# How do you find the zeros of #x^3-3x^2+6x-18#?

##### 1 Answer

May 29, 2016

#### Explanation:

To find the zeros , the values of x that make the expression equal to zero, we require to factorise the expression.

our aim is to solve

#x^3-3x^2+6x-18=0# group the terms in 'pairs' thus

#[x^3-3x^2]+[6x-18]# now factorise each pair.

#rArrx^2(x-3)+6(x-3)=(x-3)(x^2+6)# we now have

#(x-3)(x^2+6)=0# solve x - 3 = 0 → x = 3

solve

#x^2+6=0rArrx^2=-6rArrx=±isqrt6 # Thus there is 1 real and 2 complex zeros.