# How do you find all the zeros of f(x)=4x^3-4x^2-7x-2?

Jul 10, 2016

$f \left(x\right)$ has zeros: $x = - \frac{1}{2}$ with multiplicity $2$ and $x = 2$.

#### Explanation:

$f \left(x\right) = 4 {x}^{3} - 4 {x}^{2} - 7 x - 2$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 2$ and $q$ a divisor of the coefficient $4$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm 1 , \pm 2$

We can quickly see that $f \left(1\right)$ and $f \left(- 1\right)$ cannot be zero, since $7$ is odd and all the other coefficients are even.

We find:

$f \left(2\right) = 32 - 16 - 14 - 2 = 0$

So $x = 2$ is a zero and $\left(x - 2\right)$ a factor:

$4 {x}^{3} - 4 {x}^{2} - 7 x - 2 = \left(x - 2\right) \left(4 {x}^{2} + 4 x + 1\right)$

$\textcolor{w h i t e}{}$
To find the remaining zeros I could just try the other rational values we identified and factor by the next one we find, but there's a shortcut:

Do you see the pattern $4 , 4 , 1$ of the coefficients of the remaining quadratic?

Did you know that $441 = {21}^{2}$ ?

As a result, we can spot that $4 {x}^{2} + 4 x + 1$ is a perfect square trinomial:

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

This shows us that $4 {x}^{2} + 4 x + 1$ has a repeated zero $x = - \frac{1}{2}$

graph{4x^3-4x^2-7x-2 [-4.877, 5.123, -2.4, 2.6]}