Question

How do you factor `5a^2-6a+5`?

Answer 1

Use the quadratic formula to find:

`5a^2-6a+5 = 1/5(5a-3-4i)(5a-3+4i)`

Explanation:

To try to avoid confusion, I will factor `5x^2-6x+5`, which is of the form `ax^2+bx+c`, with `a = 5`, `b = -6` and `c = 5`.

This has discriminant `Delta` given by the formula:

`Delta = b^2 - 4ac = (-6)^2 - (4xx5xx5) = 36 - 100`

`= -64 = -8^2`

Since this is negative, `5x^2-6x+5 = 0` only has Complex roots, given by the quadratic formula:

`x = (-b +-sqrt(Delta))/(2a) = (6+-sqrt(-8^2))/(2xx5) = (6+-8i)/10 = (3+-4i)/5`

Hence:

`5x^2-6x+5 = 1/5(5x-3-4i)(5x-3+4i)`

So:

`5a^2-6a+5 = 1/5(5a-3-4i)(5a-3+4i)`