How do you find all zeros with multiplicities of #f(x)=x^4+2x^3-12x^2-40x-32#?

1 Answer
Feb 25, 2017

Answer:

#f(x)# has zeros #4# (with multiplicity #1#) and #-2# with multiplicity #3#.

Explanation:

Given:

#f(x) = x^4+2x^3-12x^2-40x-32#

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divosor of the constant term #-32# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-4, +-8, +-16, +-32#

Note also that the signs of the coefficients are in the pattern #+ + - - -#. By Descartes' Rule of Signs, sinc this has one change, we can deduce that #f(x)# has one positive real zero. So let us look for that one first.

#f(4) = 4^4+2(4^3)-12(4^2)-40(4)-32 = 256+128-192-160-32 = 0#

So #x=4# is a zero and #(x-4)# a factor:

#x^4+2x^3-12x^2-40x-32 = (x-4)(x^3+6x^2+12x+8)#

Let us try substituting #-2# in the remaining cubic factor:

#(-2)^3+6(-2)^2+12(-2)+8 = -8+24-24+8 = 0#

So #x=-2# is a zero and #(x+2)# a factor:

#x^3+6x^2+12x+8 = (x+2)(x^2+4x+4)#

The remaining quadratic factor is a perfect square trinomial, as we might spot by noticing #144 = 12^2# and similarly:

#x^2+4x+4 = (x+2)^2#

So #x=-2# is a zero twice more.

graph{x^4+2x^3-12x^2-40x-32 [-10, 10, -160, 80]}