What must be subtracted from $x^{3} - 6 x^{2} - 15 x + 80$ so that the result is exactly divisible by $x^{2} + x - 12$ ?

Open in App

Solution

Let ax + b be subtracted from $p (x) = x^{3} - 6 x^{2} - 15 x + 80$
so that it is exactly divisible by $x^{2} + x - 12$ .
$∴ s (x) = x^{3} - 6 x^{2} - 15 x + 80 - (a x + b)$
$= x^{3} - 6 x^{2} - (15 + a) x + (80 - b)$
Dividend = Divisor $\times$ quotient + remainder
But remainder will be zero.
$∴$ Dividend = Divisor $\times$ quotient
$\Rightarrow s (x) = (x^{2} + x - 12) \times q u o t i e n t$
$\Rightarrow s (x) = x^{3} - 6 x^{2} - (15 + a) x + (80 - b)$
Hence, $x (4 - a) + (- 4 - b) = 0, x \neq 0$
$\Rightarrow 4 - a = 0$ & $(- 4 - b) = 0$
$\Rightarrow a = 4$ and $b = - 4$
Hence, if in p(x) we subtract 4x - 4 then it is exactly divisible by $x^{2} + x - 12$

Suggest Corrections

Similar questions

Respected Sir,

Please help me in solving my below mentioned doubt:

What must be subtracted from x³- 6x² - 15x + 80 so that the resulting polynomial is exactly divisible by x²+ x + 22?

4 x^{4} - 2 x^{3} - 6 x^{2} + x - 5

2 x^{2} + x - 1

(x^{3} + 5 x^{2} + 5 x + 8)

(x^{2} + 3 x - 2)

x^{3} + 13 x^{2} + 35 x + 25

x^{2} + 11 x + 10

Join BYJU'S Learning Program