# How do you find all the zeros of #f(x) = x^4 - 9x^3 + 24x^2 - 6x - 40#?

##### 1 Answer

#### Answer:

#### Explanation:

First notice that if you reverse the signs of the coefficients on the terms of odd degree then the sum is

#1+9+24+6-40 = 0#

So

#x^4-9x^3+24x^2-6x-40 = (x+1)(x^3-10x^2+34x-40)#

Let

By the rational root theorem, any rational zeros of

That means that the only possible rational zeros are:

#+-1# ,#+-2# ,#+-4# ,#+-5# ,#+-8# ,#+-10# ,#+-20# ,#+-40#

In addition, note the

So the only possible rational zeros of

#1, 2, 4, 5, 8, 10, 20, 40#

We find:

#g(4) = 64-160+136-40 = 0#

So

#x^3-10x^2+34x-40#

#=(x-4)(x^2-6x+10)#

The remaining quadratic factor has negative discriminant, so its zeros are Complex, but we can factor it as a difference of squares:

#x^2-6x+10#

#=(x-3)^2-9+10#

#=(x-3)^2+1#

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

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

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

So the remaining zeros are:

#x = 3+-i#