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

May 29, 2016

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 ${x}^{3} - 3 {x}^{2} + 6 x - 18 = 0$

group the terms in 'pairs' thus

$\left[{x}^{3} - 3 {x}^{2}\right] + \left[6 x - 18\right]$

now factorise each pair.

$\Rightarrow {x}^{2} \left(x - 3\right) + 6 \left(x - 3\right) = \left(x - 3\right) \left({x}^{2} + 6\right)$

we now have $\left(x - 3\right) \left({x}^{2} + 6\right) = 0$

solve x - 3 = 0 → x = 3

solve x^2+6=0rArrx^2=-6rArrx=±isqrt6

Thus there is 1 real and 2 complex zeros.