# Thread: solving equations with syntax: y(x+2)(x+2)

1. ## solving equations with syntax: y(x+2)(x+2)

solving equations with syntax: y(x+2)(x+2)

Do I foil the parentheses first? distribute the y on both? multiply y with one, the distribute to the second parentheses?

2. Originally Posted by Shmuel
solving equations … y(x+2)(x+2)

Do I foil the parentheses first? distribute the y on both? multiply y with one, the distribute to the second parentheses?
This is not an equation. It's called an expression. (All equations contain an equal sign.)

We do not solve expressions; we simplify them.

You can foil first. Then you will have:

y*(quadratic polynomial) and you can now distribute the y, to finish.

You can distribute first. Then you will have:

(linear binomial containing both x and y)*(x + 2) and you can now foil, to finish.

But do not distribute the y twice!

If this expression is, in fact, part of an equation to solve, then you might not need to foil or distribute. Please post the entire exercise.

3. Let us say the equation is r(a+b)(a+b) = 0. How would I go about solving for a?

4. Originally Posted by Shmuel
Let us say the equation is r(a+b)(a+b) = 0. How would I go about solving for a?
For $r \ne 0$

a = -b

5. Originally Posted by Shmuel
Let us say the equation is r(a+b)(a+b) = 0. How would I go about solving for a?
Ooo, they were supposed to have covered this before assigning homework on it. To learn what the Zero Product Property is, and how to use it to solve factored-polynomial equations, try here.

6. Divide both sides of the equation by r to get (a+ b)^2= 0. What the square root of 0? So what is a+ b equal to? And then what is a equal to?

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