How do you find all the zeros of f(x)=x4+3x3−4x2−12x with its multiplicities?

1 Answer
George C.
Jul 16, 2016

Zeros: 0,2,−2,−3

Explanation:

f(x)=x4+3x3−4x2−12x

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...

x4+3x3−4x2−12x

=x((x3+3x2)−(4x+12))

=x(x2(x+3)−4(x+3))

=x(x2−4)(x+3)

=x(x2−22)(x+3)

=x(x−2)(x+2)(x+3)

Hence zeros:

0,2,−2,−3

Related questions
See all questions in Zeros
Impact of this question
4004 views around the world
You can reuse this answer
Creative Commons License