Find the values of a and b so that the polynomial (x3−10x2ax+b) is exactly divisible by (x -1) as well as (x -2).
Let:
f(x)=x3−10x2+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(1)=13−10×12+a×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(2)=23−10×22+a×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