Question

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

Answer 1

`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.