<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Zero Product Principle

## The zero product principle states that anything multiplied by zero is zero.

Estimated10 minsto complete
%
Progress
Practice Zero Product Principle
Progress
Estimated10 minsto complete
%
Zero Product Principle

### Zero Product Principle

The most useful thing about factoring is that we can use it to help solve polynomial equations.

#### Solving for X

Consider an equation like \begin{align*}2x^2 + 5x - 42 = 0\end{align*}. How do you solve for \begin{align*}x\end{align*}?

There’s no good way to isolate \begin{align*}x\end{align*} in this equation, so we can’t solve it using any of the techniques we’ve already learned. But the left-hand side of the equation can be factored, making the equation \begin{align*}(x + 6)(2x - 7)=0\end{align*}.

How is this helpful? The answer lies in a useful property of multiplication: if two numbers multiply to zero, then at least one of those numbers must be zero. This is called the Zero-Product Property.

What does this mean for our polynomial equation? Since the product equals 0, then at least one of the factors on the left-hand side must equal zero. So we can find the two solutions by setting each factor equal to zero and solving each equation separately.

Setting the factors equal to zero gives us:

\begin{align*}(x + 6) = 0 && \text{OR} && (2x - 7) = 0\end{align*}

Solving both of those equations gives us:

\begin{align*}& x + 6 = 0 && && 2x - 7 =0\\ & \underline{\underline{x = -6}} && \text{OR} && 2x = 7\\ & && && \underline{\underline{x = \frac{7}{2}}}\end{align*}

Notice that the solution is \begin{align*}x = -6\end{align*} OR \begin{align*}x = \frac{7}{2}\end{align*}. The OR means that either of these values of \begin{align*}x\end{align*} would make the product of the two factors equal to zero. Let’s plug the solutions back into the equation and check that this is correct.

\begin{align*}& Check: x = - 6; && Check: x = \frac{7}{2}\\ & (x + 6)(2x - 7)= && (x + 6)(2x - 7)=\\ & (-6 +6)(2(-6) -7)= && \left ( \frac{7}{2} + 6 \right ) \left (2 \cdot \frac{7}{2} - 7 \right ) =\\ & (0)(-19) = 0 && \left (\frac{19}{2} \right ) (7 - 7) = \\ & && \left (\frac{19}{2} \right ) (0) = 0\end{align*}

Both solutions check out.

Factoring a polynomial is very useful because the Zero-Product Property allows us to break up the problem into simpler separate steps. When we can’t factor a polynomial, the problem becomes harder and we must use other methods that you will learn later.

As a last note in this section, keep in mind that the Zero-Product Property only works when a product equals zero. For example, if you multiplied two numbers and the answer was nine, that wouldn’t mean that one or both of the numbers must be nine. In order to use the property, the factored polynomial must be equal to zero.

#### Solving for an Unknown Value

Solve each equation:

Since all the polynomials are in factored form, we can just set each factor equal to zero and solve the simpler equations separately

a) \begin{align*}(x - 9)(3x + 4)=0\end{align*}

\begin{align*}(x - 9)(3x + 4) = 0\end{align*} can be split up into two linear equations:

\begin{align*}& x - 9 = 0 && && 3x + 4 = 0\\ & \underline{\underline{x = 9}} && \text{or} && 3x = -4\\ & && && \underline{\underline{x = - \frac{4}{3}}}\end{align*}

b) \begin{align*}x(5x - 4) = 0\end{align*}

\begin{align*}x(5x - 4) = 0\end{align*} can be split up into two linear equations:

\begin{align*}& && && 5x - 4 = 0\\ & \underline{\underline{x = 0}} && \text{or} && 5x = 4\\ & && && \underline{\underline{x = \frac{4}{5}}}\end{align*}

c) \begin{align*}4x (x+6)(4x - 9)=0\end{align*}

\begin{align*}4x(x + 6)(4x - 9) =0\end{align*} can be split up into three linear equations:

\begin{align*}& 4x = 0 && && x + 6 = 0 && && 4x - 9 =0\\ & x= \frac{0}{4} && \text{or} && x = -6 && \text{or} && 4x = 9\\ & \underline{\underline{x = 0}} && && && && \underline{\underline{x = \frac{9}{4}}}\end{align*}

#### Solve Simple Polynomial Equations by Factoring

Now that we know the basics of factoring, we can solve some simple polynomial equations. We already saw how we can use the Zero-Product Property to solve polynomials in factored form—now we can use that knowledge to solve polynomials by factoring them first. Here are the steps:

a) If necessary, rewrite the equation in standard form so that the right-hand side equals zero.

b) Factor the polynomial completely.

c) Use the zero-product rule to set each factor equal to zero.

d) Solve each equation from step 3.

#### Solving Polynomial Equations

Solve the following polynomial equations.

a) \begin{align*}x^2 - 2x =0\end{align*}

\begin{align*}x^2 - 2x = 0\end{align*}

Rewrite: this is not necessary since the equation is in the correct form.

Factor: The common factor is \begin{align*}x\end{align*}, so this factors as \begin{align*}x(x-2)=0\end{align*}.

Set each factor equal to zero:

\begin{align*}x = 0 && \text{or} && x - 2 = 0\end{align*}

Solve:

\begin{align*}\underline{x = 0} && \text{or} && \underline{x = 2}\end{align*}

Check: Substitute each solution back into the original equation.

\begin{align*}x & = 0 \Rightarrow (0)^2 - 2(0) = 0 && \text{works out}\\ x & = 2 \Rightarrow (2)^2 - 2(2) = 4 - 4 = 0 && \text{works out}\end{align*}

Answer: \begin{align*}x = 0, x = 2\end{align*}

b) \begin{align*}2x^2 = 5x\end{align*}

\begin{align*}2x^2 = 5x\end{align*}

Rewrite: \begin{align*}2x^2 = 5x \Rightarrow 2x^2 - 5x = 0\end{align*}

Factor: The common factor is \begin{align*}x\end{align*}, so this factors as \begin{align*}x(2x - 5)=0\end{align*}.

Set each factor equal to zero:

\begin{align*}x = 0 && \text{or} && 2x - 5 = 0\end{align*}

Solve:

\begin{align*}& \underline{x = 0} && \text{or} && 2x = 5\\ & &&&& \underline{x = \frac{5}{2}}\end{align*}

Check: Substitute each solution back into the original equation.

\begin{align*}x & = 0 \Rightarrow 2(0)^2 = 5(0) \Rightarrow 0 = 0 && \text{works out}\\ x & = \frac{5}{2} \Rightarrow 2 \left ( \frac{5}{2} \right )^2 = 5 \cdot \frac{5}{2} \Rightarrow 2 \cdot \frac{25}{4} = \frac{25}{2} \Rightarrow \frac{25}{2} = \frac{25}{2} && \text{works out}\end{align*}

Answer: \begin{align*}x = 0, x =\frac{5}{2}\end{align*}

c) \begin{align*}9x^2 y - 6xy = 0\end{align*}

\begin{align*}9x^2y - 6xy = 0\end{align*}

Rewrite: not necessary

Factor: The common factor is \begin{align*}3xy\end{align*}, so this factors as \begin{align*}3xy(3x - 2)=0\end{align*}.

Set each factor equal to zero:

\begin{align*}3 = 0\end{align*} is never true, so this part does not give a solution. The factors we have left give us:

\begin{align*}x = 0 && \text{or} && y = 0 && \text{or} && 3x - 2 = 0\end{align*}

Solve:

\begin{align*}& \underline{x = 0} && \text{or} && \underline{y = 0} && \text{or} && 3x = 2\\ & &&&& \underline{x = \frac{2}{3}}\end{align*}

Check: Substitute each solution back into the original equation.

\begin{align*}& x = 0 \Rightarrow 9(0)y - 6(0)y = 0 - 0 = 0 && \text{works out}\\ & y = 0 \Rightarrow 9x^2 (0) - 6x(0) = 0 - 0 = 0 && \text{works out}\\ & x = \frac{2}{3} \Rightarrow 9 \cdot \left ( \frac{2}{3} \right)^2 y - 6 \cdot \frac{2}{3} y = 9 \cdot \frac{4}{9} y - 4y = 4y - 4y = 0 && \text{works out}\end{align*}

Answer: \begin{align*}x = 0, y = 0, x = \frac{2}{3}\end{align*}

### Example

#### Example 1

Solve the following polynomial equation.

\begin{align*}9x^2-3x=0\end{align*}

Solution: \begin{align*}9x^2-3x=0\end{align*}

Rewrite: This is not necessary since the equation is in the correct form.

Factor: The common factor is \begin{align*}3x\end{align*}, so this factors as: \begin{align*}3x(3x-1)=0\end{align*}.

Set each factor equal to zero.

\begin{align*}3x=0 \quad \text{ or } \quad 3x-1=0\end{align*}

Solve:

\begin{align*}x=0 \quad \text{ or } \ \quad x=\frac{1}{3}\end{align*}

Check: Substitute each solution back into the original equation.

\begin{align} x=0: \quad &9(0)^2-3(0)=0\\ x=\frac{1}{3}: \quad &9 \left( \frac{1}{3} \right) ^2 - 3 \left( \frac{1}{3} \right) = 0\\ \ \quad &9 \left( \frac{1}{9} \right) - 3 \left( \frac{1}{3} \right) = 0\\ \ \quad &1 - 1 = 0\\ \end{align}

Answer \begin{align*}x=0, \ x=\frac{1}{3}\end{align*}

### Review

Solve the following polynomial equations.

1. \begin{align*}x(x + 12) = 0\end{align*}
2. \begin{align*}(2x + 1)(2x - 1) = 0\end{align*}
3. \begin{align*}(x - 5)(2x + 7)(3x - 4) = 0\end{align*}
4. \begin{align*}2x(x + 9)(7x - 20) = 0\end{align*}
5. \begin{align*}x(3 + y) = 0\end{align*}
6. \begin{align*}x(x - 2y) = 0\end{align*}
7. \begin{align*}18y - 3y^2 = 0\end{align*}
8. \begin{align*}9x^2 = 27x\end{align*}
9. \begin{align*}4a^2 + a = 0\end{align*}
10. \begin{align*}b^2 - \frac{5}{3}b = 0\end{align*}
11. \begin{align*}4x^2 = 36\end{align*}
12. \begin{align*}x^3 - 5x^2 = 0\end{align*}

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English

Zero Product Rule

The zero product rule states that if the product of two expressions is equal to zero, then at least one of the original expressions much be equal to zero.