**Identifying Zeroes of Polynomials **

As you might remember from a previous math study tips post for the TASC Test Assessing Secondary Completion™, a **polynomial **is an algebraic expression of more than one term, constructed from variables and constants using only the operations of addition, subtraction, multiplication and non-negative whole-number exponents.

In more simple terms, a polynomial is a math expression that has many terms. These terms may be numbers, variables, and operators.

However, polynomials can be complicated – especially if you need to identify zeroes in the polynomial. There is a process that can help you find all rational zeroes of a polynomial that is of a degree greater than two, but Elizabeth Stapel of *PurpleMath* warns that it can be time-consuming.

At McGraw-Hill Education CTB, we know that the Mathematics subtest can seem daunting. There are many formulas you need to know, and some of them can be difficult or time-consuming. Learn the formula for identifying zeroes of polynomials in our walk-through of this example:

**An Example**

The polynomial function we will be working through is:

f(x) = x^{3} + 5x^{2} – 9x – 45

According to Emily Johnson of Chesapeake College, the first step is to group the terms and factor out a common factor:

f(x) = x^{3} + 5x^{2} – 9x – 45 = 0

(x^{3} + 5x^{2}) + (–9x – 45) = 0

At this point, we have set the polynomial to zero, and have separated the long equation into two equal parts using parentheses. When you have an equation that is separated at the subtraction sign like this one, that sign goes with the number or variable and becomes a negative. The equations are always being added.

x^{2} (x + 5) – 9(x – 5) = 0

Factor out common variables. For example, x^{2} is factored from x^{3} (leaving x) and from 5x^{2} (leaving 5). In the second parenthetical, (-9) is factored from both variables.

(x^{2} – 9)(x + 5) = 0

Here, we’ve factored the common factor of (x + 5) from the expression. What we have left after we have factored out the common factor are two factors: (x^{2} – 9) and (x + 5). Our next step is to set these factors to zero, and solve.

x^{2} – 9 = 0

9 + (x^{2} – 9) = (0) + 9

x^{2} = 9

√x^{2} = ±√9

x = ±3

The ± sign stands for “positive or negative,” and is simply an addition sign sitting on top of a subtraction sign. This means that the answer, 3, could be positive or negative in this instance.

x + 5 = 0

5 - (x + 5) = (0) – 5

x = -5

The real zeros of this polynomial function are (-5), (-3), and 3.

The next step is to determine the multiplicity for each zero. To do this, we look at each factor to see how many times it occurs.

f(x) = x^{3} + 5x^{2} – 9x – 45 =(x^{2 }– 9)(x + 5) = (x – 3)(x + 3)(x + 5)

As Johnson concludes, each factor only occurs once. Therefore, each has a multiplicity of 1. This can help you determine how to use the **multiplicity rule** to determine a graph of this polynomial function because we know that the graph will cross the x-axis at (-5), (-3), and 3.