Factorise: $48 + 8 a - a^{2}$

(12 + a) (4 - a)

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(a - 12) (a + 4)

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(12 - a) (4 + a)

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(a + 12) (a + 4)

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Solution

The correct option is C $(12 - a) (4 + a)$
We need to find two integers whose product is $48$ and sum is $8$ .
i.e. $12 \times 4 = 48$ and $12 - 4 = 8$

$∴ 48 + 8 a - a^{2} = 48 + 12 a - 4 a - a^{2}$
$= 12 (4 + a) - a (4 + a)$
$= (12 - a) (4 + a)$

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\frac{3 x^{2} - 2 x + 4}{(x - 1)^{2}} = \frac{A_{1}}{x + 1} + \frac{A_{2}}{(x + 1)^{2}} + \frac{A_{3}}{(x + 1)^{3}} + \frac{A_{4}}{(x + 1)^{4}} + \frac{A_{5}}{(x + 1)^{5}} + \frac{A_{6}}{(x + 1)^{6}}

(A_{1} + A_{3} + A_{5}, A_{2} + A_{4} + A_{6}) =

Factorise : $a^{4} - (a^{2} - 3 b^{2})^{2}$

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