How do you factor #8x^6 + 7x^3 - 1#?

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Answer 1 · 2016-09-01T10:32:10

#(1 x^3 +1)(8 x^3-1)#

This is in the form of a quadratic because.

#8(x^3)^2 + 7(x^3)^1 -1#

Find the factors of 8 (and 1) which subtract to give 7

#8-1 = 7#

The signs will be different (because of -1)
There must be more positives (because of +7)

The factors of -1 can only be +1 and -1

#(? x^3 +1)(? x^3-1)#

Place the 8 and the 1 so that we will get +8 and -1. when we multiply.

#(1 x^3 +1)(8 x^3-1)#

Answer 2 · 2016-09-01T10:39:52

Set #t=x^3# then the given equation is transformed to

#8t^2+7t-1#

#8t^2+(8-1)t-1#

#8t^2+8t-t-1#

#8t(t+1)-(t+1)#

#(t+1)*(8*t-1)#

Reverse to x we have that

#(1+x^3)*(8*x^3-1)#

Now #x^3+1# can be factored further as follows

#x^3+1=(x+1)(x^2-x+1)#

and #8*x^3-1# can be factored as follows

#8x^3-1=(2 x-1) (4 x^2+2 x+1)#

Hence finally we get

#8x^6+7x^3-1=(x+1)*(x^2-x+1)*(2*x-1)*(4*x^2+2 x+1)#