Solving Polynomial Equations

We want to find the zeros and sketch the graph of the given polynomial function. $$\begin{array}{c}f(x)=x^4+x^3-6x^2\end{array}$$ Let's do these things one at a time.

Zeros of the Function

To find the zeros, we need to find the values of $x$ for which $f(x)=0.$ $$\begin{array}{c}f(x)=0\iff x^4+x^3-6x^2=0\end{array}$$ Since the function is not written in factored form, we will begin by factoring the equation. Now, we can apply the Zero Product Property. We found that the zeros of the function are $0,$ $\text{-}3,$ and $2.$

Graph

To draw the graph of the function, we will find some additional points and consider the end behavior. Let's use a table to find additional points.

$x$	$x^4+x^3-6x^2$	$f(x)=x^4+x^3-6x^2$
$\text{-}2$	$(\text{-}2{)}^4+(\text{-}2{)}^3-6(\text{-}2{)}^2$	$\text{-}16$
$\text{-}1$	$(\text{-}1{)}^4+(\text{-}1{)}^3-6(\text{-}1{)}^2$	$\text{-}6$
$1$	$1^4+1^3-6(1{)}^2$	$\text{-}4$

The points $(\text{-}2,\text{-}16),$ $(\text{-}1,\text{-}6),$ and $(1,\text{-}4)$ are on the graph of the function. Now, we will determine the leading coefficient and degree of the polynomial function. $$\begin{array}{c}f(x)=x^4+x^3-6x^2\\ \Updownarrow \\ f(x)=1x^4+x^3-6x^2\end{array}$$ We can see now that the leading coefficient is $1,$ which is a positive number. Also, the degree is $4,$ which is an even number. Therefore, the end behavior is up and up. With this in mind, we will plot and connect the zeros and the obtained points to draw the graph.