SOLUTION: Solve for x by factoring. Show all real solutions. {{{x^6+7x^3-8=0}}}

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Question 1127134: Solve for x by factoring. Show all real solutions.
x%5E6%2B7x%5E3-8=0

Found 4 solutions by mhusband, josgarithmetic, ikleyn, greenestamps:
Answer by mhusband(3) About Me  (Show Source):
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
Start by recognizing equation is in quadratic form.

%28x%5E3%29%5E2%2B7%28x%5E3%29-8=0

%28x%5E3%2B8%29%28x%5E3-1%29=0
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Answer by ikleyn(52754) About Me  (Show Source):
You can put this solution on YOUR website!
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In this case, factoring works this way


    x%5E6+%2B+7x%5E3+-+8 = %28x%5E3%2B8%29%2A%28x%5E3-1%29.     (1)


The person who has some minimal experience in factoring, will find this decomposition easy.



    But if you have no such experience, then you can introduce new variable  y = x%5E3,

    and then your polynomial becomes


        x%5E6+%2B+7x%5E3+-+8 = y%5E2+%2B+7y+-8.


    It is very well familiar quadratic polynomial of "y", and therefore you can factor it quickly


        y%5E2+%2B+7y%5E3+-8 = (y+8)*(y-1).


    Returning then from "y" to x%5E3,  you get this decomposition (1).



In any case, when you get this decomposition


    x%5E6+%2B+7x%5E3+-+8 = %28x%5E3%2B8%29%2A%28x%5E3-1%29,


you easily get the roots:


a)  The factor  x%5E3%2B8 gives you the root  x%5E3 = -8,   or  x= root%283%2C-8%29 = -2,    and


b)  The factor  x%5E3-1 gives you the root  x%5E3 = 1,   or  x= root%283%2C1%29 = 1.

Solved.

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Introducing new variables, as I showed you in my post, often can help you in many other complicated cases.

It is very useful trick and a method.

As mathematicians say, introducing new variable/variables allows reducing the equation degree, which always facilitates the solution.


Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


The problem says "Solve for x by factoring. Show all real solutions."

While some of the responses you have received so far show the correct real solutions, none of them shows the complete factorization of the polynomial to show that the other 4 roots are not real.

I think a complete answer to problem as posed needs to do that.

x%5E6%2B7x%5E3-8+=+0
%28x%5E3%2B8%29%28x%5E3-1%29+=+0
%28x%2B2%29%28x%5E2-2x%2B4%29%28x-1%29%28x%5E2%2Bx%2B1%29+=+0

The factors (x+2) and (x-1) give you real roots, x=-2 and x=1.

For both of the quadratic factors, the discriminant is negative, so the roots are not real.

So the answer to the QUESTION that is asked in the problem is that the real roots are -2 and 1.