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

1 Answer
Jul 16, 2016

Answer:

Zeros: #0, 2, -2, -3#

Explanation:

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

Note that all of the terms are divisible by #x# and the ratio of the first and second terms is the same as that between the third and fourth terms. So this will factor by grouping...

#x^4+3x^3-4x^2-12x#

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

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

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

#=x(x^2-2^2)(x+3)#

#=x(x-2)(x+2)(x+3)#

Hence zeros:

#0, 2, -2, -3#