# How do you find all zeros of f(x) = 2x^4 - 5x^3 +3x^2 + 4x - 6?

Jul 30, 2015

$\textcolor{red}{x = - 1 , x = \frac{3}{2} , x = 1 - i , x = 1 + i}$

#### Explanation:

You use the rational root theorem and the quadratic formula to find the roots.

$f \left(x\right) = 2 {x}^{4} - 5 {x}^{3} + 3 {x}^{2} + 4 x - 6$

According to the rational root theorem, the rational roots of $f \left(x\right) = 0$ must all be of the form $\frac{p}{q}$, with $p$ a divisor of the constant term $- 6$ and $q$ a divisor of the coefficient $2$ of ${x}^{4}$.

So the only possible rational roots are:

±1, ±2, ±3, ±6, ±1/2, ±3/2

By trial and error, we find:

$f \left(- 1\right) = 2 + 5 + 3 - 4 - 6 = 0$

$f \left(\frac{3}{2}\right) = 2 {\left(\frac{3}{2}\right)}^{4} - 5 {\left(\frac{3}{2}\right)}^{3} + 3 {\left(\frac{3}{3}\right)}^{2} + 4 \left(\frac{3}{2}\right) - 6$

$= 2 \left(\frac{81}{16}\right) - 5 \left(\frac{27}{8}\right) + 3 \left(\frac{9}{4}\right) + 4 \left(\frac{3}{2}\right) - 6 = \frac{81}{8} - \frac{135}{8} + \frac{27}{4} + \frac{12}{2} - 6 = \frac{81 - 135 + 54 + 48 - 48}{8} = 0$

So $x = - 1$ and $x = \frac{3}{2}$ are roots of $f \left(x\right) = 0$, and $\left(x + 1\right)$ and $\left(x - \frac{3}{2}\right)$ are factors of $f \left(x\right)$.

Divide $f \left(x\right)$ by $\left(x + 1\right) \left(x - \frac{3}{2}\right) = {x}^{2} + \frac{5}{2} x + \frac{3}{2}$ to find:

$\frac{2 {x}^{4} - 5 {x}^{3} + 3 {x}^{2} + 4 x - 6}{{x}^{2} + \frac{5}{2} x + \frac{3}{2}} = 2 {x}^{2} - 4 x + 4 = 2 \left({x}^{2} - 2 x + 2\right)$

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

Using the quadratic formula, we find that the roots of ${x}^{2} - 2 x + 2 = 0$ are:

x = (-(-2)±sqrt((-2)^2-4×1×2))/(2×1) = (2±sqrt(4-8))/2 =(2±sqrt(-4))/2 = (2±2i)/2 = 1±i

The zeroes of $f \left(x\right) = 2 {x}^{4} - 5 {x}^{3} + 3 {x}^{2} + 4 x - 6$ are $x = - 1 , x = \frac{3}{2} , x = 1 - i$, and $x = 1 + i$.