Need a tutorial

Mar 2013
96
1
Texas
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.
 
Jan 2008
588
242
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:
 
Last edited:
Jan 2008
588
242
Ok, feel free to post here if you have problems with those.
 
Mar 2013
96
1
Texas
I stumbled apon another problem that I cant see past. I figured out a,b,c no problem but I dont know how to simplify d. can someone point me in the right direction?
help.jpg
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
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.
 
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Mar 2013
96
1
Texas
I wish I could wrap my head around that... guess I just got to go study more. Thanks for the help.