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Find the value of m for which $(2 x - 1)$ is a factor of $p (x) = x^{3} + 2 x^{2} + 3 x + m .$

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\frac{- 17}{8}

\frac{17}{8}

\frac{3}{7}

\frac{3}{8}

answer is D.

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Detailed Solution

Given that

(2 x - 1)

is a factor of the given polynomial

p (x) = x^{3} + 2 x^{2} + 3 x + m .

We know by Factor theorem that if

(x - k)

is a factor of the given polynomial

p (x)

, then

p (k) = 0

and if it is not a factor then

p (k) = Remainder .

Equate

(2 x - 1) = 0.

\Rightarrow x = \frac{1}{2}

Here,

p (\frac{1}{2}) = 0.

\begin{array}{l} \Rightarrow \frac{1}{8} + 2 \times \frac{1}{4} + \frac{3}{2} + m = 0 \\ \Rightarrow \frac{1 + 4 + 12 + 8 m}{8} = 0 \end{array}

\begin{array}{l} \Rightarrow 8 m + 17 = 0 \\ \Rightarrow 8 m = - 17 \\ \Rightarrow m = \frac{- 17}{8} \end{array}

Therefore the value of

m = \frac{- 17}{8} .

Therefore, option (4) is the correct answer.

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