How do you find all the zeros of #f(x) = x^3-4x^2+9x-36#?

1 Answer
May 23, 2018

Answer:

The zeros are:

#x=4" "# and #" "x=+-3i#

Explanation:

Given:

#f(x) = x^3-4x^2+9x-36#

Note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

#x^3-4x^2+9x-36 = (x^3-4x^2)+(9x-36)#

#color(white)(x^3-4x^2+9x-36) = x^2(x-4)+9(x-4)#

#color(white)(x^3-4x^2+9x-36) = (x^2+9)(x-4)#

#color(white)(x^3-4x^2+9x-36) = (x^2+3^2)(x-4)#

#color(white)(x^3-4x^2+9x-36) = (x^2-(3i)^2)(x-4)#

#color(white)(x^3-4x^2+9x-36) = (x-3i)(x+3i)(x-4)#

So the zeros are:

#x=4" "# and #" "x=+-3i#