# How do you find all the zeros of f(x)=4x^3-12x^2-x+3?

Aug 9, 2016

$f \left(x\right)$ has zeros: $\frac{1}{2}$, $- \frac{1}{2}$, $3$

#### Explanation:

Notice that the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

$f \left(x\right) = 4 {x}^{3} - 12 {x}^{2} - x + 3$

$= \left(4 {x}^{3} - 12 {x}^{2}\right) - \left(x - 3\right)$

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

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

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

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

Hence the zeros of $f \left(x\right)$ are: $\frac{1}{2}$, $- \frac{1}{2}$, $3$