Given that $x^{2} + x - 6$ is a factor of $2 x^{4} + x^{3} - a x^{2} + b x + a + b - 1$ , then the value of $a$ and $b$ is

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Solution

We have,
$x^{2} + x - 6 = (x + 3) (x - 2)$
Let, $f (x) = 2 x^{4} + x^{3} - a x^{2} + b x + a + b - 1$
Now, $f (- 3) = 2 (- 3)^{4} + (- 3)^{3} - a (- 3)^{2} + b (- 3) + a + b - 1 = 0$
$\Rightarrow 134 - 8 a - 2 b = 0$
$\Rightarrow 4 a + b = 67..... (i)$
$f (2) = 2 (2)^{4} + (2)^{3} - a (2)^{2} + b (2) + a + b - 1 = 0$
$\Rightarrow 39 - 3 a + 3 b = 0$
$\Rightarrow a - b = 13....... (i i)$
solving $(i)$ and $(i i)$ , we get
$a = 16$ and $b = 3$

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