Question

Is it possible to factor `y=2x^2 + 9x - 5`? If so, what are the factors?

Answer 1

`y = (2x-1)(x+5)`

Explanation:

As it so happens, any quadratic can be factored into the form

`ax^2 + bx + c`
`= a(x - (-b+sqrt(b^2-4ac))/(2a))(x-(-b-sqrt(b^2-4ac))/(2a))`

That big block comes from the quadratic formula, along with the fact that given a polynomial `P(x)`, if `P(a)=0`, then `(x-a)` is a factor of `P(x)`.

However, sometimes (when `b^2-4ac < 0`) the factors involve complex numbers, and there is also a lot to calculate there. Let's see if we can find some real factors without resorting to that right away.

If we have the factorization
`(Ax + B)(Cx + D) = 2x^2 +9x - 5`
then expanding the left hand side and comparing coefficients gives us the system

`{(AC = 2), (AD+BC = 9), (BD = -5):}`

If we limit ourselves to integer solutions, then we quickly see from the first and third equations that `A` and `C` have to be `+-1` and `+-2` and that `B` and `D` have to be `+-1` and ∓`5`. A quick check will show that

`{(A = 2), (B = -1), (C = 1), (D = 5):}`

is a solution. Thus, substituting back, we have

`(2x-1)(x+5) = 2x^2+9x-5`