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$A = 5 x^{2} - 16 x - 21$
$B = 3 (x + 2)^{2} - 5 (x + 2) + 2$
Factors of $B$ are

(3 x + 4) (x - 1)

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(3 x - 4) (x - 1)

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(3 x + 4) (x + 1)

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(3 x - 4) (x + 1)

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Solution

The correct option is D $(3 x + 4) (x + 1)$
$A = 5 x^{2} - 16 x - 21$
$B = 3 (x + 2)^{2} - 5 (x + 2) + 2$
Factor of A
$5 x^{2} - 16 x - 21$
$\Rightarrow 5 x^{2} - 21 x + 5 x - 21$
$\Rightarrow x (5 x - 21) + 1 (5 x - 21)$
$\Rightarrow (5 x - 21) (x + 1)$ Ans
factor of B
$\Rightarrow 3 (x + 2)^{2} - 5 (x + 2) + 2$
$\Rightarrow 3 (x^{2} + 4 + 4 x) - 5 x - 10 + 2$
$\Rightarrow 3 x^{2} + 12 x + 12 - 5 x - 8$
$\Rightarrow 3 x^{2} + 7 x + 4$
$\Rightarrow 3 x^{2} + 4 x + 3 x + 4$
$\Rightarrow x (3 x + 4) + 1 (3 x + 4)$
$\Rightarrow (3 x + 4) (x + 1)$ Ans

Suggest Corrections

Similar questions

A = 5 x^{2} - 16 x - 21

B = 3 (x + 2)^{2} - 5 (x + 2) + 2

A

B

A = 5 x^{2} + 3 x^{2} - 4 x + 1, B = 3 x^{3} + 5 x^{2} + 3 x - 8

C = 6 x^{3} - 4 x^{2} - 7 x + 3

(A + B) = C

3 x (5 x - 2) = - 4 x + 1

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

\frac{d y}{d x}

$I f 2 x^{2} - 3 x y + y^{2} + x + 2 y - 8 = 0 t h e n \frac{d y}{d x} =$

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