1. ## Factoring

solve each equation by factoring and using the zero product property

1/12x^2 - 2/3x = 4

( the x^2 and x are in the middle between the numbers. It is not 12x^2 or 3x )

2. ## Re: Factoring

Originally Posted by catherinee
solve each equation by factoring and using the zero product property

1/12x^2 - 2/3x = 4

( the x^2 and x are in the middle between the numbers. It is not 12x^2 or 3x )
so we're talking

$$\frac{1}{12}x^2-\frac{2}{3}x=4$$

start by multiplying both sides by 12 and getting all the terms on the left side and 0 on the right side.

Then factor that quadratic polynomial on the left via one of the methods you've learned. Here if you need a review.

Then just apply the zero product property to your two factors to obtain your two solutions.

3. ## Re: Factoring

so step 1 would be ---> x^2-8x=48 ?

4. ## Re: Factoring

Yes, and then x^2- 8x- 48= 0. I presume you know that will factor as (x- a)(x- b) where ab= -48 and a+ b= 8. What are possible integer values for a and b?

(x-12)(x+4)

x-12=0 x+4=0
x=12 x=-4

6. ## Re: Factoring

Originally Posted by catherinee
(x-12)(x+4)

x-12=0 x+4=0
x=12 x=-4
bravissimo!

7. ## Re: Factoring

Perhaps "bravssima"?