Question

If $(x^3+a{x}^2-bx+10)$ is divisible by $x^2-3x+2$, find the values of $a$ and $b$

Answer 1

The Factor Theorem states that if $a$ is the root of any polynomial $p\left(x\right)$ that is if $p\left(a\right)=0$, then $(x-a)$ is the factor of the polynomial $p\left(x\right)$.

Let $p\left(x\right)=x^3+a{x}^2-bx+10$ and $g\left(x\right)=x^2-3x+2$

Factorise $g\left(x\right)=x^2-3x+2$:

$x^2-3x+2=x^2-2x-x+2=x(x-2)-1(x-2)=(x-2)(x-1)$

Therefore, $g\left(x\right)=(x-2)(x-1)$

It is given that $p\left(x\right)$ is divisible by $g\left(x\right)$, therefore, by factor theorem $p\left(2\right)=0$ and $p\left(1\right)=0$. Let us first find $p\left(2\right)$ and $p\left(1\right)$ as follows:

$p\left(1\right)=1^3+(a\times 1^2)-(b\times 1)+10=1+(a\times 1)-b+10=a-b+11p\left(2\right)=2^3+(a\times 2^2)-(b\times 2)+10=8+(a\times 4)-2b+10=4a-2b+18$

Now equate $p\left(2\right)=0$ and $p\left(1\right)=0$ as shown below:

$a-b+11=0\Rightarrow a-b=-11.......\left(1\right)4a-2b+18=0\Rightarrow 2(2a-b+9)=0\Rightarrow 2a-b+9=0\Rightarrow 2a-b=-9.......\left(2\right)$

Now subtract equation 1 from equation 2:

$(2a-a)+(-b+b)=(-9+11)$

$\Rightarrow$$a=2$

Substitute $a=2$ in equation 1:

$2-b=-11$

$\Rightarrow$$-b=-11-2$

$\Rightarrow$$-b=-13$

$\Rightarrow$$b=13$

Hence, $a=2$ and $b=13$.