# How do you find all zeros with multiplicities of #f(x)=2x^4+x^3-7x^2-3x+3#?

##### 1 Answer

The zeros are:

#### Explanation:

Given:

#f(x) = 2x^4+x^3-7x^2-3x+3#

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

That means that the only possible *rational* zeros are:

#+-1/2, +-1, +-3/2, +-3#

We find:

#f(-1) = 2-1-7+3+3 = 0#

So

#2x^4+x^3-7x^2-3x+3 = (x+1)(2x^3-x^2-6x+3)#

Then note that the ratio of the first and second terms of the remaining cubic is the same as that between the third and fourth terms. So this cubic will factor by grouping:

#2x^3-x^2-6x+3 = (2x^3-x^2)-(6x-3)#

#color(white)(2x^3-x^2-6x+3) = x^2(2x-1)-3(2x-1)#

#color(white)(2x^3-x^2-6x+3) = (x^2-3)(2x-1)#

#color(white)(2x^3-x^2-6x+3) = (x^2-(sqrt(3))^2)(2x-1)#

#color(white)(2x^3-x^2-6x+3) = (x-sqrt(3))(x+sqrt(3))(2x-1)#

So the other zeros are