How do you find all zeros with multiplicities of #f(x)=x^4-9x^2-4x+12#?

1 Answer
Feb 25, 2017

Answer:

#f(x)# has zeros:

#1# with multiplicity #1#

#3# with multiplicity #1#

#-2# with multiplicity #2#

Explanation:

Given:

#f(x) = x^4-9x^2-4x+12#

First note that the sum of the coefficients is #0#.

That is:

#1-9-4+12 = 0#

So #x=1# is a zero and #(x-1)# a factor:

#x^4-9x^2-4x+12 = (x-1)(x^3+x^2-8x-12)#

By the rational roots theorem, any rational zeros of #x^3+x^2-8x-12# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-12# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational zeros of this cubic are:

#+-1, +-2, +-3, +-4, +-6, +-12#

We find:

#color(blue)(3)^3+color(blue)(3)^2-8*color(blue)(3)-12 = 27+9-24-12 = 0#

So #x=3# is a zero and #(x-3)# a factor:

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

The remaining quadratic factor can be recognised as a perfect square trinomial. Like #144=12^2# we find:

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

Hence the other root is #-2# with multiplicity #2#.