# Zero Product Property 3

Alignments to Content Standards: A-REI.A.1

The Zero Product Property (ZPP) states that if the product of two numbers is zero, then at least one of the numbers is zero. In symbols, if $ab=0$, then $a=0$ or $b=0$. We can use this property when we solve equations where a product is 0. For each equation below, use the ZPP to find all solutions. Explain each step in your reasoning.

1. $x(13−4x)=0,$
2. $7(y+12)=0,$
3. $(x−19)(x+3)=0,$
4. $(y−6)(3z−4)=0.$

## IM Commentary

This task is part of a series of tasks that lead students to understand and apply the zero product property to solving quadratic equations. The emphasis is on using the structure of a factorable expression to help find its solutions (rather than memorizing steps without understanding). Teachers should feel free to skip any tasks in the series that students have already mastered.

In previous tasks, students stated and proved the ZPP. In this particular task, we are trying to get students to use the property to reason about solutions to equations given in factored form. In tasks that follow in this series, students will first need to factor expressions before applying the ZPP to solve quadratic equations.

Note: the use of "or" in the ZPP is the mathematician's inclusive "or." That is, when we say, "$a=0$ or $b=0$," the "or both $a$ and $b$ are $0$" is implied. Alternatively, one could state this verbally: "If the product of two numbers is zero, then at least one of them has to be zero." The last equation in the problem is a good place to highlight the meaning of "or" in the ZPP, since in order for this equation to be true, $y$ could be 6, $z$ could be $\frac{4}{3}$, but also both could be true at the same time.

1. If $x(13-4x)=0$, then by the Zero Product Property (ZPP), either $x=0$ or $13-4x=0$. So $x=0$ is one solution to this equation. Solving $13-4x=0$, we get $x=\frac{13}{4}$ for the other solution.
2. If $7(y+12)=0$, then by the ZPP, either $7=0$, which is not true, or $y+12=0$. Hence the only solution to this equation is $y=-12$.
3. If $(x-19)(x+3)=0$, then by the ZPP, either $x-19=0$ or $x+3=0$. So the solutions to the equation are $x=19$ and $x=-3$.
4. If $(y-6)(3z-4)=0$, then by the ZPP, $y-6=0$ or $3z-4=0$. So the solutions to the equation are $y=6$ (and $z$ can be any value) or $z=\frac{4}{3}$ (and $y$ can be any value).