# How do you solve 4x^3-12x^2-3x+9=0?

May 6, 2018

$x = 3 , - \frac{\sqrt{3}}{2}$ or $\frac{\sqrt{3}}{2}$

#### Explanation:

According to factor theorem , if $f \left(a\right) = 0$, then (x–a) is a factor of the polynomial $f \left(x\right)$. Converse of this theorem is also true i.e. if $\left(x - a\right)$ is a factor of the polynomial $f \left(x\right)$, then $f \left(a\right) = 0$.

Here for $x = 3$, $f \left(x\right) = 4 {x}^{3} - 12 {x}^{2} - 3 x + 9$, we have

$f \left(3\right) = 4 \cdot {3}^{3} - 12 \cdot {3}^{2} - 3 \cdot 3 + 9 = 108 - 108 - 9 + 9 = 0$

Hence $x - 3$ is a factor of $4 {x}^{3} - 12 {x}^{2} - 3 x + 9$ and hence

$4 {x}^{3} - 12 {x}^{2} - 3 x + 9 = 0$

$\implies 4 {x}^{2} \left(x - 3\right) - 3 \left(x - 3\right) = 0$

or $\left(x - 3\right) \left(4 {x}^{2} - 3\right) = 0$

i.e. $\left(x - 3\right) \left(2 x + \sqrt{3}\right) \left(2 x - \sqrt{3}\right) = 0$

Hence $x = 3 , - \frac{\sqrt{3}}{2}$ or $\frac{\sqrt{3}}{2}$