# Solving for x in a trinomial that has a GCF

1. Jul 17, 2007

### danielle36

What I am trying to do here is to factor so i can come up with the solutions for x (which are 4 and -3).
4x$$^{2}$$ + 4x - 48 = 0

Here's what I've done to solve so far:
4(x$$^{2}$$ - 1x - 12) = 0
4(x-4)(x+3)=0

Now I'm not even sure if what I'm doing is right here, but if it is my problem is comming to the solution set itself - I'm really just not sure what to do with that initial 4.

2. Jul 17, 2007

### cepheid

Staff Emeritus
You have basically solved the problem. The question is, what values of x cause the expression on the left hand side to be zero (i.e. what values of x *satisfy* the equation)? If you ask yourself the question in this way, you will realize that the presence of the additional factor of 4 in front makes NO difference at all. The solution set is still {4,-3}, just as it would be if the 4 were not there. Take a look:

If x = 4, the expression becomes 4*0*7 = 0
If x = -3, it becomes 4*(-7)*0 = 0

Anything multiplied by zero is zero. Therefore, if any ONE of the three factors is zero, the entire product must be zero.

3. Jul 17, 2007

### CompuChip

If it makes you feel better, you can divide out the factor four.
You know that performing the same operation on both sides of the equals sign, does not change the equation. Dividing the left hand side by four gives (x - 4)(x + 3). The right hand side gives 0/4 = 0. So the solutions to
(x - 4)(x + 3) = 0
are the same as those to
4 (x - 4)(x + 3) = 0.