# How do you find all the zeros of 4x^3-4x^2-9x+9 with 1 as a zero?

##### 1 Answer
May 26, 2016

The 3 roots are $x = - \frac{3}{2} , 1 , \frac{3}{2}$

Note I can't find the long division symbol so I will use the square root symbol in it's place.

#### Explanation:

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

$f \left(1\right) = 4 \cdot {1}^{3} - 4 \cdot {1}^{2} - 9 \cdot 1 + 9 = 4 - 4 - 9 + 9 = 0$

This means that x=1 is a root and $\left(x - 1\right)$ is a factor of this polynomial.

We need to find the other factors, we do this by dividing f(x) by $\left(x - 1\right)$ to find other factors.

$\frac{4 {x}^{3} - 4 {x}^{2} - 9 x + 9}{x - 1}$

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

Since $\left(x \cdot 4 {x}^{2}\right) = 4 {x}^{3}$ we get $4 {x}^{2}$ as a term in the factor

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

we need to find the remainder to find what else need to be found.
we do $4 {x}^{2} \cdot \left(x - 1\right) = 4 {x}^{3} - 4 {x}^{2}$
                                    $4 {x}^{2}$
$\left(x - 1\right) \sqrt{4 {x}^{3} - 4 {x}^{2} - 9 x + 9}$
                              $4 {x}^{3} - 4 {x}^{2}$

We subtract this to get 0

                                    $4 {x}^{2}$
$\left(x - 1\right) \sqrt{4 {x}^{3} - 4 {x}^{2} - 9 x + 9}$
                        $\underline{- \left(4 {x}^{3} - 4 {x}^{2}\right)}$
                                 $0 {x}^{3} - 0 {x}^{2}$

this zero means that there is NO linear term we and bring down the next terms.

                                    $4 {x}^{2} + 0 x$
$\left(x - 1\right) \sqrt{4 {x}^{3} - 4 {x}^{2} - 9 x + 9}$
                        $\underline{- \left(4 {x}^{3} - 4 {x}^{2}\right)}$
                                    $0 {x}^{3} - 0 {x}^{2}$ $- 9 x + 9$

$x \cdot - 9 = - 9 x$ so the next term is $- 9$

                                    $4 {x}^{2} + 0 x - 9$
$\left(x - 1\right) \sqrt{4 {x}^{3} - 4 {x}^{2} - 9 x + 9}$
                        $\underline{- \left(4 {x}^{3} - 4 {x}^{2}\right)}$
                                    $0 {x}^{3} - 0 {x}^{2}$ $- 9 x + 9$

$- 9 \cdot \left(x - 1\right) = - 9 x + 9$

                                    $4 {x}^{2} + 0 x - 9$
$\left(x - 1\right) \sqrt{4 {x}^{3} - 4 {x}^{2} - 9 x + 9}$
                        $\underline{- \left(4 {x}^{3} - 4 {x}^{2}\right)}$
                                    $0 {x}^{3} - 0 {x}^{2}$ $- 9 x + 9$
                                                            $\underline{- \left(- 9 x + 9\right)}$

                                    $4 {x}^{2} + 0 x - 9$
$\left(x - 1\right) \sqrt{4 {x}^{3} - 4 {x}^{2} - 9 x + 9}$
                        $\underline{- \left(4 {x}^{3} - 4 {x}^{2}\right)}$
                                    $0 {x}^{3} - 0 {x}^{2}$ $- 9 x + 9$
                                                            $\underline{- \left(- 9 x + 9\right)}$
                                                                                        $0$

So $\left(x - 1\right) \left(4 {x}^{2} - 9\right)$ with no reminder (you can check this by doing the expansion). We need to factor this completely.

$\left(4 {x}^{2} - 9\right)$ is a difference of squares with factors $\left(2 x - 3\right) \cdot \left(2 x + 3\right)$

We have $\left(x - 1\right) \cdot \left(2 x - 3\right) \cdot \left(2 x + 3\right) = 0$

$2 x - 3 = 0$ gives us a root at $x = \frac{3}{2}$ and $2 x + 3 = 0$ gives us a root at $x = - \frac{3}{2}$

The roots are $x = - \frac{3}{2} , 1 , \frac{3}{2}$