Factorise:
(i) $x^{2} + 8 x + 15$
(ii) $x^{4} + 4$

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Solution

We have,
(i) $x^{2} + 8 x + 15 = (x^{2} + 8 x + 16) - 1$ [Replacing $15$ by $16 - 1$ ]
$= {(x)^{2} + 2 \times x \times 4 + 4^{2}} - 1 = (x + 4)^{2} - 1^{2}$
$= {x + 4 + 1} {(x + 4) - 1}$
$= (x + 5) (x + 3)$

(ii) $x^{4} + 4 = x^{4} + 4 x^{2} + 4 - 4 x^{2}$ [Adding and subtracting $4 x^{2}$ ]
$= {(x^{2})^{2} + 2 x^{2} \times 2 + 2^{2}} - 4 x^{2}$
$= (x^{2} + 2)^{2} - (2 x)^{2}$
$= {(x^{2} + 2) + 2 x} {(x^{2} + 2) - 2 x}$
$= (x^{2} + 2 x + 2) (x^{2} - 2 x + 2)$

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Factorise: (i) x2+8x+15 (ii) x4+4

We have, (i) x2+8x+15=(x2+8x+16)−1 [Replacing 15 by 16−1] ={(x)2+2×x×4+42}−1=(x+4)2−12 ={x+4+1}{(x+4)−1} =(x+5)(x+3) (ii) x4+4=x4+4x2+4−4x2 [Adding and subtracting 4x2] ={(x2)2+2x2×2+22}−4x2 =(x2+2)2−(2x)2 ={(x2+2)+2x}{(x2+2)−2x} =(x2+2x+2)(x2−2x+2)

Factorise:
(i) $x^{2} + 8 x + 15$
(ii) $x^{4} + 4$