How do you find all zeros of the function #2x^6-3x^2-x+1#?

1 Answer
Apr 2, 2016

Answer:

See explanation...

Explanation:

By the rational roots theorem, the only possible rational zeros are expressible in the form #p/q# with integers #p# and #q# with #p# a divisor of the constant term #1# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2#, #+-1#

If we write #f(x) = x^6-3x^2-x+1# and evaluate #f(x)# for these values, we find that none work. So #f(x)# has no rational zeros.

Since #f(x)# has degree #6# it is not surprising to find that its roots have no simple closed algebraic formulation.

We can find them numerically, for example by using Newton's method.

#f'(x) = 6x^5-6x-1#

If we choose an initial approximation #a_0# for a zero of #f(x)#, then we can find better approximations by iterating using the formula:

#a_(i+1) = a_i - f(x)/(f'(x)) = a_i - (x^6-3x^2-x+1)/(6x^5-6x-1)#

The two Real zeros can readily be found by putting these formulae into a spreadsheet. To find the four Complex zeros is a little more complicated. If your spreadsheet application (like mine) does not handle Complex numbers directly, then you need to separate out Real and imaginary parts in separate columns.

If I have sufficient time, I may create such a spreadsheet.