Factorize :
$x^{2} + 3 x + 2 + a x + 2 a$

(3 x - 2) (x + a + 2)

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(x + 2) (x + a + 1)

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(3 x + 2) (x - a + 1)

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(x - 2) (x + a - 1)

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Solution

The correct option is D $(x + 2) (x + a + 1)$
$x^{2} + 3 x + 2 + a x + 2 a$
$= x^{2} + a x + x + 2 x + 2 + 2 a$
$= x (x + a + 1) + 2 (x + a + 1)$
$= (x + a + 1) (x + 2)$

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Similar questions

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a x - 12

a

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 {2 x}^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a x - 12,

a^{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

A x^{n} + B

n

\frac{x^{2} - 2 x + 1}{x - 1} \times \frac{2 x - 4}{x^{2} - 3 x + 2}

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