# Math Help - Factor Question

1. ## Factor Question

How did this factorization come to be?

I see where 4x and 2x came from, and I under stand that -3*3=-9 for the C position, but how did the 6 get there in the B position? Isn't it supposed to be what you get when you add -3 +3? That would be 0, maybe something Im missing?

2. ## Re: Factor Question

The coefficients of x inside the brackets are 4 and two - the expansion in the b position is then:
4x*3+2x*(-3)=6x, as shown.

3. ## Re: Factor Question

Originally Posted by itgl72
How did this factorization come to be?

I see where 4x and 2x came from, and I under stand that -3*3=-9 for the C position, but how did the 6 get there in the B position? Isn't it supposed to be what you get when you add -3 +3? That would be 0, maybe something Im missing?
Multiply your a and c values, to give 8 x (-9) = -72. You are looking for two numbers that multiply to give -72 and add to give the b value, 6. The numbers are -6 and 12. So we break up the middle term according to these two numbers and factorise by grouping.

\displaystyle \begin{align*} 8x^2 + 6x - 9 &= 8x^2 + 12x - 6x - 9 \\ &= 4x(2x + 3) - 3(2x + 3) \\ &= (2x + 3)(4x - 3) \end{align*}

4. ## Re: Factor Question

Im good on this one, thanks all.

5. ## Re: Factor Question

8x^2+6x-9
Factors of this equations are:
=> (4x-3)(2x+3)

Since the equation is related to factoring trinomials method i.e. ax^2+bx+c. Solve it by factorizing each term as :
8 can be factorize as 4 & 2. 6 can be factorize as 2 , 3 & -9 can be factorize as 3, -3. Now them accordingly to make equation balanced.
8x^2+6x-9
=> 8x^2+12x-6x-9
=>4x(2x+3)-3(2x+3)
=>(4x-3)(2x+3)
Thus the equation get its factors.