On dividing $2 x^{2} + 3 x + 1$ by a linear polynomial g(x), the quotient is 2x – 1 and remainder is ‘r’, where $r ϵ R$ , then g(x) is

x – 1

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x + 1

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x + 2

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x + 4

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Solution

The correct option is C x + 2
$f (x) = 2 x^{2} + 3 x + 1 = (2 x - 1) g (x) + r,$ where g (x) is the divisor.

Let $g (x) = (x - a)$

When $f (x)$ is divided by $x - a$ , the remainder is $f (a)$

Thus, $r = 2 a^{2} + 3 a + 1$

Hence, $2 x^{2} + 3 x + 1 = (2 x - 1) (x - a) + 2 a^{2} + 3 a + 1$

$\Rightarrow 2 x^{2} + 3 x + 1 = 2 x^{2} - 2 a x - x + a + 2 a^{2} + 3 a + 1$

$\Rightarrow 2 x^{2} + 3 x + 1 = 2 x^{2} - (2 a + 1) x + 2 a^{2} + 4 a + 1$

Comparing the coefficient of x on both sides, we get:

$3 = - (2 a + 1)$
$\Rightarrow 4 = - 2 a$
$\Rightarrow a = - 2$

Thus, $g (x) = (x + 2)$

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