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Division of Polynomials

Factorise : i...

Question

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

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Solution

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

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