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

1 Answer
May 29, 2016

Answer:

#x=3,x=±isqrt6 #

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.