# solving equations by factoring principals of zero produ

#### Mindy

##### New member
Can anyone help with a problem such as the following:

1. (7X-28)(28x-7)=0

2. (1/3=3x)(1/5-2x)=0

Thanks,

Mindy

#### Loren

##### Senior Member
1. (7X-28)(28x-7)=0

2. (1/3=3x)(1/5-2x)=0

I think you have a typo in the second entry. Maybe you can correct that.
The main idea in solving an equation such as #1 is that when you multiply two numbers together and get a product of zero, at least one of those two numbers is zero. You can't get any other answer than zero if one or the other or both of the factors is/are zero. So, if you have an equation such as (x-2)(2x+3)= 0 you know that either (x-2)=0 or (2x+3)=0 or both are equal to zero. So you solve each of these equations and get x=2 OR x=-3/2. Both of those values of x satisfy the original statement. Now, you should be able to solve the two equations posted (after making the one typo correction).

#### wjm11

##### Senior Member
1. (7X-28)(28x-7)=0

2. (1/3=3x)(1/5-2x)=0
Hi, Mindy,

Do you understand what the Zero Product Theorem is saying? It means that if we have a bunch of factors on one side of the equation and zero on the other, we can set each individual factor equal to zero, then solve.

7x-28 = 0
x = ?

28x-7 = 0
x = ?

You’ll get two different answers for x from the above equations. If you plug either one back into the original equation, you’ll find they (should) work.