Question

Find the values of a and b so that the polynomial $(x^3-10x^{2}ax+b)$ 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

⇒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:
$f\left(1\right)=1^3-10\times 1^2+a\times 1+b=1-10+a+b=-9+a+b$
∴ f(1)=0

⇒a+b=9 ...(1)
Also,
x-2=0

⇒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:
$f\left(2\right)=2^3-10\times 2^2+a\times 2+b=8-40+2a+b=-32+2a+b$
∴ f(2)=0

⇒2a+b=32 ...(2)
Subtracting (1) from (2), we get:

a=23
Putting the value of a, we get the value of b, i.e., -14.
∴ a = 23 and b = -14