Question

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

Answer 1

`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)`