I used the U-substitution
6x^6 + 13x^3 - 5
x^3 = u x^6 = u^2
6u^2 + 13u - 5
(6u^2 + 2u) + (15u - 5)
u(6u + 2) + 3 (6u + 2)
(6u + 2) (u + 3)
(6x^6 + 2) (x^6 + 3)
Is this correct?
no, if you foil out your factor it will come out as 6x^12 + 20x^6 + 6. i am not familiar with the U substitution but if you try again i can tell you if you are right.
Your step 3:
6u^2 + 13u - 5
What number multiplied give -5?
One possibility...
(?u - 1) ( ?u + 5)
What numbers multiplied give 6?
One possibility...
(3u - 1)(2u + 5)
Now substitute back for u.
ok I get
(3x^3 - 1) (2x^3 + 5)
Is this now correct?
Yes.
Also, lyne mentioned checking using FOIL (First Outer Inner Last). That is a good acronym for multiplying binomials.
i tried to foil this and im not getting the problem
OK,
(3x^3 - 1) (2x^3 + 5)
First
Multiply the first terms of the two factors:
(3x^3)(2x^3) = 6x^6
Outer
Multiply the outer terms of the factors:
(3x^3)(5) = 15x^3
Inner
Multiply the inner terms of the factors:
(-1)(2x^3) = -2x^3
Last
Multiply the last terms of the factors:
(-1)(5) = -5
Adding all the terms:
6X^6 + 15x^3 - 2x^3 - 5
= 6x^6 + 13x^3 - 5
Hopefully no typos.
6x^6 + 13x^3 - 5
let u = x^3
6u^2 + 13u -5 = (3u -1)(2u +5)
Which is the same as
(3x^3 -1)(2x^3 + 5)
That leaves you with two cubics to factor. One of the factors of
(3x^3 -1) is (3^1/3)*x -1
(I am using "3^1/3)" to denote "cube root of"). The other factor is (
[3^2/3)x^2 + (3^1/3)x + 1].
I will leave the rest to you. Other factors may require imaginary numbers.