Find the values of a and b so that the polynomial

Solution:

Let:

$f(x)=x^{3}-10 x^{2}+a x+b$

Now,

$x-1=0 \Rightarrow x=1$

By the factor theorem, we can say:

$f(x)$ will be exactly divisible by $(x-1)$ if $f(1)=0$.

Thus, we have:

$f(1)=1^{3}-10 \times 1^{2}+a \times 1+b$

$=(1-10+a+b)$

$=-9+a+b$

$\therefore f(1)=0 \Rightarrow a+b=9 \quad \ldots(1)$

Also,

$x-2=0 \Rightarrow x=2$

By the factor theorem, we can say:

$f(x)$ will be exactly divisible by $(x-2)$ if $f(2)=0$.

Thus, we have:

$f(2)=2^{3}-10 \times 2^{2}+a \times 2+b$

$=(8-40+2 a+b)$

$=-32+2 a+b$

$\therefore f(2)=0 \Rightarrow 2 a+b=32 \quad \ldots(2)$

Subtracting (1) from (2), we get:

$a=23$

Putting the value of $a$, we get the value of $b$, i.e., $-14$.

$\therefore a=23$ and $b=-14$