# How do you find all the zeros of #f(x)=x^3-2x^2-11x+52#?

##### 1 Answer

#### Answer:

#### Explanation:

#f(x) = x^3-2x^2-11x+52#

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

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

#+-1, +-2, +-4, +-13, +-26, +-52#

In addition, note that the signs of the coefficients are in the pattern

The pattern of signs of coefficients of

So check for negative rational zeros first:

#f(-1) = -1-2+11+52 = 60#

#f(-2) = -8-8+22+52 = 58#

#f(-4) = -64-32+44+52 = 0#

So

#x^3-2x^2-11x+52 = (x+4)(x^2-6x+13)#

The discriminant of the remaining quadratic factor is negative, so it has no Real zeros, but we can factor it with Complex coefficients:

#x^2-6x+13#

#=(x-3)^2-9+13#

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

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

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

Hence zeros: