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

Given:

Given expression is $x^3 - 6x^2 - 15x + 80$.

To do:

We have to find the expression that has to be subtracted from $x^3 - 6x^2 - 15x + 80$ so that the result is exactly divisibly by $x^2 + x - 12$.

Solution:

We know that,

If $(x-m)$ is a root of $f(x)$ then $f(m)=0$.

Let $f(x)=x^{3}-6 x^{2}-15 x+80$ and $g(x)=x^{2}+x-12$.

Dividing $f(x)$ by $g(x)$, let the remainder be $r(x)$

Let by subtracting $a x+b$ from $f(x)$ the resultant value is divisible by $g(x)$.

$p(x)=f(x)-(ax+b)$

$=x^{3}-6 x^{2}-15 x+80-ax-b$

$g(x)=x^{2}+x-12$

$=x^{2}+4 x-3 x-12$

$=x(x+4)-3(x+4)$

$=(x+4)(x-3)$

$x+4$ and $x-3$ are factors of $p(x)$.

This implies,

$p(-4)=(-4)^{3}-6(-4)^{2}-15(-4)+80-a(-4)-b=0$

$-64-96+60+80+4 a-b=0$

$-160+140+4 a-b=0$

$4 a-b-20=0$

$b=4a-20$..........(i)

$p(3)=(3)^{3}-6(3)^{2}-15(3)+80-a(3)-b=0$

$27-54-45+80-3 a-b=0$

$107-99-3 a-b=0$

$8-3 a-b=0$

$8-3a-(4a-20)=0$                     [From (i)]

$8-3a-4a+20=0$

$7a=28$

$a=4$

$b=4(4)-20$

$=16-20$

$=-4$

Therefore, $4x-4$ must be subtracted from $x^3 - 6x^2 - 15x + 80$ so that the result is exactly divisibly by $x^2 + x - 12$. 

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Updated on: 10-Oct-2022

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