
Subject: Adv. Algebra crisis
Name: Dakota
Who is asking: Student
Level: Secondary
Question:
Find ALL zeros of P(x) = x^{ 6}  64
Hi,
There are a couple of things you can do with this:

Look at as [x^{3}]^{2}  [8]^{2}.
This then comes apart like a^{ 2}  b^{ 2} (or any other difference of squares).
[x^{ 3}  8][x^{ 3} + 8].
Now I presume you can do something with a difference of cubes.

Can you GUESS a couple of roots (zeros)?
e.g. x=2, ... ?
There are rules (e.g. Descarte's Rule of Signs) that predict
the possible number of roots of such an equation.
Descartes Rule goes like this:
Look at the polynomial  and see the number of times the
SIGN of the coefficients change as you go from the top to the bottom:
x^{ 6}  64
= 1x^{ 6}+ 0x^{ 5} + 0x^{ 4} + 0x^{ 3} + 0x^{ 2} +0x  64
+ 0 0 0 0 0 
This is ONE sign change.
This number is the MAXIMUM number of zeros in the interval 0 to + infinity.
Similarly, if you replace (x) by (x), you turn the real x axis around,
and the sign changes of the new formula predict the zeros of the original
formula in the interval infinity to zero.
(x)^{ 6}  64 = x^{ 6} 64.
Again there is one sign chance at one zero predicted between  infinity
and zero.
Of course, you can tell if x = 0 is a zero, by seing if x is a factor!
Now if you have guessed (or otherwise) found enough zeros
to match the maxima predicted by Descartes rule, then you have found
ALL zeros.
For this problem, you can find enough solutions (from (a) or by guessing).

If you are having trouble guessing, you could also try a a good
graph, using a graphing calculator or other computer programs.
This picture will give you a solid idea of what is likely to happen.
For y= x^{ 6}  64, the graph will look roughly like a parabola:
 with both ends pointing up into large values of Y,
 with the middle points, x = 0, y= 64 down below the xaxis.
 with to points where it crosses the xaxis (one positive, one negative
as predicted by Descartes Rule).
This also confirms directly that you can only expect two zeros.
It also tells you roughly what they are.
The reason you can turn such rough information into an exact guess
in this problem is that somebody 'rigged' the solutions to be
whole numbers, so that once you see the picture or start guessing
(with whole numbers which are factors of 64) you quickly get
answers which check.
Finally, you know you have a zero if it gives zero when you plug it in.
Nothing about the method used is finally important in your claim
e.g. x=2 is a zero. Since 2^{ 6}  64 = 0, you have PROVEN
that x=2 is a zero!
What is hard here is knowing that you have ALL the zeros.
That is what the picture and Descartes Rule (or detailed knowledge
about how to factor cubics, and then quadratics) gives: a guarantee
that there are no more.
By the way, Descartes Rule really is about how many times the
'shape' of the graph can turn 'upwards' or 'downwards' in the interval.
Walter Whiteley
Go to Math Central

