# Solving for x in a trinomial that has a GCF

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.

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cepheid
Staff Emeritus
Gold Member
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.

CompuChip