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

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Solution

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

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

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