# Thread: How to factor complex trinomials?

1. ## How to factor complex trinomials?

How do you factor trinomials where the coeffecient a is something other than one?

such as: 6x^2 + x -1 and 2x^2 - 7x + 3 ?

i also have a bit of problem with trigonometry and rather than making a second thread i thought it would be more efficient to just include it in this thread?

Solve each equation for 0 < theta (Θ) < 2 pi (2π)

2sin^2 θ + sin θ - 6 = 0

(1 + 1/tan^2 θ)(1 - sin^2 θ) = 1/tan^2 θ

i hope y'all don't think i'm a dummy, i'm actually doing pretty good in summer school.

2. You should really make different threads for different questions, especially when the threads would normally be in different forums...

Anyway, for the trinomials question, for a trinomial $ax^2 + bx + c$, multiply your $a$ and $c$ values together. Whatever two numbers you pick has to multiply to give this number, and add to give the $b$ value.

So for the first...

$6x^2 + x - 1$

multiplying the $a$ and $c$ values gives $-6$.

So your two numbers have to multiply to give $-6$ and add to be $1$. So choose $3, -2$.

Now we break up the middle term using these two values and factorise by grouping.

$6x^2 + x - 1 = 6x^2 + 3x - 2x - 1$

$= 3x(2x + 1) - 1(2x + 1)$

$= (2x + 1)(3x - 1)$.

Try the second one yourself

For the trig questions:

The first one is a quadratic in $\sin{\theta}$. So let $x = \sin{\theta}$ to give

$2x^2 + x - 6 = 0$

which you should be able to factorise using the same method I showed you and solve using NFL. Once you have $x$, you know what $\sin{\theta}$ is, so you should be able to find $\theta$.

For the second trig question, try converting everything to sines and cosines, using $\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$. Everything should fall into place. The Pythagorean Identity $\sin^2{\theta} + \cos^2{\theta} = 1$ will also come in handy.

3. thanks a bundle.

4. You should also understand that most trinomials (in a very precise sense "almost all" trinomials) cannot be factored using integers as coefficients.

For example $(x- 3+ \sqrt{5})(x- 3+ \sqrt{5}= (x- 3)^2- 5= x^2- 6x+ 4$ cannot be factored with integer coefficients. And there are simply too many possibilities to use Prove It's method if the coefficients are allowed to be irrational.

5. Originally Posted by HallsofIvy
You should also understand that most trinomials (in a very precise sense "almost all" trinomials) cannot be factored using integers as coefficients.

For example $(x- 3+ \sqrt{5})(x- 3+ \sqrt{5}= (x- 3)^2- 5= x^2- 6x+ 4$ cannot be factored with integer coefficients. And there are simply too many possibilities to use Prove It's method if the coefficients are allowed to be irrational.
True, but to avoid the rigmarole of completing the square, it's always best to check if it can be factorised over the integers using the method I've explained.