How do you factor completely #9x^2 + 20x + 100#?

1 Answer
Feb 2, 2017

Answer:

#9x^2+20x+100 = (3x+10/3-20/3sqrt(2)i)(3x+10/3+20/3sqrt(2)i)#

Explanation:

Given:

#9x^2+20x+100#

This is in the form:

#ax^2+bx+c#

with #a=9#, #b=20# and #c=100#.

This quadratic has discriminant #Delta# given by the formula:

#Delta = b^2 - 4ac = 20^2-4(9)(100) = 400-3600 = -3200#

Since #Delta < 0# this quadratic has no Real zeros and no linear factors with Real coefficients.

We can factor it using Complex coefficients by completing the square as follows:

#9x^2+20x+100 = (3x)^2+2(3x)(10/3)+(10/3)^2-(10/3)^2+100#

#color(white)(9x^2+20x+100) = (3x+10/3)^2+800/9#

#color(white)(9x^2+20x+100) = (3x+10/3)^2-(20/3sqrt(2)i)^2#

#color(white)(9x^2+20x+100) = ((3x+10/3)-20/3sqrt(2)i)((3x+10/3)+20/3sqrt(2)i)#

#color(white)(9x^2+20x+100) = (3x+10/3-20/3sqrt(2)i)(3x+10/3+20/3sqrt(2)i)#