How do you find all the zeros of #-x^5+3x^4+16x^3-2x^2-95x-44#?

1 Answer
Feb 27, 2016

Answer:

Use Newton's method to find numeric approximations for the three Real zeros, then divide by the corresponding factors to get a quadratic for the Complex zeros.

Explanation:

#f(x) = -x^5+3x^4+16x^3-2x^2-95x-44#

You could try the rational root theorem first, which would allow you to infer that the only possible rational zeros of #f(x)# are the factors of #44#, viz:

#+-1#, #+-2#, #+-4#, #+-11#, #+-22#, #+-44#.

None of these work, so #f(x)# has no rational zeros, but in the process of trying you might find:

#f(-1) = 37#

#f(1) = -123#

#f(4) = 312#

#f(11) = -97163#

So #f(x)# changes sign at least #3# times and has at least #3# Real zeros.

We can use Newton's method to find good approximations for the Real roots by choosing suitable starting approximations #a_0# and iterating using the formula:

#a_(i+1) = a_i - f(a_i)/(f'(a_i))#

In our example, #f'(x) = -5x^4+12x^3+48x^2-4x-95#

Putting the iteration formula into a spreadsheet and using initial values #a_0 = -1#, #a_0 = 2# and #a_0 = 11#, I found the following approximations after a few iterations:

#-0.485335316717177#

#2.624730249302921#

#5.259365512110042#

To find the Complex zeros, you can either put together a more complicated spreadsheet, with separate columns for Real and Imaginary parts, or you can divide #f(x)# by the factors corresponding to the zeros we have found to get a quadratic and use the quadratic formula.