Model and solve problems using the zeros of a polynomial function.

Assignment

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  • p168 6–12, 15–30

Using Zeros to Graph Functions

After you’ve found the zeros of a polynomial, you can next determine what the function is doing in between those zeros. The polynomial $f(x)=x(x-4)(x+3)$ is already factored, so we can see its zeros are $0$, $4$, and $-3$. This means you have four intervals. Substituting a value from each interval and then checking the resulting sign will tell you if that portion of the graph is above or below the $x$-axis.

For the polynomial above, the values $1$, $-1$, one million and negative one million work well for quickly arriving at what sign is produced.

Multiplicity of Zeros

Sometimes a zero occurs multiple times, like in the case of $f(x)=(x-1)(x+2)^2$. Written without the exponent we would have $f(x) = (x-1)(x+2)(x+2)$, so our zeros are $1$, $-2$, and $-2$ again.

Graphically, this has impact as a zero with an even multiplicity will not cross the $x$-axis. It instead will “bounce” off of it and not change signs. You can use this to your advantage when sketching a graph. If you know one interval, you automatically know the rest without the need for testing.

The graph of $f(x)=(x+2)^2(x-1)^3$ has two zeros at $-2$ and $1$, meaning three intervals. I’ll plug in a massively large positive number to determine the sign there and get a positive. Working to the left, our first zero is $1$ which has an odd multiplicity, meaning the graph crosses there. So, the next interval must be the opposite, which would be negative. The next zero has an even multiplicity, meaning it doesn’t cross. So, the sign stays negative. That means, in left-to-right increasing order, our intervals are negative, negative, positive, and the graph confirms that.

Finding Zeros

Assuming no calculator, your first step would be to factor as much as possible. Keep in mind that the goal is degree one terms. The function $f(x)=2x(x-1)(x+3)^2$ has all degree one terms, so the zeros are easy to spot ($0$, $1$ and $-3$ in this case).

Some of the problems you’ll see also give you a graph that shows a potential zero. Your potential zero $k$ is associated with the binomial $x-k$. If you divide by that binomial (I recommend synthetic to speed things up) a remainder of zero will tell you that it actually is a zero, and not just something that looks like a zero.

Also, don’t forget that once you get your polynomial down to a quadratic, then you can use the quadratic formula.

For a quick example, let’s look at $f(x) = x^3 + x^2 - 3x - 6$. This can’t be factored with what we know, but a graph comes with this (not shown) and it looks like there is a zero at $x=2$. You can use long division and divide the whole thing by $x-2$, but I will use synthetic.

\[\begin{array}{cc} \begin{array}{c} \\ 2 \\ \\ \end{array} & \begin{array}{|rrrr} 1 & 1 & -3 & -6 \\ & 2 & 6 & 6 \\ \hline 1 & 3 & 3 & 0 \end{array} \end{array}\]

We get a remainder of zero, so $x-2$ is a factor meaning we can rewrite our polynomial as

\[f(x)=(x-2)(x^2+3x+3x)\]

Still can’t factor with what we know, but it is a quadratic. Using the quadratic formula we get our two remaining zeros.

\[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-3\pm\sqrt{3^2-4(1)(3)}}{2(1)} = -\frac{3}{2} \pm \frac{\sqrt{3}}{2}i\]

Complex zeros, but that’s fine. Our final list of zeros includes $2$, $-\frac{3}{2} - \frac{\sqrt{3}}{2}i$, and $-\frac{3}{2} + \frac{\sqrt{3}}{2}i$. Since only one of those zeros is a real number, teh graph of our polynomial only crosses the $x$-axis once at $x=2$.

Solving Polynomial Equations and Inequalities

Given an equation or an inequality involving polynomials, combine like terms so that one side of the equation is zero. Then you can work with the one polynomial, factoring and dividing as needed.

With inequalities, you’ll need a sign chart to determine where the polynomial will be either above or below zero.