# Algebra factoring equations

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• Feb 7th 2008, 07:24 PM
Belanova
Algebra factoring equations
2x^2+8x+8=0

25e^2+90e+81=0

The answers should be in x=? x=?
• Feb 7th 2008, 08:03 PM
coach2uf
2x^2+8x+8 = 0

2(x^2+4x+4) = 0

x^2+4x+4 factors into (x+2)(x+2)

2(x+2)^2= 0

X = -2

I don't know about the second one.
• Feb 7th 2008, 08:04 PM
mr fantastic
Quote:

Originally Posted by Belanova
2x^2+8x+8=0

25e^2+90e+81=0

The answers should be in x=? x=?

In general, a^2 + 2ab + b^2 = (a + b)^2. So:

2(x^2 + 4x + 4) = 0 => 2(x + 2)^2 = 0 .......

(5e + 9)^2 = 0 ........
• Feb 7th 2008, 08:19 PM
Belanova
Thanks very much. What about these? :D
48q^2+168q+147=0
147r^2+126r+27=0
• Feb 7th 2008, 08:25 PM
mr fantastic
Quote:

Originally Posted by Belanova
Thanks very much. What about these? :D

48q^2+168q+147=0 Mr F says: 3(4q + 7)^2 = 0

147r^2+126r+27=0 Mr F says: Take out a common factor (hint: 3) and write as a perfect square. Same as the others.

..
• Feb 7th 2008, 08:33 PM
coach2uf
Wow, that makes those problems incredibly simple. I was never taught that, thanks. (Clapping)

Quote:

Originally Posted by mr fantastic
In general, a^2 + 2ab + b^2 = (a + b)^2. So:

2(x^2 + 4x + 4) = 0 => 2(x + 2)^2 = 0 .......

(5e + 9)^2 = 0 ........

• Feb 7th 2008, 08:35 PM
mr fantastic
Quote:

Originally Posted by coach2uf
Wow, that makes those problems incredibly simple. I was never taught that, thanks. (Clapping)

It only works for a quadratic trinomial that's a perfect square. That is, something that has the form $a^2 \pm 2ab + b^2$.