# How do you find the roots of #x^3-5x^2+4x+10=0#?

##### 1 Answer

#### Answer:

The roots are:

#### Explanation:

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

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

#+-1, +-2, +-5, +-10#

We find:

#f(-1) = -1-5-4+10 = 0#

So

#x^3-5x^2+4x+10 = (x+1)(x^2-6x+10)#

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

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

with

#x^2-6x+10 = x^2-6x+9+1#

#color(white)(x^2-6x+10) = (x-3)^2-i^2#

#color(white)(x^2-6x+10) = ((x-3)-i)((x-3)+i)#

#color(white)(x^2-6x+10) = (x-3-i)(x-3+i)#

Hence the other two zeros are: