Question

Find the values of a and b so that the polynomial (x³ − 10x² + ax + b) is exactly divisible by (x −1) as well as (x − 2).

Answer 1

Let:
$f\left(x\right)=x^3-10x^2+ax+b$
Now,
$x-1=0\Rightarrow x=1$
By the factor theorem, we can say:
$f\left(x\right)\text{will be exactly divisible by}\left(x-1\right)\text{if}f\left(1\right)=0$.
Thus, we have:
$$\begin{gathered} f\left(1\right)=1^3-10\times 1^2+a\times 1+b \\ =\left(1-10+a+b\right) \\ =-9+a+b \end{gathered}$$
∴ $f\left(1\right)=0\Rightarrow a+b=9...\left(1\right)$
Also,
$x-2=0\Rightarrow x=2$
By the factor theorem, we can say:
$f\left(x\right)\text{will be exactly divisible by}\left(x-2\right)\text{if}f\left(2\right)=0$.
Thus, we have:
$$\begin{gathered} f\left(2\right)=2^3-10\times 2^2+a\times 2+b \\ =\left(8-40+2a+b\right) \\ =-32+2a+b \end{gathered}$$
∴ $f\left(2\right)=0\Rightarrow 2a+b=32...\left(2\right)$

$$\begin{gathered} \mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get}: \\ a=23 \end{gathered}$$
Putting the value of a, we get the value of b, i.e., $-$14.
∴ a = 23 and b = $-$14