Question

3x^2-2x-5 factor the trump all out of that problem

Answer 1

Will do!

A few things to note about this problem... : 3x^2 - 2x - 5
The coefficient in front of x^2 (3) can only be expressed as 3, thus the start of the factor will look like this: 

(3x...

The constant (-5) is negative so that means the two numbers which we choose must be a negative and a positive. Meaning two possible sign orientations are as followed: (The _ are blanks)

(3x - _)(x + _)
(3x + _)(x - _)

The x on the second factor is need to make 3x^2. Now to keep going we need two numbers that multiplied together equal -5. 

The only numbers that make sense are ±1, and ±5.

Now we have to somehow arrange it so that one of the two multiplied by 3 and added to the other equals -2. 

Let's see our options: 
3(-5), 1
5, 3(-1)
3(5), -1
-5, 3(1)

Let's try each of these out. If they work then they'll equal -2.
3(-5) + 1 = -15 + 1 = -14 
NOPE!

5 + 3(-1) = 5 - 3 = 2
NOPE! Close though! 

3(5) - 1 = 15 - 1 = 14
NOPE!

-5 + 3(1) = -5 + 3 = -2 
Checks out!

So now we plug it in!
Remember the 5 needs to be negative, and 1 has to be multiplied by 3 (It has to be in the second factor... below is what I mean)

(3x - 5)(x + 1) 

Let's check it by multiplying it all out again.

3x*x + 3x*1 - 5x - 5
3x^2 + 3x - 5x - 5 
3x^2 - 2x - 5

Checks out. Hope this helped! 

Answer 2

Trial and error
(3x-5)(x+1)
woo, good first guess


anywa, the real legit way is

3x^2-2x-5

find
-2=a+b such that a is a factor of 3 and b is a factor of -5
-2=3+-5

so split it up
3x^2+3x-5x-5
factor
(3x)(x+1)+(-5)(x+1)
undistribute
(3x-5)(x+1)

tada