Question

How do you factor `5x^2-3x+4`?

Answer 1

`5x^2-3x+4 = 1/20(10x-3-sqrt(71)i)(10x-3+sqrt(71)i)`

Explanation:

Given:

`5x^2-3x+4`

We can factor with non-real complex coefficients by completing the square and using the difference of squares identity:

`A^2-B^2 = (A-B)(A+B)`

with `A=10x-3` and `B=sqrt(71)i`

I will first multiply by `2^2 * 5 = 20`, then divide by `20` at the end - in order to avoid much arithmetic with fractions.

`20(5x^2-3x+4) = 100x^2-60x+80`

`color(white)(20(5x^2-3x+4)) = (10x)^2-2(10x)(3)+3^2+71`

`color(white)(20(5x^2-3x+4)) = (10x-3)^2+(sqrt(71))^2`

`color(white)(20(5x^2-3x+4)) = (10x-3)^2-(sqrt(71)i)^2`

`color(white)(20(5x^2-3x+4)) = ((10x-3)-sqrt(71)i)((10x-3)+sqrt(71)i)`

`color(white)(20(5x^2-3x+4)) = (10x-3-sqrt(71)i)(10x-3+sqrt(71)i)`

So:

`5x^2-3x+4 = 1/20(10x-3-sqrt(71)i)(10x-3+sqrt(71)i)`