How do you find the zeroes of #f(x)=x^4+x^3+x^2-9x-10#?

1 Answer
Jul 23, 2015

Answer:

Use the rational root theorem and the quadratic formula to find roots:

#x = -1#, #x = 2#, #x = -1+-2i#

Explanation:

By the rational root theorem any rational roots of #f(x) = 0# must be of the form #p/q# where #p# and #q# are integers, #q != 0#, #p# a divisor of the constant term #-10# and #q# a divisor of the coefficient #1# of the term of highest degree #x^4#.

So the only possible rational roots are:

#+-1#, #+-2#, #+-5#, #+-10#

We find:

#f(1) = 1 + 1 + 1 - 9 - 10 = -16#

#f(-1) = 1 - 1 + 1 + 9 - 10 = 0#

#f(2) = 16 + 8 + 4 - 18 - 10 = 0#

So at least #x = -1# and #x = 2# are roots of #f(x) = 0# and #(x+1)# and #(x-2)# are factors of #f(x)#.

Divide #f(x)# by #(x+1)(x-2) = x^2-x-2# to find:

#f(x) = x^4+x^3+x^2-9x-10#

#= (x^2-x-2)(x^2+2x+5)#

Using the quadratic formula, the roots of #x^2+2x+5 = 0# are:

#x = (-2+-sqrt(2^2-(4xx1xx5)))/(2*1)#

#=(-2+-sqrt(4-20))/2#

#=(-2+-sqrt(-16))/2#

#=-1+-2i#