Factorise $a^{2} + 7 a + 10$ + pa + 6p

(a+5) + (a+ 2+ p)

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(a+5) + (a+ p)

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(a+2) + (a+ 5+ p)

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(a+7+ p) + (a+ p)

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Solution

The correct option is A
(a+5) + (a+ 2+ p)

$a^{2} + 7 a + 10 + p a + 5 p$
= $a^{2} + 5 a + 2 a + 10 + p a + 5 p$ ( Factorising $a^{2} + 7 a + 10$ )
= $a (a + 5) + 2 (a + 5) + p a + 5 p$
= $(a + 2) (a + 5) + p a + 5 p$
= $(a + 2) (a + 5) + p (a + 5)$
= $(a + 5) (a + 2 + p)$

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Similar questions

Factorise $a^{2} + 7 a + 10$ + pa + 5p

Factorisation of $a^{2} + 7 a + 10$ gives:

Factorising and dividing $a^{2} + 7 a + 10$ by (a + 5) gives:

6 p - 12 q

(i) p^{2} + 6 p + 8 (i i) p^{2} - 10 q + 21 (i i i) p^{2} + 6 p - 16

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