f(x)=4x^3-4x^2-9x+9
f(1)=4*1^3-4*1^2-9*1+9=4-4-9+9=0
This means that x=1 is a root and (x-1) is a factor of this polynomial.
We need to find the other factors, we do this by dividing f(x) by (x-1) to find other factors.
{4x^3-4x^2-9x+9}/{x-1}
(x-1)sqrt(4x^3-4x^2-9x+9)
Since (x*4x^2)=4x^3 we get 4x^2 as a term in the factor
4x^2
(x-1)sqrt(4x^3-4x^2-9x+9)
we need to find the remainder to find what else need to be found.
we do 4x^2*(x-1)=4x^3-4x^2
4x^2
(x-1)sqrt(4x^3-4x^2-9x+9)
4x^3-4x^2
We subtract this to get 0
4x^2
(x-1)sqrt(4x^3-4x^2-9x+9)
ul {- (4x^3-4x^2)}
0x^3-0x^2
this zero means that there is NO linear term we and bring down the next terms.
4x^2 + 0 x
(x-1)sqrt(4x^3-4x^2-9x+9)
ul {- (4x^3-4x^2)}
0x^3-0x^2 -9x+9
x*-9=-9x so the next term is -9
4x^2 + 0 x -9
(x-1)sqrt(4x^3-4x^2-9x+9)
ul {- (4x^3-4x^2)}
0x^3-0x^2 -9x+9
-9*(x-1)=-9x + 9
4x^2 + 0 x -9
(x-1)sqrt(4x^3-4x^2-9x+9)
ul {- (4x^3-4x^2)}
0x^3-0x^2 -9x+9
ul{-(-9x + 9)}
4x^2 + 0 x -9
(x-1)sqrt(4x^3-4x^2-9x+9)
ul {- (4x^3-4x^2)}
0x^3-0x^2 -9x+9
ul{-(-9x + 9)}
0
So (x-1)(4x^2 -9) with no reminder (you can check this by doing the expansion). We need to factor this completely.
(4x^2 -9) is a difference of squares with factors (2x-3)*(2x+3)
We have (x-1) * (2x-3) * (2x+3)=0
2x-3=0 gives us a root at x=3/2 and 2x+3=0 gives us a root at x=-3/2
The roots are x=-3/2, 1, 3/2
