3.4 Graphs of polynomial functions  (Page 2/13)

We can attempt to factor this polynomial to find solutions for $f\left(x\right)=0.$

This gives us five $x\text{-}$ intercepts: $(0,0),(1,0),(-1,0),(\sqrt{2},0),$ and $(-\sqrt{2},0).$ See [link] . We can see that this is an even function.

Find solutions for $f(x)=0$ by factoring.

There are three $x\text{-}$ intercepts: $(-1,0),(1,0),$ and $\left(5,0\right).$ See [link] .

The y -intercept can be found by evaluating $g\left(0\right).$

So the y -intercept is $(0,12).$

The x -intercepts can be found by solving $g\left(x\right)=0.$

So the $x\text{-}$ intercepts are $(2,0)$ and $\left(-\frac{3}{2},0\right).$

This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.

Looking at the graph of this function, as shown in [link] , it appears that there are x -intercepts at $x=\mathrm{-3},\mathrm{-2},$ and $1.$

We can check whether these are correct by substituting these values for $x$ and verifying that

Since $h(x)=x^3+4x^2+x-6,$ we have:

Each $x\text{-}$ intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form.