The factors of p-2 - 6p - 16- are -p-2-p-8-p-2-p-8-p-2-p-8-p-2-p-8-

The correct option is A (p 2)(p+8)p^2 + 6p 16 = p^2 + 8p 2p 16 [∵ 6p = 8p 2p and 16p^2 = 8p × 2p] = p(p+8) 2(p+8) [Regrouping the terms] = (p + 8)( ...

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Standard VIII

Mathematics

Rational Algebraic Expression

The factors o...

Question

The factors of $p^{2} + 6 p - 16$ are __________.

(p-2)(p+8)

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(p+2)(p-8)

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(p+2)(p+8)

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(p-2)(p-8)

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Solution

The correct option is A (p-2)(p+8)
$p^{2} + 6 p - 16$
= $p^{2} + 8 p - 2 p - 16$
$[∵ 6 p = 8 p - 2 p a n d - 16 p^{2} = 8 p \times - 2 p]$
= $p (p + 8) - 2 (p + 8)$ [Regrouping the terms]
= (p + 8)(p - 2)

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Similar questions

If $s e c θ + t a n θ = p$ , prove that

(i) $s e c θ = \frac{1}{2} (p + \frac{1}{p})$

(ii) $t a n θ = \frac{1}{2} (p - \frac{1}{p})$

(iii) $s i n θ = \frac{p^{2} - 1}{p^{2} + 1}$

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p^{2} + 6 p - 16

are __________.

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The factors of p2+6p−16 are __________.

The correct option is A (p-2)(p+8)p2+6p−16 = p2+8p−2p−16 [∵6p=8p−2p and −16p2=8p×−2p] = p(p+8)−2(p+8) [Regrouping the terms] = (p + 8)(p - 2)

The factors of $p^{2} + 6 p - 16$ are __________.

The correct option is A (p-2)(p+8)
$p^{2} + 6 p - 16$
= $p^{2} + 8 p - 2 p - 16$
$[∵ 6 p = 8 p - 2 p a n d - 16 p^{2} = 8 p \times - 2 p]$
= $p (p + 8) - 2 (p + 8)$ [Regrouping the terms]
= (p + 8)(p - 2)