If $p (x) = a x^{3} + 3 x - 13$ and $q (x) = 2 x^{3} - 5 x + 1$ are divided by $x + 2$ remainder is same in each case. find the value of $a$ .

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Solution

$p (x) = a x^{3} + 3 x - 13, q (x) = 2 x^{3} - 5 x + 1$
Given When $p (x)$ & $q (x)$ are divided by $x + 2$ then remainder is same
To find $a$
$x + 2 | ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ {a x}^{3} + 3 x - 13 ∣ ∣ {a x}^{3} - 2 a x + (44 + 3) {a x}^{3} + {2 a x}^{2}_________- {2 a x}^{2} + 3 x - 13 - {2 a x}^{2} - 4 a x + +_(4 a + 3) x - 13 (4 a + 3) x + 2 (4 a + 3)______- 8 a - 19$

$x + 2 | ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ {2 x}^{3} - 5 x + 1 ∣ ∣ {2 x}^{2} - 4 x + 3 {2 x}^{3} + {4 x}^{2}_- {4 x}^{2} - 5 x + 1 - {4 x}^{2} - 8 x + +_3 x + 1 3 x + 6______________- 5$

$∵$ Remainder are same
$- 8 a - 19 = - 5$
$8 a + 19 = 5$
$8 a = 5 - 19$
$a = \frac{- 14}{8} = \frac{- 7}{4}$
$a = \frac{- 7}{4}$

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