Zero Product Property 4

Task

The Zero Product Property 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 $b=0$. Sometimes, we can take advantage of this property to help us find solutions to equations. Explain how the property can be used to find both solutions to each of the following equations, and explain each step in your reasoning.

1. $(x-1)(x-3)=0$
2. $2x(x-1)+3x-3=0$
3. $x+4=x(x+4)$
4. $x^2=6x$
5. $x^2+10=7x$

IM Commentary

This task is the fourth in a series of tasks that leads students to understand The Zero Product Property (ZPP) and apply it to solving quadratic equations. The emphasis is on using the structure of a factorable expression to justify the solution method (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: If the product of two numbers is zero, then at least one of the two numbers must be zero. In symbols, where $a$ and $b$ represent numbers, if $ab=0$, then $a=0$ or $b=0$.

In the earlier tasks in the sequence, students applied the ZPP to equations in factored form. In this task, students can still solve using ZPP, but after the first equation (which can serve as a warm-up/refresher), they must consider the structure of each equation MP.7 and rewrite it first. There should be a strong emphasis on explaining each step in solving the equation.

Links to other tasks in this series:

• Zero Product Property 1
• Zero Product Property 2
• Zero Product Property 3

Solution

1. The solution is $1$ or $3$. This equation is written as the product of two expressions. Since it is equal to zero, the ZPP says that at least one of those factors must equal zero. Either $x-1=0$, so $1$ is a solution or $x-3=0$, so $3$ is a solution.
2. The solution is $-\frac{3}{2}$ or $1$. This equation is not of the form, "a product of two factors equals zero." Using the structure on the left hand side, I can apply the distributive property to rewrite the equation as $2x(x-1)+3(x-1)=0$, and then use the distributive property again to write $(x-1)(2x+3)=0$. Now using ZPP I can say $x-1=0$ or $2x+3=0$, so $x=1$ or $x=-\frac{3}{2}$. 
3. The solution is $-4$ or $1$. Since I hope to use the ZPP, I need to rewrite the expression as some polynomial of $x$ equals 0. So I'll subtract everything on the right hand side from both sides and rewrite the equation as $x+4-x(x+4)=0$. If I view the first $x+4$ as the product of $1$ and $x+4$  I can apply the distributive property and rewrite the equation as $(x+4)(1-x)=0$. Now using ZPP I can say $x+4=0$ or $1-x=0$, so $x=-4$ or $x=1$.
4. The solution is $0$ or $6$. This could be figured out by inspecting the operations involved, but if I want to use ZPP, I rewrite the equation as $x^2-6x=0$ and then apply the distributive property to find $x(x-6)=0$. Now I can use ZPP to say $x=0$ or $x=6$.
5. The solution is $2$ or $5$. I first subtract $7x$ from both sides to write $x^2-7x+10=0$. Then I factor the left hand side to get $(x-2)(x-5)=0$. Applying the ZPP to this equation gives $x-2=0$ or $x-5=0$, so $x=2$ or $x=5$.