Question

If x^2-3x+2 divides x^3-6x^2+ax+b exactly, then find the value of 'a' and 'b'.

Answer 1

Let g(x) be (ax^4 + 2x^3 - 3x^2 + bx) 
and f(x) be (x^2 - 4).

If f(x) is a factor of g(x), then the solution to f(x) should also be the solution to g(x). 
This is because if g(x) is a factor of f(x), then g(x) can be written as f(x) * h(x) (where h(x) is some other factor of g(x)) and when f(x) becomes zero, then g(x) also evaluates to zero.

Now solution to f(x): 
x^2 - 4 = 0;
Using (a^2 - b^2) = (a+b)(a-b), we get
(x+2)(x-2)=0;
So x is either 2 or -2.

Now g(2) = a(2^4) + 2(2^3) - 3(2^2) + b(2) = 0
16a + 16 - 12 + 2b = 0
16a + 2b + 4 =0
8a + b + 2 = 0 - (Equation 1)

And g(-2) = a(-2^4) + 2(-2^3) - 3(-2^2) + b(-2) = 0
16a - 16 - 12 - 2b = 0
16a - 2b - 28 =0
8a - b - 14 = 0 - (Equation 2)

Adding both the equations you get:
16a - 12 = 0
a= 12/16 = 3/4

Substituting the value of a in Equation 1 you get:
8(3/4) + b + 2 =0
6 + b + 2 =0
b = -8

Thus the answer is a= 3/4 and b = -8.

Verification: 

If you divide (ax^4 + 2x^3 - 3x^2 + bx) by (x^2 - 4), you get quotient as ((3/4)x^2 + 2x) and remainder as 0, thus proving that g(x) is a factor of f(x) and we can say hence verified! :)