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# Common Core Standards: Math

# Math.CCSS.Math.Content.HSA-APR.B.3

**3. Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.**

In plain English, this standard is all about factoring polynomials and finding their zeros. It's about time too, because it can be frustrating dealing with students who claim that *x*^{3} – 10*x*^{2} – 2*x* + 24 has 1 zero because of 10*x*^{2}. So frustrating that you may or may not want to quit this teaching gig and get a job at the local circus. At least they'd take you more seriously there.

First, your students should know that finding the zeros of a polynomial doesn't mean counting how many zeros they can find. That's not algebra. That's counting.

The **zeros** of a polynomial are the *x* values when we set the polynomial itself to equal zero. In other words, when we plug in any of the zeros of a polynomial in for *x*, our answer should be 0. So the zeros of *x*^{3} – 10*x*^{2} – 2*x* + 24 are the *x* values that make the equation *x*^{3} – 10*x*^{2} – 2*x* + 24 = 0 true.

Why are these zero values important? On the coordinate plane, they're the places where the function crosses the *x*-axis. The zeros of the polynomial (also called the solutions or "roots") are the *x*-intercepts of the graph.

To find the zeros, students should factor polynomials into linear factors (where *x* is only to the first power). Then, setting each linear factor equal to 0 and solving for the variable will give us the *x*-intercepts, which can be used to graph the function.

Students should also have a rough idea of how different polynomials should look when graphed. Linear functions make lines, quadratic functions make parabolas, and so on. They don't have to know exactly which term affects which aspect of the graph, but their answers shouldn't look like a game of connect-the-dots.

For example, we're told to graph the function defined by the polynomial *x*^{2} + 5*x* + 6. Students should feel comfortable factoring this into a product of two linear factors: (*x* + 2)(*x* + 3). Solving the equation *x*^{2} + 5*x* + 6 = 0 would be difficult, but (*x* + 2)(*x* + 3) = 0 is much easier.

The entire polynomial equals 0 when each individual factor equals 0. That gives us *x* + 2 = 0 and *x* + 3 = 0, or *x* = -2 and *x* = -3. So our *x*-intercepts are at *x* = -2 and *x* = -3. Knowing these intercepts makes it easier to graph the function.

Make sure to give your students polynomials they can factor. Giving them something like *x*^{5} + 9*x*^{4} – 17*x*^{3} + 2*x*^{2} + 1 is just cruel if they've only learned how to factor quadratic equations.