# Algebra

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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 -

12 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 -

x^4-1 is difference of two squares
(x^2-1)(x^2+1)
then
(x-1)(x+1)(x^2+1)

• Algebra -

for 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 -

-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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