Recently, we learned strategies for transforming expressions from standard form to factored form. In earlier lessons, we have also seen that when a quadratic expression is in factored form, we can find values of the variable that make the expression equal zero. Suppose we are solving the equation $x(x+4)=0$, which says that the product of $x$ and $x+4$ is 0. By the zero product property, we know this means that either $x=0$ or $x+4=0$, which then tells us that 0 and -4 are solutions.

Together, these two skills—writing quadratic expressions in factored form and using the zero product property when a factored expression equals 0—allow us to solve quadratic equations given in other forms. Here is an example:

$\displaystyle \begin{aligned} n^2-4n &= 140 &\qquad& \text{Original equation}\\n^2-4n-140 & =0 &\qquad& \text{Subtract 140 from each side so the right side is 0.}\\ (n-14)(n+10) &= 0 &\qquad& \text{Rewrite in factored form.} \\ \\n-14=0 \quad \text{or} \quad &n+10=0 &\qquad& \text {Apply the zero product property.} \\ n=14 \quad \text{or} \quad &n=\text-10 &\qquad& \text{Solve each equation.}\end{aligned}$

When a quadratic equation is written as as an expression in factored form equal to 0, we can also see the number of solutions the equation has.

In the previous example, it is not obvious how many solutions there are when the equation is in the form $n^2-4n-140=0$. When the equation is rewritten as $(n-14)(n+10) =0$, we can see that there are two numbers that could make the expression equal 0: 14 and -10.

How many solutions does the equation $x^2-20x+100=0$ have?

Let’s rewrite it in factored form: $(x-10)(x-10)=0$. The two factors are identical, which means that there is only one value of $x$ that makes the expression $(x-10)(x-10)$ equal 0. The equation has only one solution: 10.