If $x^{2} + x - 12$ divides $P (x) = x^{3} + a x^{2} + b x - 84$ exactly, find $a$ and $b$ .

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Solution

$x^{2} + x - 12 = 0 \Rightarrow x^{2} + 4 x - 3 x - 12 = 0 \Rightarrow (x - 3) (x + 4) = 0$
so $x = 3$ or $- 4$
$p (x) \div g (x) = q (x) + r \Rightarrow p (x) \div g (x) = q (x) + 0$
$x^{3} + a x^{2} + b x - 84 \div x^{2} + x - 12$
after dividing remainder will be $a x^{2} - x^{2} + b x + 12 x - 84 = 0$
Let $x = 3 \Rightarrow 27 a - 9 + 3 b + 36 - 84 = 0$
$9 a + 3 b = 57 \Rightarrow 3 (3 a + b) = 57$
$3 a + b = 19 . . . (1)$
Let $x = - 4$
$16 a - 16 - 4 b - 48 - 84 = 0 \Rightarrow 16 a - 4 b = 148$
$4 (4 a - b) = 148 \Rightarrow 4 a - b = 37 . . . (2)$
Add (1) & (2)
$3 a + b + 4 a - b = 19 + 37$
$7 a = 56$
$a = 8$
Now $a = 8$ in (1)
$3 a + b = 19$
$24 + b = 19$
$b = - 5$

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