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Study Guides > Intermediate Algebra

Read: Solve Polynomial Equations

Learning Objectives

  • Use factoring methods to factor polynomial equations
  • Use the principle of zero products to solve polynomial equations
Not all of the techniques we use for solving linear equations will apply to solving polynomial equations. In this section we will introduce a method for solving polynomial equations that combines factoring and the zero product principle.

The Principle of Zero Products

The number zero Zero
What if we told you that we multiplied two numbers together and got an answer of zero? What could you say about the two numbers? Could they be [latex]2[/latex] and [latex]5[/latex]? Could they be [latex]9[/latex] and [latex]1[/latex]? No! When the result (answer) from multiplying two numbers is zero, that means that one of them had to be zero. This idea is called the zero product principle, and it is useful for solving polynomial equations that can be factored.  

Principle of Zero Products

The Principle of Zero Products states that if the product of two numbers is [latex]0[/latex], then at least one of the factors is [latex]0[/latex]. If [latex]ab=0[/latex], then either [latex]a=0[/latex] or [latex]b=0[/latex], or both a and b are [latex]0[/latex].
Let's start with a simple example.  We will factor a GCF from a binomial and apply the principle of zero products to solve a polynomial equation.

Example

Solve: [latex-display]-t^2+t=0[/latex-display]

Answer: Each term has a common factor of [latex]t[/latex], so we can factor and use the zero product principle. Rewrite each term as the product of the GCF and the remaining terms. [latex-display]\begin{array}{c}-t^2=t\left(-t\right)\\t=t\left(1\right)\end{array}[/latex-display] Rewrite the polynomial equation using the factored terms in place of the original terms.

[latex]\begin{array}{c}-t^2+t=0\\t\left(-t\right)+t\left(1\right)\\t\left(-t+1\right)=0\end{array}[/latex]

Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the zero product principle.

[latex]\begin{array}{c}t=0\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,\,\,-t+1=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-1}\,\,\,\underline{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-t=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{-t}{-1}=\frac{-1}{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=1\end{array}[/latex]

Answer

[latex-display]t=0\text{ OR }t=1[/latex-display]

In the following video we show two more examples of using both factoring and the principle of zero products to solve a polynomial equation. https://youtu.be/gIwMkTAclw8 In the next video we show that you can factor a trinomial using methods previously learned to solve a quadratic equation. https://youtu.be/bi7i_RuIGl0 You and I both know that it is rare to be given an equation to solve that has zero on one side, so let's try another one.

Example

Solve: [latex]s^2-4s=5[/latex]

Answer: First, move all the terms to one side.  The goal is to try and see if we can use the zero product principle, since that is the only tool we know for solving polynomial equations.

[latex]\begin{array}{c}\,\,\,\,\,\,\,s^2-4s=5\\\,\,\,\,\,\,\,s^2-4s-5=0\\\end{array}[/latex]

We now have all the terms on the left side, and zero on the other side. The polynomial[latex]s^2-4s-5[/latex] factors nicely, which makes this equation a good candidate for the zero product principle. (imagine that)

[latex]\begin{array}{c}s^2-4s-5=0\\\left(s+1\right)\left(s-5\right)=0\end{array}[/latex]

We separate our factors into two linear equations using the principle of zero products.

[latex-display]\begin{array}{c}\left(s-5\right)=0\\s-5=0\\\,\,\,\,\,\,\,\,\,s=5\end{array}[/latex-display] OR [latex-display]\begin{array}{c}\left(s+1\right)=0\\s+1=0\\s=-1\end{array}[/latex-display]

Answer

[latex-display]s=-1\text{ OR }s=5[/latex-display]

  We will work through one more example that is similar to the one above, except this example has fractions, yay!

Example

Solve [latex]y^2-5=-\frac{7}{2}y+\frac{5}{2}[/latex]

Answer: We can solve this in one of two ways.  One way is to eliminate the fractions like you may have done when solving linear equations, and the second is to find a common denominator and factor fractions. Eliminating fractions is easier, so we will show that way. Start by multiplying the whole equation by [latex]2[/latex] to eliminate fractions: [latex-display]\begin{array}{ccc}2\left(y^2-5=-\frac{7}{2}y+\frac{5}{2}\right)\\\,\,\,\,\,\,2(y^2)+2(-5)=2\left(-\frac{7}{2}\right)+2\left(-\frac{5}{2}\right)\\2y^2-10=-7y+5\end{array}[/latex-display] Now we can move all the terms to one side and see if this will factor so we can use the principle of zero products. [latex-display]\begin{array}{c}2y^2-10=-7y-5\\2y^2-10+7y-5=0\\2y^2-15+7y=0\\2y^2+7y-15=0\end{array}[/latex-display] We can now check whether this polynomial will factor, using a table we can list factors until we find two numbers with a product of [latex]2\cdot-15=-30[/latex] and a sum of 7.

Factors of [latex]2\cdot-15=-30[/latex] Sum of Factors
[latex]1,-30[/latex] [latex]-29[/latex]
[latex]-1,30[/latex] [latex]29[/latex]
[latex]2,-15[/latex] [latex]-13[/latex]
[latex]-2,15[/latex] [latex]13[/latex]
[latex]3,-10[/latex] [latex]-7[/latex]
[latex]-3,10[/latex] [latex]7[/latex]
We have found the factors that will produce the middle term we want,[latex]-3,10[/latex]. We need to place the factors in a way that will lead to a term of [latex]10y[/latex]: [latex-display]\left(2y-3\right)\left(y+5\right)=0[/latex-display] Now we can set each factor equal to zero and solve: [latex-display]\begin{array}{ccc}\left(2y-3\right)=0\text{ OR }\left(y+5\right)=0\\2y=3\text{ OR }y=-5\\y=\frac{3}{2}\text{ OR }y=-5\end{array}[/latex-display] You can always check your work to make sure your solutions are correct: Check [latex]y=\frac{3}{2}[/latex] [latex-display]\begin{array}{ccc}\left(\frac{3}{2}\right)^2-5=-\frac{7}{2}\left(\frac{3}{2}\right)+\frac{5}{2}\\\frac{9}{4}-5=-\frac{21}{4}+\frac{5}{2}\\\text{ common denominator = 4}\\\frac{9}{4}-\frac{20}{4}=-\frac{21}{4}+\frac{10}{4}\\-\frac{11}{4}=-\frac{11}{4}\end{array}[/latex-display] [latex-display]y=\frac{3}{2}[/latex] is indeed a solution, now check [latex]y=-5[/latex-display] [latex-display]\begin{array}{ccc}\left(-5\right)^2-5=-\frac{7}{2}\left(-5\right)+\frac{5}{2}\\25-5=\frac{35}{2}+\frac{5}{2}\\20=\frac{40}{2}\\20=20\end{array}[/latex-display] [latex]y=-5[/latex] is also a solution, so we must have done something right!

 Answer

[latex]y=\frac{3}{2}\text{ OR }y=-5[/latex]

In our last video, we show how to solve another quadratic equation that contains fractions. https://youtu.be/kDj_qdKW-ls

Licenses & Attributions

CC licensed content, Original

  • Solve a Quadratic Equations with Fractions by Factoring (a not 1). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.

CC licensed content, Shared previously

  • Ex: Factor and Solve Quadratic Equation - Greatest Common Factor Only. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex: Factor and Solve Quadratic Equations When A equals 1. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Unit 12: Factoring, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology Located at: https://www.nroc.org/. License: CC BY: Attribution.