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

##### 1 Answer

#### Answer:

#### Explanation:

Given:

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

By the rational roots theorem, any *rational* zeros of

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

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

So

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

Let us try substituting

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

So

#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

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

So

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