# How do you find all the zeros of f(x)=4x^3-20x^2-3x+15?

Feb 27, 2016

Factor by grouping and by using the difference of squares identity to find:

$f \left(x\right) = \left(2 x - \sqrt{3}\right) \left(2 x + \sqrt{3}\right) \left(x - 5\right)$

hence has zeros $x = \pm \frac{\sqrt{3}}{2}$ and $x = 5$

#### Explanation:

Factor by grouping, then use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = 2 x$ and $b = \sqrt{3}$, as follows:

$f \left(x\right) = 4 {x}^{3} - 20 {x}^{2} - 3 x + 15$

$= \left(4 {x}^{3} - 20 {x}^{2}\right) - \left(3 x - 15\right)$

$= 4 {x}^{2} \left(x - 5\right) - 3 \left(x - 5\right)$

$= \left(4 {x}^{2} - 3\right) \left(x - 5\right)$

$= \left({\left(2 x\right)}^{2} - {\left(\sqrt{3}\right)}^{2}\right) \left(x - 5\right)$

$= \left(2 x - \sqrt{3}\right) \left(2 x + \sqrt{3}\right) \left(x - 5\right)$

So the zeros of $f \left(x\right)$ are $x = \pm \frac{\sqrt{3}}{2}$ and $x = 5$