Question

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

Answer 1

`2x^2+x-5=color(red)(2(x+(1+sqrt(41))/4)(x+(1-sqrt(41))/4))`

Explanation:

Apply the quadratic formula
`color(white)("XXX") x=(-b+-sqrt(b^2-4ac))/(2a)`
`color(white)("XXXXXXXXX")` where `ax^2+bx+c=0`

to `2x^2+x-5=0`

obtaining the solutions
`color(white)("XXX") x=(-1+-sqrt(41))/4`
and thus the factors
`color(white)("XXX") (x-(-1+sqrt(41))/4) and (x-(-1-sqrt(41))/4)`

Since
`color(white)("XXX") (x+(1-sqrt(41))/4)(x+(1+sqrt(41))/4)=x^2+x/2-5/2`
we will have an additional constant factor of `2`

Answer 2

`f(x) = 2(x + 1/4 - sqrt41/4)(x + 1/4 + sqrt41/4)`

Explanation:

Use the quadratic function in intercept form (Google Search):
`f(x) = a(x - x_1)(x - x_2).`

`x_1 and x_2` are the `2` real roots of `f(x) = 0`.

They are given by the improved quadratic formula (Socratic Search):

`x = - b/(2a) +- d/(2a)`

In this case:

`D = b^2 - 4ac = 1 + 40 = 41 rarr d = +- sqrt41`
`x = - 1/4 +- sqrt41/4`
`x_1 = - 1/4 + sqrt41/4`
`x_2 = - 1/4 - sqrt41/4`

`f(x) = a(x - x_1)(x - x_2)`

`f(x) = 2(x + 1/4 - sqrt41/4)(x + 1/4 + sqrt41/4)`