Question

How do you factor `5a^2-3a+15`?

Answer 1

`5a^2-3a+15 = 1/20(10a-3-sqrt(291)i)(10a-3+sqrt(291)i)`

Explanation:

Given:

`5a^2-3a+15`

This is of the form:

`Aa^2+Ba+C`

with `A=5`, `B=-3` and `C=15`

It has discriminant `Delta` given by the formula:

`Delta = B^2-4AC = (-3)^2-4(5)(15) = 9-300 = -291`

Since `Delta < 0` this quadratic has no real zeros and no linear factors with real coefficients.

We can still factor it, but we need complex coefficients.

We can complete the square, then use the difference of squares identity:

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

with `A=10a-3` and `B=sqrt(291)i` as follows:

`20(5a^2-3a+15) = 100a^2-60a+300`

`color(white)(20(5a^2-3a+15)) = (10a)^2-2(10a)(3)+9+291`

`color(white)(20(5a^2-3a+15)) = (10a-3)^2+(sqrt(291))^2`

`color(white)(20(5a^2-3a+15)) = (10a-3)^2-(sqrt(291)i)^2`

`color(white)(20(5a^2-3a+15)) = ((10a-3)-sqrt(291)i)((10a-3)+sqrt(291)i)`

`color(white)(20(5a^2-3a+15)) = (10a-3-sqrt(291)i)(10a-3+sqrt(291)i)`

So:

`5a^2-3a+15 = 1/20(10a-3-sqrt(291)i)(10a-3+sqrt(291)i)`