# How do you factor given that f(1)=0 and f(x)=4x^3-4x^2-9x+9?

Sep 1, 2016

$4 {x}^{3} - 4 {x}^{2} - 9 x + 9 = \left(x - 1\right) \left(2 x + 3\right) \left(2 x - 3\right)$

#### Explanation:

As in $f \left(x\right) = 4 {x}^{3} - 4 {x}^{2} - 9 x + 9$, $f \left(1\right) = 0$, one of the factors is $\left(x - 1\right)$.

Now dividing $4 {x}^{3} - 4 {x}^{2} - 9 x + 9$ by $\left(x - 1\right)$, as

$4 {x}^{3} - 4 {x}^{2} - 9 x + 9$

= $4 {x}^{2} \left(x - 1\right) - 9 \left(x - 1\right)$

= $\left(x - 1\right) \left(4 {x}^{2} - 9\right)$

= $\left(x - 1\right) \left({\left(2 x\right)}^{2} - {3}^{2}\right)$

Now using identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, the above can be factorized as

$\left(x - 1\right) \left(2 x + 3\right) \left(2 x - 3\right)$