How do you find all the zeros of #f(x) = x^3-4x^2+9x-36#?
1 Answer
May 23, 2018
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#
