Factorise:
(i) $x^{2} + 8 x + 16$ (ii) $4 a^{2} - 4 a + 1$

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Solution

We have,
(i) $x^{2} + 8 x + 16 = x^{2} + 2 \times x \times 4 = 4^{2}$
$= (x + 4)^{2}$ [Using: $a^{2} + 2 a b + b^{2} = (a + b)^{2}$ ]
$= (x + 4) (x + 4)$
(ii) $4 a^{2} - 4 a + 1 = (2 a)^{2} - 2 \times 2 a \times 1 + (1)^{2}$
$= (2 a - 1)^{2}$ [Using: $a^{2} - 2 a b + b^{2} = (a - b)^{2}$ ]
$= (2 a - 1) (2 a - 1)$

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Q. Factorise :

(i) x^{2} + 8 x + 16

(i i) 4 a^{2} - 4 a + 1

Q. Factorise:
(i)

25 - 4 x^{2} - 12 x y - 9 y^{2}

(ii)

x^{2} - a^{2} + 10 a b - 25 b^{2}

(iii)

6 x y + x^{2} + 9 y^{2} - 4 a^{2}

(iv)

9 x^{2} + 4 + 12 x - a^{2} + 2 a b - b^{2}

Q. Factorise:
(i)

x^{2} + 8 x + 15

(ii)

x^{4} + 4

Q. The area included between the parabolas

y = \frac{x^{2}}{4 a}

and

y = \frac{8 a^{3}}{x^{2} + 4 a^{2}}

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Factorise: (i) x2+8x+16 (ii) 4a2−4a+1

We have, (i) x2+8x+16=x2+2×x×4=42 =(x+4)2 [Using: a2+2ab+b2=(a+b)2] =(x+4)(x+4) (ii) 4a2−4a+1=(2a)2−2×2a×1+(1)2 =(2a−1)2 [Using: a2−2ab+b2=(a−b)2] =(2a−1)(2a−1)

Factorise:
(i) $x^{2} + 8 x + 16$ (ii) $4 a^{2} - 4 a + 1$