Find the value of $a$ if $2 x - 3$ is a factor of $2 x^{4} - x^{3} - 3 x^{2} - 2 x + a$ .

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Solution

If $2 x - 3$ is the factor of the polynomial $2 x^{4} - x^{3} - 3 x^{2} - 2 x + a$ then $2 x - 3 = 0$ that is:

$2 x = 3 \Rightarrow x = \frac{3}{2}$

Substitute the value of $x$ in the given polynomial as follows:

$2 x^{4} - x^{3} - 3 x^{2} - 2 x + a = 0 \Rightarrow 2 {(\frac{3}{2})}^{4} - {(\frac{3}{2})}^{3} - 3 {(\frac{3}{2})}^{2} - (2 \times \frac{3}{2}) + a = 0 \Rightarrow (2 \times \frac{81}{16}) - \frac{27}{8} - (3 \times \frac{9}{4}) - 3 + a = 0 \Rightarrow \frac{81}{8} - \frac{27}{8} - \frac{27}{4} - 3 + a = 0 \Rightarrow \frac{81 - 27 - 54 - 24 + 8 a}{8} = 0$
$\Rightarrow 81 - 27 - 54 - 24 + 8 a = 0 \Rightarrow - 24 + 8 a = 0 \Rightarrow 8 a = 24 \Rightarrow a = \frac{24}{8} \Rightarrow a = 3$

Hence, $a = 3$ .

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