Solve Quadratic Equations by factoring

[tlcourtney]

I've been on this one problem for about an hour now, will someone please help me. I truly do not understand this problem. I feel like crying but crying won't help me solve the problem

7-7x=(3x+2)(x-1)

 

[Loren]

Careful on this one. It appears that you could divide both sides by x-1. DON'T. If you do, you will will be eliminating one of the possible roots. Your best policy is to eliminate the parenthesis first, gather like terms, get all terms on the left side of the equation, then solve by factoring. It turns out that if you divide both sides by x-1, you will have been dividing by zero, which is not undefined. It's usually a good policy to NEVER divide both sides of an equation by an expression involving the unknown.

 

[tlcourtney]

I still don't get it. My book gives a basic example but this is complicated to me.

 

[Loren]

7-7x=(3x+2)(x-1)
Simplify...
7-7x = 3x[sup:3qw5bv0u]2[/sup:3qw5bv0u] - x - 2
Get all terms on one side of the equation...
7 - 7x - 3x[sup:3qw5bv0u]2[/sup:3qw5bv0u] + x + 2 = 0
Simplify...
-3x[sup:3qw5bv0u]2[/sup:3qw5bv0u] - 6x + 9 = 0
Factor out a monomial...
-3(x[sup:3qw5bv0u]2[/sup:3qw5bv0u] + 2x - 3) = 0
Factor the trinomial and solve for x.
Hope you can take it from here. You will get two possible solutions that should be checked back in the original equation.

 

[tlcourtney]

Thanks Loren that helps a great deal. I have to remember to do 1 step at a time.