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October 24, 2014

October 24, 2014

Posted by **Rachelle** on Monday, November 10, 2008 at 4:38pm.

My first one is 12x^2+x=6. How would you facor that?

My second is how would x^4-1 equal the answer 1(x^4-1)(x^2=1)(x^2-1)?

Finally, how would you factor something such as -3x(x+2)-8(x+2), when there is no common factor other than one?

Thank you so much for your assistance! :)

- Algebra -
**Damon**, Monday, November 10, 2008 at 4:51pm12 x^2 + x - 6 = 0

I solved the quadratic and got 2/3 and -3/4 so then tried

(3x+2)(4x-3)=0

- Algebra -
**Damon**, Monday, November 10, 2008 at 4:57pmx^4-1 is difference of two squares

(x^2-1)(x^2+1)

then

(x-1)(x+1)(x^2+1)

- Algebra -
**Reiny**, Monday, November 10, 2008 at 4:59pmfor 12x^2+x=6 write is as

12x^2+x-6 = 0

then (3x-2)(4x+3) = 0

x^4-1 factors to

(x^2 -1)(x^2 + 1)

=(x-1)(x+1)(x^2+1)

for -3x(x+2)-8(x+2) you should see that ((x+2) is a common factor, so

-3x(x+2)-8(x+2)

= (x+2)(-3x-8)

= -(x+2)(3x+8)

- Algebra -
**Damon**, Monday, November 10, 2008 at 4:59pm-3x(x+2)-8(x+2)

sure there is a common factor -- (x+2)

(x+2)(-3x-8)

(x+2)(-1)(3x+8)

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