Chapter 14: Factorisation

(i) a²+8a+16

= a²+2×4×a+4²

= (a+4)²

Using the identity (x+y)² = x²+2xy+y²

(ii) p²–10p+25

= p²-2×5×p+5²

= (p-5)²

Using the identity (x-y)² = x²-2xy+y²

(iii) 25m²+30m+9

= (5m)²+2×5m×3+3²

= (5m+3)²

Using the identity (x+y)² = x²+2xy+y²

(iv) 49y²+84yz+36z²

=(7y)²+2×7y×6z+(6z)²

= (7y+6z)²

Using the identity (x+y)² = x²+2xy+y²

(v) 4x²–8x+4

= (2x)²-2×4x+2²

= (2x-2)²

Using the identity (x-y)² = x²-2xy+y²

(vi) 121b²-88bc+16c²

= (11b)²-2×11b×4c+(4c)²

= (11b-4c)²

Using the identity (x-y)² = x²-2xy+y²

(vii) (l+m)²-4lm (Hint: Expand (l+m)² first)

Expand (l+m)² using the identity (x+y)² = x²+2xy+y²

(l+m)²-4lm = l²+m²+2lm-4lm

= l²+m²-2lm

= (l-m)²

Using the identity (x-y)² = x²-2xy+y²

(viii) a⁴+2a²b²+b⁴

= (a²)²+2×a^2×b²+(b²)²

= (a²+b²)²

Using the identity (x+y)² = x²+2xy+y²

(i) 4p²–9q²

= (2p)²-(3q)²

= (2p-3q)(2p+3q)

Using the identity x²-y² = (x+y)(x-y)

(ii) 63a²–112b²

= 7(9a² –16b²)

= 7((3a)²–(4b)²)

= 7(3a+4b)(3a-4b)

Using the identity x²-y² = (x+y)(x-y)

(iii) 49x²–36

= (7x)² -6²

= (7x+6)(7x–6)

Using the identity x²-y² = (x+y)(x-y)

(iv) 16x⁵–144x³

= 16x³(x²–9)

= 16x³(x²–9)

= 16x³(x–3)(x+3)

Using the identity x²-y² = (x+y)(x-y)

(v) (l+m) ²-(l-m) ²

= {(l+m)-(l–m)}{(l +m)+(l–m)}

Using the identity x²-y² = (x+y)(x-y)

= (l+m–l+m)(l+m+l–m)

= (2m)(2l)

= 4 ml

(vi) 9x²y²–16

= (3xy)²-4²

= (3xy–4)(3xy+4)

Using the identity x²-y² = (x+y)(x-y)

(vii) (x²–2xy+y²)–z²

= (x–y)²–z²

Using the identity (x-y)² = x²-2xy+y²

= {(x–y)–z}{(x–y)+z}

= (x–y–z)(x–y+z)

Using the identity x²-y² = (x+y)(x-y)

(viii) 25a²–4b²+28bc–49c²

= 25a²–(4b²-28bc+49c² )

= (5a)²-{(2b)²-2(2b)(7c)+(7c)²}

= (5a)²-(2b-7c)²

Using the identity x²-y² = (x+y)(x-y) , we have

= (5a+2b-7c)(5a-2b+7c)

(i) ax²+bx = x(ax+b)

(ii) 7p²+21q² = 7(p²+3q²)

(iii) 2x³+2xy²+2xz² = 2x(x²+y²+z²)

(iv) am²+bm²+bn²+an² = m²(a+b)+n²(a+b) = (a+b)(m²+n²)

(v) (lm+l)+m+1 = lm+m+l+1 = m(l+1)+(l+1) = (m+1)(l+1)

(vi) y(y+z)+9(y+z) = (y+9)(y+z)

(vii) 5y²–20y–8z+2yz = 5y(y–4)+2z(y–4) = (y–4)(5y+2z)

(viii) 10ab+4a+5b+2 = 5b(2a+1)+2(2a+1) = (2a+1)(5b+2)

(ix) 6xy–4y+6–9x = 6xy–9x–4y+6 = 3x(2y–3)–2(2y–3) = (2y–3)(3x–2)

(i) a⁴–b⁴

= (a²)²-(b²)²

= (a²-b²) (a²+b²)

= (a – b)(a + b)(a²+b²)

(ii) p⁴–81

= (p²)²-(9)²

= (p²-9)(p²+9)

= (p²-3²)(p²+9)

=(p-3)(p+3)(p²+9)

(iii) x⁴–(y+z) ⁴ = (x²)²-[(y+z)²]²

= {x²-(y+z)²}{ x²+(y+z)²}

= {(x –(y+z)(x+(y+z)}{x²+(y+z)²}

= (x–y–z)(x+y+z) {x²+(y+z)²}

(iv) x⁴–(x–z) ⁴ = (x²)²-{(x-z)²}²

= {x²-(x-z)²}{x²+(x-z)²}

= { x-(x-z)}{x+(x-z)} {x²+(x-z)²}

= z(2x-z)( x²+x²-2xz+z²)

= z(2x-z)( 2x²-2xz+z²)

(v) a⁴–2a²b²+b⁴ = (a²)²-2a²b²+(b²)²

= (a²-b²)²

= ((a–b)(a+b))²

= (a – b)² (a + b)²

(i) p²+6p+8

We observed that 8 = 4×2 and 4+2 = 6

p²+6p+8 can be written as p²+2p+4p+8

Taking Common terms, we get

p²+6p+8 = p²+2p+4p+8 = p(p+2)+4(p+2)

Again, p+2 is common in both the terms.

= (p+2)(p+4)

This implies that p²+6p+8 = (p+2)(p+4)

(ii) q²–10q+21

We observed that 21 = -7×-3 and -7+(-3) = -10

q²–10q+21 = q²–3q-7q+21

= q(q–3)–7(q–3)

= (q–7)(q–3)

This implies that q²–10q+21 = (q–7)(q–3)

(iii) p²+6p–16

We observed that -16 = -2×8 and 8+(-2) = 6

p²+6p–16 = p²–2p+8p–16

= p(p–2)+8(p–2)

= (p+8)(p–2)

So, p²+6p–16 = (p+8)(p–2)