Factorise : $16 m n + 8 + 4 n + 32 m$

(4 n + 1) (m + 2)

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4 (4 m + 1) (n + 2)

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(4 m + 1) (n + 2)

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4 (4 n + 1) (m + 2)

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Solution

The correct option is B $4 (4 m + 1) (n + 2)$
$16 m n + 8 + 4 n + 32 m$
$= 16 m n + 4 n + 32 m + 8$ ... (Regrouping terms)
$= 4 n (4 m + 1) + 8 (4 m + 1)$
$= (4 n + 8) (4 m + 1)$
$= 4 (4 m + 1) (n + 2)$

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Similar questions

\sum_{m = 1}^{n} (\sum_{k = 1}^{m} (\sum_{p = k}^{m}^{n} C_{m} .^{m} C_{p} .^{p} C_{k})) =

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(4 m)^{2} \times (4 n)^{2} = (16 m n)^{2}

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