Learning Outcomes
- Use the zero product principle to solve polynomial equations
Key words
- Zero product principle: if two or more factors are multiplied to a product of zero, at least one of the factors is zero
The Zero Product Principle
If we multiply two numbers together and get an answer of zero, what can we say about the two numbers? The only way to get a product of zero is if we multiply by [latex]0[/latex]. This means that one of the factors has to be zero. This idea is called the zero product principle, and it is useful for solving certain kinds of equations.
Zero Product PRINCIPLE
The Zero Product Principle states that if the product of two or more factors is [latex]0[/latex], then at least one of the factors must be [latex]0[/latex].
If [latex]ab=0[/latex], then either [latex]a=0[/latex] or [latex]b=0[/latex], or [latex]\text{both } a\text{ and }b=0[/latex].
When we say [latex]a=0[/latex] or [latex]b=0[/latex], [latex]\text{both } a\text{ and }b=0[/latex] is implied. i.e. at least one of the factors must equal zero.
The zero product principle can be used to solve factored equations that are equal to zero.
Example
Use the zero product principle to solve [latex]5y=0[/latex]
Solution
By the zero product principle, when two factors are multiplied and the result is zero at least one of them is equal to zero. Therefore, either [latex]5=0[/latex], or [latex]y=0[/latex].
In this case, we know that [latex]5[/latex] is not equal to zero, so [latex]y[/latex] must be equal to zero.
We can verify this with algebra.
[latex]\begin{array}{c}5y=0\\\text{}\\\frac{5y}{5}=\frac{0}{5}\\\text{}\\y=0\end{array}[/latex]
Answer
[latex]y=0[/latex]
We can extend this idea to products of more than just two factors.
Example
Solve [latex]5x(x-4)(3x+2)=0[/latex]
Solution
There are three factors with a product of zero, so at least one of the factors must equal zero:
[latex]\begin{equation}\begin{aligned}5x(x-4)(3x+2)&=0 \\ 5x=0\text{ or }x-4=0\text{ or }3x+2&=0 \\ x=0\text{ or }x=4\text{ or }3x&=-2 \\x=0\text{ or }x=4\text{ or }x&=-\frac{2}{3} \end{aligned}\end{equation}[/latex]
Answer
[latex]x=0\text{ or }x=4\text{ or }x=-\frac{2}{3}[/latex]
Try It
1. Solve the equation: [latex]-7y(3y-2)(4y+1)=0[/latex]
2. Solve the equation: [latex]x^2(2x+1)(5x-2)(x+2)=0[/latex]
Let’s consider the equation [latex]t(5-t)=0[/latex]. If we set each factor to zero and solve, we get two solutions [latex]t=0[/latex] or [latex]t=5[/latex].
Why don’t we just use the distributive property and the properties of equations to solve this kind of equation? Let’s try using the distributive property on this example to explain why this can be problematic.
[latex]\begin{equation}\begin{aligned}t\left(5-t\right)&=0 \\ 5t-t^2&=0 \\5t&=t^2 \\ \frac{5t}{t}&=\frac{t^2}{t} \\ 5&=t \end{aligned}\end{equation}[/latex]
Wait, our original solution was [latex]t=0\text{ or }t=5[/latex]. How did we lose one of the solutions? The problem occurred when we divided both sides of the equation by [latex]t[/latex]. Remember, that we can divide both sides of an equation by the same non-zero term. So if we want to divide by [latex]t[/latex], we must ensure that [latex]t\ne 0[/latex]. SInce we don’t know whether[latex]t=0[/latex] or not, we cannot divide by [latex]t[/latex]. We lost the solution [latex]t=0[/latex] when we divided by [latex]t[/latex].
When we are solving polynomial equations, we need to use some different methods than we used to solve linear equations to make sure we get all of the correct answers. The zero product principle is one tool that allows us to do this.
Caution! It is important to remember that the principle of zero products only works when we have an equation with zero on one side, and only a product on the other side. The table below gives examples of equations for which we can and cannot apply the principle of zero products.
| YES Zero Product Principle Works to Solve | NO Zero Product Principle Does Not Work to Solve | WHY NOT? |
| [latex]\frac{1}{2}\left(x-2\right)=0[/latex] | [latex]\frac{1}{2}\left(x-2\right)=28[/latex] | There is a product on the left, but it is not equal to zero. |
| [latex]s\left(9+s\right)=0[/latex] | [latex]s^2+9=0[/latex] | There is a sum equal to zero but no product equal to zero. |
The following video presents more examples of how to use the zero product principle to solve polynomial equations that are in factored form.
Solving Polynomial Equations
In this section we will solve polynomial equations that can be factored using the zero product principle.
We will begin with an example where the polynomial is already equal to zero.
Example
Solve: [latex]-t^2+t=0[/latex]
Solution
To solve this equation, we need to factor the left side. Each term has a common factor of [latex]t[/latex] and the leading term is negative, so we can factor out [latex]-t[/latex] and use the zero product principle:
[latex]\begin{equation}\begin{aligned}-t^2+t&=0 \\-t(t-1)&=0\end{aligned}\end{equation}[/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{equation}\begin{aligned}-t(t-1)&=0 \\ -t&=0\text{ OR }t-1=0 \\ t&=0\text{ OR }t=1\end{aligned}\end{equation}[/latex]
Answer
[latex]t=0\text{ OR }t=1[/latex]
The following video presents more examples of solving equations by factoring. A polynomial of degree two is often referred to as a quadratic. i.e. a quadratic is any polynomial of the form [latex]ax^2+bx+c[/latex], where [latex]a, b, c[/latex] are real numbers.
When we don’t have a zero on one side of the equation, we can subtract those terms from both sides of the equation to force a zero on one side.
Example
Solve: [latex]6t=3t^2-12t[/latex]
Solution
Our first goal is to try and see if we can use the zero product principle, since that is currently the only tool we know for solving polynomial equations. So, let’s move all the terms to one side, leaving zero on the other side.
[latex]\begin{equation}\begin{aligned}6t&=3t^2-12t \\0&=3t^2-12t-6t \\ 0&=3t^2-18t\end{aligned}\end{equation}[/latex]
We now have all the terms on the right side, and zero on the other side. Each term on the right side has a common factor of [latex]3t[/latex], so we can factor and use the zero product principle:
[latex]\begin{equation}\begin{aligned} 0&=3t^2-18t \\ 0&=3t(t-6)\end{aligned}\end{equation}[/latex]
Set each factor to zero and solve the equations:
[latex]\begin{equation}\begin{aligned}0&=3t(t-6) \\3t&=0\text{ OR }t-6=0 \\ t&=0\text{ OR }t=6\end{aligned}\end{equation}[/latex]
Answer
[latex]t=6\text{ OR }t=0[/latex]
The video that follows provides another example of solving a polynomial equation using the zero product principle and factoring.
We will work through one more example that is similar to the ones above, except this example has fractions. If we were asked to factor an expression containing fractions, we would have no choice but to work with the fractions and pull out a GCF that is a fraction. But, since an equation has two sides, we can eliminate the fractions by multiplying the whole equation by the least common multiple. Let’s do the same example both ways to see the difference.
Example
Solve [latex]\frac{1}{2}y=-4y-\frac{1}{2}y^2[/latex]
Solution
To work with the fractions, we first find a common denominator, then factor:
[latex]\begin{equation}\begin{aligned}\frac{1}{2}y&=-4y-\frac{1}{2}y^2\\0&=-\frac{1}{2}y-4y-\frac{1}{2}y^2\end{aligned}\end{equation}[/latex]
To combine like terms, we use the common denominator [latex]2[/latex]:
[latex]\begin{equation}\begin{aligned}0&=-\frac{1}{2}y-4y-\frac{1}{2}y^2\\\text{ }\\0&=-\frac{1}{2}y-\frac{8y}{2}-\frac{1}{2}y^2\\\text{}\\0&=-\frac{9}{2}y-\frac{1}{2}y^2\end{aligned}\end{equation}[/latex]
Find the greatest common factor of the terms of the polynomial:
Factors of [latex]-\frac{9}{2}y[/latex] are [latex]-\frac{1}{2}\cdot{3}\cdot{3}\cdot{y}[/latex]
Factors of [latex]-\frac{1}{2}y^2[/latex] are [latex]-\frac{1}{2}\cdot{y}\cdot{y}[/latex]
Both terms have [latex]-\frac{1}{2}\text{ and }y[/latex] in common.
Rewrite each term as the product of the GCF and the remaining terms:
[latex]-\frac{9}{2}y=-\frac{1}{2}y\left(3\cdot{3}\right)=-\frac{1}{2}y\left(9\right)[/latex]
[latex]-\frac{1}{2}y^2=-\frac{1}{2}y\left(y\right)[/latex]
Rewrite the polynomial equation using the factored terms in place of the original terms. Remember to pay attention to signs when we factor. Notice that we end up with a sum as a factor because the common factor is a negative number. [latex]\left(9+y\right)[/latex]
[latex]\begin{equation}\begin{aligned}0&=-\frac{9}{2}y-\frac{1}{2}y^2\\\text{}\\0&=-\frac{1}{2}y\left(9\right)-\frac{1}{2}y\left(y\right)\\\text{}\\0&=-\frac{1}{2}y\left(9+y\right)\end{aligned}\end{equation}[/latex]
Solve the two equations.
[latex]\begin{equation}\begin{aligned}-\frac{1}{2}y&=0\text{ or }y+9=0\\ y&=0\text{ or }y=-9\end{aligned}\end{equation}[/latex]
Answer
[latex]y=0\text{ or }y=-9[/latex]
In this last example, we used many skills to solve one equation. Let’s summarize them:
- We needed a common denominator to combine the like terms [latex]-4y\text{ and }-\frac{1}{2}y[/latex], after we moved all the terms to one side of the equation
- We found the GCF of the terms [latex]-\frac{9}{2}y\text{ and }-\frac{1}{2}y^2[/latex]
- We used the GCF to factor the polynomial [latex]-\frac{9}{2}y-\frac{1}{2}y^2[/latex]
- We used the zero product principle to solve the polynomial equation [latex]0=-\frac{1}{2}y\left(9+y\right)[/latex]
Sometimes solving an equation requires the combination of many algebraic principles and techniques. The last facet of solving the polynomial equation [latex]\frac{1}{2}y=-4y-\frac{1}{2}y^2[/latex] that we should talk about is negative signs.
What follows is the same example completed by first eliminating the fractions by multiplying the whole equation by the least common multiple.
ExAMPLE
Solve the equation: [latex]\frac{1}{2}y=-4y-\frac{1}{2}y^2[/latex]
Solution
First multiply both sides of the equation by the LCD [latex]2[/latex]: [latex]2\cdot\frac{1}{2}y=2\cdot \left (-4y-\frac{1}{2}y^2\right )\\y=-8y-y^2[/latex]
Add [latex]y^2+8y[/latex] to both sides of the equation: [latex]y^2+8y+y=0[/latex]
Combine like terms: [latex]y^2+9y=0[/latex]
Factor: [latex]y(y+9)=0[/latex]
Set each factor to zero and solve the equations: [latex]y=0\text{ or }y+9=0\\y=0\text{ or }y=-9[/latex]
Answer
[latex]y=0\text{ or }y=-9[/latex]
We get the same solution whether we work with the fractions or eliminate the fractions. We get to choose which way we prefer to complete such problems with fractions, but generally speaking, it is more efficient to eliminate the fractions as a first step.
The following video presents another example of solving an equation with fractional coefficients using factoring and the zero product principle.
If the GCF of a polynomial is negative, pay attention to the signs that are left when we factor it from the terms of a polynomial.Try It
Solve the equation: 5x^2=4+2x^2
Often, there is no GCF to pull out of an equation.
Example
Solve the equation: [latex]3x^2-4x-4=0[/latex]
Solution
Factor the left side of the equation:
[latex]\begin{equation}\begin{aligned}3x^2-4x-4 &=0\\ac=-12\text{ and }b=-4\text{ so split }-4x\text{ into }-6x+2x\\3x^2-6x+2x-4 &=0\\3x(x-2)+2(x-2)&=0\\(3x+2)(x-2)&=0\end{aligned}\end{equation}[/latex]
Set each factor to zero and solve:
[latex]\begin{equation}\begin{aligned}3x+2&=0\text{ or }x-2=0\\3x&=-2\text{ or }x=2\\ x &=-\frac{2}{3}\text{ or }x=2\end{aligned}\end{equation}[/latex]
Answer
[latex]x=-\frac{2}{3}\text{ or }x-2=0[/latex]
Example
Solve the equation: [latex]9x^4-24x^2=-30x^3[/latex]
Solution
Get all non-zero terms on the same side of the equation:
[latex]\begin{equation}\begin{aligned}9x^4-24x^2&=15x^3\\9x^4+30x^3-24x^2 &=0\end{aligned}\end{equation}[/latex]
Look for a GCF in all three terms: [latex]3x^2[/latex]
Factor:
[latex]\begin{equation}\begin{aligned}3x^2\left (3x^2+10x-8 \right ) &=0\;\;\;\;\;ac=-24\text{ and }b=10\text{ so split }10x\text{ into }12x-2x\\3x^2\left ( 3x^2+12x-2x-8\right ) &=0\\3x^2\left[ 3x(x+4)-2(x+4)\right ] &=0\\ 3x^2(x+4)(3x-2)&=0\end{aligned}\end{equation}[/latex]
Set each factor to zero and solve:
[latex]\begin{equation}\begin{aligned}3x^2(x+4)(3x-2)&=0\\3x^2=0\text{ or }x+4 &=0\text{ or }3x-2=0\\x=0\text{ or }x &=-4\text{ or }3x=2\\x=0\text{ or }x &=-4\text{ or }x=-\frac{2}{3}\end{aligned}\end{equation}[/latex]
Answer
[latex]x=0\text{ or }x=-4\text{ or }x=-\frac{2}{3}[/latex]
Try It
Solve the equation: [latex]42x^4-49x^3=35x^2[/latex]
