Factoring a Quadratic Trinomial

Conaanaa

New member
Joined
Oct 3, 2006
Messages
6
I need help factoring ax² + bx + c

Here's an example problem I have...

2x² - 15x + 7

I used http://www.helpalgebra.com/onlinebook/f ... ouping.htm to try and solve it but here's what I got...

WORK -------------
2x² - 15x + 7
2x² - 14x - 1x + 7
(2x² - 14x) + (-x + 7)
x(2x - 14) + (-x + 7)

WORK -------------

Now I am stuck and I don't know what to do. Help, please!

I've tried several other problems on my homework and I can't figure out any of them! Please help, thanks!
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, Conaanaa!

Your factoring is incomplete . . .


\(\displaystyle 2x^2\,-\,15x\,+\,7\)

Work:
\(\displaystyle 2x^2\,-\,15x\,+\,7\)
\(\displaystyle 2x^2\,-\,14x\,-\,1x\,+\,7\)
\(\displaystyle (2x^2\,-\,14x) \,+\,(-x\,+\,7)\)
\(\displaystyle x(2x\,-\,14)\,+\.(-x\,+\,7)\;\) <-- here!

The first two terms are: \(\displaystyle \,2x^2\,-\,14x\)

. . Factor completely: \(\displaystyle \,2x(x\,-\,7)\)


The last two terms are: \(\displaystyle \,-x\,+\,7\)

. . It doesn't factor, but we can take out -1: \(\displaystyle \,-1(x\,-\,7)\)


And we have: \(\displaystyle \,2x\underbrace{(x\,-\,7)}\,-\,1\underbrace{(x\,-\,7)}\)
. . . . . . . . . . . . . . . \(\displaystyle \nwarrow\)
There!\(\displaystyle \nearrow\;\)

Take out the common factor: \(\displaystyle \,(x\,-\,7)(2x\,-\,1)\)

 

Conaanaa

New member
Joined
Oct 3, 2006
Messages
6
Oh, wow I see! Thanks for the help!
 
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