# Need a tutorial

#### thatsmessedup

Hi guys and gals,
I am reading schaums outline of precalculus 2nd ed. and I am having some issues solving a few of the problems.

Can someone point me to a tutorial that can help me factor equations like this:

4x^6-37x^3+9

I understand how to do the more simple ones like x^2-12x+27 but the one above is throwing me for a curve.
Thanks you very much.

#### Gusbob

There is no general formula for solving polynomials of degree 5+ (and even the ones for degrees 3 and 4, the formulas are usually to complicated to implement efficiently), so often you look for tricky factorisations. You don't need it in this case.

Set $$\displaystyle x^3=y$$ to get a quadratic $$\displaystyle 4y^2-37y+9$$. Solve for $$\displaystyle y$$, and then solve $$\displaystyle x^3=y$$ for $$\displaystyle x$$.

For information about actual factorisations of higher degree polynomials, google 'factoring higher degree polynomials'. I took a look at the first few links and they all seem pretty good, so the choice is yours. Also this youtube video looks good for your case:

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#### thatsmessedup

Thank you so much. I am googleing now

#### Gusbob

Ok, feel free to post here if you have problems with those.

#### Prove It

MHF Helper
You should post each new question in a new thread. Anyway

\displaystyle \displaystyle \begin{align*} \frac{p\, x^{p-1}}{q \left( x^{\frac{p}{q}} \right) ^{q-1}} &= \frac{p \, x^{p-1}}{ q\, x^{\left( q - 1 \right) \frac{p}{q} } } \\ &= \frac{p \, x^{p-1}}{ q\, x^{p - \frac{1}{q}} } \\ &= \frac{p}{q} \, x^{ p - 1 - \left( p - \frac{1}{q} \right) } \\ &= \frac{p}{q} \, x^{ \frac{1}{q} - 1 } \end{align*}

I do not agree with the answer given to you.

• thatsmessedup

#### thatsmessedup

I wish I could wrap my head around that... guess I just got to go study more. Thanks for the help.