Answer

Factorising,

⇒ x2−y2−2y−1x^2 - y^2 - 2y - 1

⇒ x2−(y2+2y+1)x^2 - (y^2 + 2y + 1)

Using the identity,

[∵ (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2]

⇒ (x)2−(y+1)2(x)^2 - (y + 1)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (x+(y+1))(x−(y+1))(x + (y + 1))(x - (y + 1))

⇒ (x+y+1)(x−y−1)(x + y + 1)(x - y - 1)

Hence, x2−y2−2y−1=(x+y+1)(x−y−1)\bm{x^2 - y^2 - 2y - 1 = (x + y + 1)(x - y - 1)}

Question 10(ii)

Factorise the following expression using algebraic identities:

p2−4pq+4q2−r2p^2 - 4pq + 4q^2 - r^2

Answer

Factorising,

⇒ p2−4pq+4q2−r2p^2 - 4pq + 4q^2 - r^2

⇒ (p2−4pq+4q2)−r2(p^2 - 4pq + 4q^2) - r^2

Using the identity,

[∵ (x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2]

⇒ (p−2q)2−(r)2(p - 2q)^2 - (r)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ ((p−2q)+r)((p−2q)−r)((p - 2q) + r)((p - 2q) - r)

⇒ (p−2q+r)(p−2q−r)(p - 2q + r)(p - 2q - r)

Hence, p2−4pq+4q2−r2=(p−2q+r)(p−2q−r)\bm{p^2 - 4pq + 4q^2 - r^2 = (p - 2q + r)(p - 2q - r)}

Question 11(i)

Factorise the following expression using algebraic identities:

9x2−y2+4y−49x^2 - y^2 + 4y - 4

Answer

Factorising,

⇒ 9x2−y2+4y−49x^2 - y^2 + 4y - 4

⇒ 9x2−(y2−4y+4)9x^2 - (y^2 - 4y + 4)

Using the identity,

[∵ (x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2]

⇒ (3x)2−(y−2)2(3x)^2 - (y - 2)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (3x+(y−2))(3x−(y−2))(3x + (y - 2))(3x - (y - 2))

⇒ (3x+y−2)(3x−y+2)(3x + y - 2)(3x - y + 2)

Hence, 9x2−y2+4y−4=(3x+y−2)(3x−y+2)\bm{9x^2 - y^2 + 4y - 4 = (3x + y - 2)(3x - y + 2)}

Question 11(ii)

Factorise the following expression using algebraic identities:

4a2−4b2+4a+14a^2 - 4b^2 + 4a + 1

Answer

Factorising,

⇒ 4a2−4b2+4a+14a^2 - 4b^2 + 4a + 1

⇒ (4a2+4a+1)−4b2(4a^2 + 4a + 1) - 4b^2

⇒ [(2a)2+2×2a×1+1]−(4b2)[(2a)^2 + 2 \times 2a \times 1 + 1] - (4b^2)

Using the identity,

[∵ (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2]

⇒ (2a+1)2−(2b)2(2a + 1)^2 - (2b)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ ((2a+1)+2b)((2a+1)−2b)((2a + 1) + 2b)((2a + 1) - 2b)

⇒ (2a+2b+1)(2a−2b+1)(2a + 2b + 1)(2a - 2b + 1)

Hence, 4a2−4b2+4a+1=(2a+2b+1)(2a−2b+1)\bm{4a^2 - 4b^2 + 4a + 1 = (2a + 2b + 1)(2a - 2b + 1)}

Question 12(i)

Factorise the following expression using algebraic identities:

625−p4625 - p^4

Answer

Factorising,

⇒ 625−p4625 - p^4

⇒ (25)2−(p2)2(25)^2 - (p^2)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (25+p2)(25−p2)(25 + p^2)(25 - p^2)

⇒ (25+p2)((5)2−(p)2)(25 + p^2)((5)^2 - (p)^2)

⇒ (25+p2)(5+p)(5−p)(25 + p^2)(5 + p)(5 - p)

Hence, 625−p4=(25+p2)(5+p)(5−p)\bm{625 - p^4 = (25 + p^2)(5 + p)(5 - p)}

Question 12(ii)

Factorise the following expression using algebraic identities:

5y5−405y5y^5 - 405y

Answer

Factorising,

⇒ 5y5−405y5y^5 - 405y

⇒ 5y(y4−81)5y(y^4 - 81)

⇒ 5y((y2)2−(9)2)5y((y^2)^2 - (9)^2)

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ 5y(y2+9)(y2−9)5y(y^2 + 9)(y^2 - 9)

⇒ 5y(y2+9)((y)2−(3)2)5y(y^2 + 9)((y)^2 - (3)^2)

⇒ 5y(y2+9)(y+3)(y−3)5y(y^2 + 9)(y + 3)(y - 3)

Hence, 5y5−405y=5y(y2+9)(y+3)(y−3)\bm{5y^5 - 405y = 5y(y^2 + 9)(y + 3)(y - 3)}

Question 13(i)

Factorise the following expression using algebraic identities:

x4−y4+x2−y2x^4 - y^4 + x^2 - y^2

Answer

Factorising,

⇒ x4−y4+x2−y2x^4 - y^4 + x^2 - y^2

⇒ (x4−y4)+(x2−y2)(x^4 - y^4) + (x^2 - y^2)

⇒ ((x2)2−(y2)2)+(x2−y2)((x^2)^2 - (y^2)^2) + (x^2 - y^2)

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (x2+y2)(x2−y2)+1(x2−y2)(x^2 + y^2)(x^2 - y^2) + 1(x^2 - y^2)

⇒ (x2−y2)(x2+y2+1)(x^2 - y^2)(x^2 + y^2 + 1)

⇒ (x+y)(x−y)(x2+y2+1)(x + y)(x - y)(x^2 + y^2 + 1)

Hence, x4−y4+x2−y2=(x+y)(x−y)(x2+y2+1)\bm{x^4 - y^4 + x^2 - y^2 = (x + y)(x - y)(x^2 + y^2 + 1)}

Question 13(ii)

Factorise the following expression using algebraic identities:

64a2−9b2+42bc−49c264a^2 - 9b^2 + 42bc - 49c^2

Answer

Factorising,

⇒ 64a2−9b2+42bc−49c264a^2 - 9b^2 + 42bc - 49c^2

⇒ 64a2−(9b2−42bc+49c2)64a^2 - (9b^2 - 42bc + 49c^2)

Using the identity,

[∵ (x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2]

⇒ (8a)2−(3b−7c)2(8a)^2 - (3b - 7c)^2

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

⇒ (8a+(3b−7c))(8a−(3b−7c))(8a + (3b - 7c))(8a - (3b - 7c))

⇒ (8a+3b−7c)(8a−3b+7c)(8a + 3b - 7c)(8a - 3b + 7c)

Hence, 64a2−9b2+42bc−49c2=(8a+3b−7c)(8a−3b+7c)\bm{64a^2 - 9b^2 + 42bc - 49c^2 = (8a + 3b - 7c)(8a - 3b + 7c)}

Question 14(i)

Evaluate the following:

382−22216\dfrac{38^2 - 22^2}{16}

Answer

Using the identity,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)]

Evaluating,

⇒382−22216⇒(38+22)(38−22)16⇒60×1616⇒60×1616⇒60\Rightarrow \dfrac{38^2 - 22^2}{16}\\[1em] \Rightarrow \dfrac{(38 + 22)(38 - 22)}{16}\\[1em] \Rightarrow \dfrac{60 \times 16}{16}\\[1em] \Rightarrow \dfrac{60 \times \cancel{16}}{\cancel{16}}\\[1em] \Rightarrow 60

Hence, 382−22216=60\bm{\dfrac{38^2 - 22^2}{16} = 60}

Question 14(ii)

Evaluate the following:

(5.3)2−(3.3)2(11.2)2−(11.2)×(2.4)+(1.2)2\dfrac{(5.3)^2 - (3.3)^2}{(11.2)^2 - (11.2) \times (2.4) + (1.2)^2}

Answer

Using the identities,

[∵ (x2−y2)=(x+y)(x−y)(x^2 - y^2) = (x + y)(x - y)] and [∵ (x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2]

Evaluating,

⇒(5.3)2−(3.3)2(11.2)2−(11.2)×(2.4)+(1.2)2⇒(5.3)2−(3.3)2(11.2)2−2×(11.2)×(1.2)+(1.2)2⇒(5.3+3.3)(5.3−3.3)(11.2−1.2)2⇒8.6×2(10)2⇒17.2100⇒0.172\Rightarrow \dfrac{(5.3)^2 - (3.3)^2}{(11.2)^2 - (11.2) \times (2.4) + (1.2)^2}\\[1em] \Rightarrow \dfrac{(5.3)^2 - (3.3)^2}{(11.2)^2 - 2 \times (11.2) \times (1.2) + (1.2)^2}\\[1em] \Rightarrow \dfrac{(5.3 + 3.3)(5.3 - 3.3)}{(11.2 - 1.2)^2}\\[1em] \Rightarrow \dfrac{8.6 \times 2}{(10)^2}\\[1em] \Rightarrow \dfrac{17.2}{100}\\[1em] \Rightarrow 0.172

Hence, (5.3)2−(3.3)2(11.2)2−(11.2)×(2.4)+(1.2)2=0.172\bm{\dfrac{(5.3)^2 - (3.3)^2}{(11.2)^2 - (11.2) \times (2.4) + (1.2)^2} = 0.172}

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