What must be added to $3x^3 + x^2 - 22x + 9$ so that the result is exactly divisible by $3x^2 + 7x - 6$?

Given:

Given expression is $3x^3 + x^2 - 22x + 9$.

To do:

We have to find the expression that has to be added to $3x^3 + x^2 - 22x + 9$ so that the result is exactly divisibly by $3x^2 + 7x - 6$.

Solution:

We know that,

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

Let $f(x)=3x^3 + x^2 - 22x + 9$ and $g(x)=3x^{2}+7x-6$.

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,

$3x^3 + x^2 - 22x + 9+a x+b$ is divisible by $g(x)$

$g(x)=3 x^{2}+7 x-6$

$=3 x^{2}+9 x-2 x-6$

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

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

$x+3$ and $3x-2$ are factors of $3x^3 + x^2 - 22x + 9+a x+b$.

$p(-3)=3(-3)^{3}+(-3)^{2}-22(-3)+9+a(-3)+b=0$

$-81+9+66+9-3 a+b=0$

$-81+84-3 a+b=0$

$3-3 a+b=0$

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

$3 x-2=0$

$3 x=2$

$x=\frac{2}{3}$

$p(\frac{2}{3})=3(\frac{2}{3})^{3}+(\frac{2}{3})^{2}-22(\frac{2}{3})+9+a(\frac{2}{3})+b=0$

$3 \times \frac{8}{27}+\frac{4}{9}-\frac{44}{3}+9+\frac{2}{3} a+b=0$

$\frac{8}{9}+\frac{4}{9}-\frac{44}{3}+\frac{9}{1}+\frac{2}{3} a+b=0$

$\frac{8+4-132+81+6 a+9 b}{9}=0$

$\frac{93-132+6 a+9 b}{9}=0$

$\frac{-39+6 a+9 b}{9}=0$

$-39+6 a+9 b=0$

$6 a+9 (3a-3)=39$                   [From (i)]

$6a+27a-27=39$

$33a=39+27$

$33a=66$

$a=2$

$b=3(2)-3$

$=6-3$

$=3$

Therefore, $2x+3$ must be added to $3x^3 + x^2 - 22x + 9$ so that the result is exactly divisibly by $3x^2 + 7x - 6$.

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Updated on: 10-Oct-2022

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