# How do you find all the zeros of #f(x)=x^4+6x^3-3x^2+24x-28#?

##### 1 Answer

#### Answer:

Zeros:

#### Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this later with

#f(x) = x^4+6x^3-3x^2+24x-28#

First note that the sum of the coefficients is

#1+6-3+24-28=0#

Hence

#x^4+6x^3-3x^2+24x-28 = (x-1)(x^3+7x^2+4x+28)#

Note that the ratio between the first and second terms of the remaining cubic polynomial is the same as that between its third and fourth terms, so it can be factored by grouping:

#x^3+7x^2+4x+28#

#=(x^3+7x^2)+(4x+28)#

#=x^2(x+7)+4(x+7)#

#=(x^2+4)(x+7)#

Hence

The remaining quadratic is positive for all Real values of

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

So the last two zeros of