Question

find the value of M and n so that polynomial f(x)=x^3-6x^2+mx-n is exactly divisible by (x-1),(x-2)

Answer 1

(x-1) & (x-2) are the factors of f (x)

so we can substitute the the value of x as x=1 or x=2.

firstly x=1

x^3-6x^2+mx-n=0

(1)^3-6 (1)^2+m (1)-n=0

1-6+m-n=0

m-n=5. ...(1)

Now x=2

x^3-6x^2+mx-n=0

(2)^3-6 (2)^2+m (2)-n=0

8-24+2m-n=0

2m-n=16. ... (2)

By subtracting equation (1) from (2) we get

m=11

n=6

value of m&n are 11&6 respectively