What must be added to $x^3 - 3x^2 - 12x + 19$ so that the result is exactly pisibly by $x^2 + x - 6$?

Given:

Given expression is $x^3 - 3x^2 - 12x + 19$.

To do:

We have to find the expression that has to be added to $x^3 - 3x^2 - 12x + 19$ so that the result is exactly divisibly by $x^2 + x - 6$.

Solution:

We know that,

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

Let $f(x)=x^{3}-3 x^{2}-12 x+19$ and $g(x)=x^{2}+x-6$.

Dividing $f(x)$ by $g(x)$, we get the remainder of degree one. So, let $a x+b$ be added to $f(x)$ to get $p(x)=f(x)+r(x)$

This implies,

$x^{3}-3 x^{2}-12 x+19+a x+b$ is divisible by $g(x)$

$g(x)=x^{2}+x-6$

$=x^{2}+3 x-2 x-6$

$=x(x+3)-2(x+3)$

$=(x+3)(x-2)$

This implies,

$(x+3)$ and $(x-2)$ are factors of $x^{3}-3 x^{2}-12 x+19+a x+b$.

Therefore,

$f(-3)=(-3)^{3}-3(-3)^{2}-12(-3)+19+a(-3)+b=0$

$-27-27+36+19-3 a+b=0$

$1-3 a+b=0$

$b=3a-1$.............(i)

$f(2)=(2)^{3}-3(2)^{2}-12(20+19+a(2)+b=0$

$8-12-24+19+2 a+b=0$

$-9+2 a+b=0$

$-9+2a+3a-1=0$                [From (i)]

$5a=10$

$a=2$

$\Rightarrow b=3(2)-1$

$=6-1$

$=5$

Therefore, $2x+5$ must be added to $x^3 - 3x^2 - 12x + 19$ so that the result is exactly divisibly by $x^2 + x - 6$.

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Updated on: 10-Oct-2022

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