Question

How do you factor the trinomial `2x^2+ 11x-9`?

Answer 1

`2x^2+11x-9 = 1/8(4x+11-sqrt(193))(4x+11+sqrt(193))`

Explanation:

Given:

`2x^2+11x-9`

First note that this would factor really quickly if the constant was `+9` instead of `-9`:

`2x^2+11x+9 = 2x^2+2x+9x+9`

`color(white)(2x^2+11x+9) = 2x(x+1)+9(x+1)`

`color(white)(2x^2+11x+9) = (2x+9)(x+1)`

It is still possible to factor it using irrational coefficients.

Here's one way by completing the square and using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

`8(2x^2+11x-9) = 16x^2+88x-72`

`color(white)(8(2x^2+11x-9)) = (4x)^2+2(4x)(11)+121-193`

`color(white)(8(2x^2+11x-9)) = (4x+11)^2-(sqrt(193))^2`

`color(white)(8(2x^2+11x-9)) = ((4x+11)-sqrt(193))((4x+11)+sqrt(193))`

`color(white)(8(2x^2+11x-9)) = (4x+11-sqrt(193))(4x+11+sqrt(193))`

So:

`2x^2+11x-9 = 1/8(4x+11-sqrt(193))(4x+11+sqrt(193))`