Factoring- Pre Algebra Homework question

Question

I'm studying pre-algebra with an App called Brilliant. I'm in a lesson called "Factoring Sums with Variables". I got most of the exercises right, but there's one that I don't understand. To begin with, I don't even understand the question :( The bold part of the text is what I don't understand. Where did they get the $2ac$ from?

Question: Which of these equations does not represent a possible solution to the equation?

$ 6a^2bc+18a^2b^2c-24abc^2=0$?

1. $3ab+9ab^2-12bc=0$
2. $2a^2bc+3a^2b^2c-4abc^2=0$
3. $2a^2+6a^2b-8ac=0$
4. $a + 3ab-4c=0$

It gave me the following hint:

"First, we have $3ab+9ab^2-12bc.$ If we multiply by $2ac$ and distribute, we get:

$2ac(3ab+9ab^2-12bc)$
$=6a^2bc+18a^2b^2c-24abc^2$

So the equation is equivalent to:

$2ac(3ab+9ab^2-12bc)=0$
and by the zero-product property, we know that possible solutions are when $2ac=0$ or $3ab+9ab^2-12bc=0$

Answer 1

What they’re trying to do is find a factor to multiply the equations by in order to get $6a^2bc+18a^2b^2c-24abc^2$. In the first equation, if you multiply both sides by $2ac$ you achieve this. The way you multiply $2ac$ is by using the distributive property, like so:

$\begin{align} 2ac(3ab+9ab^2-12bc)&=2ac(3ab)+2ac(9ab^2)+2ac(-12bc)\\ &=6a^2bc+18ab^2c-24abc^2 \end{align}$

This is what they mean by “distribute” — you make sure every term gets multiplied by $2ac$. When you distribute cookies to your family for the holidays you make sure everybody in the house gets a cookie, but it’s probably not practical to ensure everyone in the world outside of the house gets a cookie too (although it would be kinda nice if you could). When you use multiplication to distribute, you multiply each term inside the parentheses touching the factor by the factor.

The zero product property tells us that if we have unknown values multiplying by each other that are equal to 0, the equation is true if and only if at least one of the unknown values is equal to 0. So in this example, either $2ac$ is equal to 0, or $3ab+9ab^2-12bc$ is equal to 0. The point is that all values of $a$, $b$, and $c$ that satisfy $3ab+9ab^2-12bc=0$ are possible solutions to the equation $6a^2bc+18a^2b^2c-24abc^2=0.$ Are the other answer choices possible solutions as well?

Answer 2

Your equation can be written as $$ 6abc(a+3ab-4c)=0. $$ In a field with $6\neq 0$, either $abc=0$, or $a+3b-4c=0$, which is equation number $4$. So this equation definitely can be a solution. To see that equation $2$ is false in general, take $a=2$, $b=5$ and $c=8$. Then the original equation is true, but not equation $2$. Actually, all others can be solutions.