Zero Product Property 4

Alignments to Content Standards: A-REI.A.1 A-REI.B.4.b

The Zero Product Property states that if the product of two numbers is zero, then at least one ofÂ the numbers is zero. In symbols, if $ab=0$, then $a=0$ or $b=0$.Â Sometimes, we can take advantage of this property to help us find solutions to equations. Explain how the property can be used to findÂ bothÂ solutions to each ofÂ 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 itÂ to solving quadratic equations. The emphasis is on using the structure of a factorable expression toÂ justify theÂ solution method (rather than memorizing steps without understanding). Teachers should feel free to skip any tasks in the series thatÂ students have already mastered.Â Â

In previousÂ 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.Â 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 eachÂ equation MP.7 andÂ rewrite it first. There should be a strong emphasis on explaining each step in solving the equation.

1. The solutionÂ is $1$ orÂ $3$. This equation is written asÂ the product of two expressions. Since it is equal to zero, theÂ ZPP says that at least one of those factorsÂ must equal zero. Either $x-1=0$, soÂ $1$ is a solution or $x-3=0$, soÂ $3$ is a solution.
2. The solutionÂ is $-\frac{3}{2}$ or $1$.Â This equation is not of the form, "aÂ productÂ of two factorsÂ equalsÂ zero." Using the structure on the left hand side, I can apply the distributive property toÂ 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 $x=-\frac{3}{2}$.Â
3. The solutionÂ is $-4$Â or $1$. Since I hope to use theÂ ZPP, I need to rewrite the expression as some polynomial of $x$ equalsÂ 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 IÂ viewÂ the first $x+4$ as the product of $1$ and $x+4$Â Â 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 isÂ $0$Â or $6$. This could be figured out by inspecting the operations involved, but if I want to use ZPP, I rewrite the equation asÂ $x^2-6x=0$ and then apply the distributive property to findÂ $x(x-6)=0$. Now I can use ZPP to say $x=0$ or $x=6$.
5. The solution is $2$ orÂ $5$. I first subtract $7x$ from both sides to write $x^2-7x+10=0$. Then I factor theÂ left hand side to getÂ $(x-2)(x-5)=0$. Applying theÂ ZPP to this equation givesÂ $x-2=0$ or $x-5=0$, so $x=2$ or $x=5$.