# What are the zeros of the function? What are their multiplicities? f(x)=x^4-4x+3x^2

Nov 25, 2017

#### Answer:

$x = 0 , x = 1 \text{ and } x = - \frac{1}{2} \pm \frac{\sqrt{15}}{2} i$

#### Explanation:

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

$\textcolor{w h i t e}{f \left(x\right)} = x \left({x}^{3} + 3 x - 4\right)$

$\text{the coefficients of } {x}^{3} + 3 x - 4$

$1 + 3 - 4 = 0 \Rightarrow x = 1 \text{ is a root and "(x-1)" is a factor}$

$\textcolor{red}{{x}^{2}} \left(x - 1\right) \textcolor{m a \ge n t a}{+ {x}^{2}} + 3 x - 4$

$= \textcolor{red}{{x}^{2}} \left(x - 1\right) \textcolor{red}{+ x} \left(x - 1\right) \textcolor{m a \ge n t a}{+ x} + 3 x - 4$

$= \textcolor{red}{{x}^{2}} \left(x - 1\right) \textcolor{red}{+ x} \left(x - 1\right) \textcolor{red}{+ 4} \left(x - 1\right) \cancel{\textcolor{m a \ge n t a}{+ 4}} \cancel{- 4}$

$= \textcolor{red}{{x}^{2}} \left(x - 1\right) \textcolor{red}{+ x} \left(x - 1\right) \textcolor{red}{+ 4} \left(x - 1\right) + 0$

$\Rightarrow x \left(x - 1\right) \left({x}^{2} + x + 4\right) = 0$

$\Rightarrow x = 0 , x = 1 \to \left(\text{with multiplicity of 1}\right)$

$\text{check the "color(blue)"discriminant "" of } {x}^{2} + x + 4$

$\text{with } a = 1 , b = 1 , c = 4$

$\Delta = {b}^{2} - 4 a c = 1 - 16 = - 15$

$\text{indicating there are 2 complex roots}$

$\text{using the "color(blue)"quadratic formula}$

$x = \frac{- 1 \pm \sqrt{- 15}}{2}$

$\Rightarrow x = - \frac{1}{2} \pm \frac{\sqrt{15}}{2} i$

$\text{there are 2 real zeros and 2 complex zeros}$