Question

How do you find the zeros of $y=x^4-7x^2+12$?

Answer 1

Find the zeros of the given polynomial:

To calculate the zeros of the given polynomial, so put $y=0$ and the equation can be written as:

$x^4-7x^2+12=0$

Assume that $\text{t}=x^2$ and the equation can be written as:

${\text{t}}^2-7\text{t}+12=0$

Now, solve the equation for $\text{t}$:

$\begin{array}{rcl}{\text{t}}^2-4\text{t}-3\text{t}+12& =& 0\left[\text{Factoring the polynomial}\right]\\ & \Rightarrow & \text{t}\left(\text{t}-4\right)-3\left(\text{t}-4\right)=0\\ & \Rightarrow & \left(\text{t}-3\right)\left(\text{t}-4\right)=0\left[\text{Simplifying}\right]\\ & \Rightarrow & \text{t}-3=0\text{or}\text{t}-4=0\\ \therefore \text{t}& =& 3\text{or}\text{t}=4\end{array}$

Since $x^2=\text{t}$, so the value of $x$ can be calculate in below:

$\begin{array}{rcl}{x}^2& =& 3\text{or}{x}^2=4\\ \therefore x& =& \pm \sqrt{3}\text{or}x=\pm 2\end{array}$

Hence, the zeros of the given polynomial are $\sqrt{\mathbf{3}}\mathbf{,}\mathbf{}\mathbf{-}\sqrt{\mathbf{3}}\mathbf{,}\mathbf{}\mathbf{2}$ and $\mathbf{-}\mathbf{2}$.