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]

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]

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]

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]

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]

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]