Question

How can this be simplified to the most basic form?

`4sec^2x - 11secx - 3`

Answer 1

`(secx-3)(4secx+1)`

Explanation:

Take a look at the expression. We have a squared term, a first-degree term, and a constant. This can be treated as a quadratic, except instead of dealing with `x,` we're dealing with `secx.`

Let's temporarily say `u=secx` and rewrite everything in terms of `u:`

`4u^2-11u-3`

Now this perfectly resembles a quadratic. Let's use the Quadratic Formula to factor:

`u=(-b+-sqrt(b^2-4ac))/(2a), a=4,b=-11,c=-3`

`u=(11+-sqrt(121-(4)(4)(-3)))/(2*4)`

`u=(11+-sqrt(169))/(2*4)=(11+-13)/8=3, -1/4`

`u=3, u=-1/4`

If `u=-1/4, 4u=-1`.
We do this because we generally don't want fractions in our factored form when it's avoidable.

Writing in factored form:

`(u-3)(4u+1)`

We simply moved our solutions for `u` over to the same side as `u`, which resulted in switching the signs.

Replacing `u` with `secx:`

`(secx-3)(4secx+1)`