# Thread: Having trouble Factoring

1. ## Having trouble Factoring

Here's my question.

Determine the x-intercepts of the polynomial function.

6x^3 + 13x^2 +2x

So I've factored it down to

x(6x^2 + 13x + 2)

Do I have to find 2 numbers that are the product of 2 and the sum of 13? I don't think it's possible. What do I do?

2. Originally Posted by Arise01 Here's my question.

Determine the x-intercepts of the polynomial function.

6x^3 + 13x^2 +2x

So I've factored it down to

x(6x^2 + 13x + 2)

Do I have to find 2 numbers that are the product of 2 and the sum of 13? I don't think it's possible. What do I do?
Good start.

To factor this won't be easy. To start with, you have a $\displaystyle 6x^2$ to consider.

This means you could have something of the form (3x+...)(2x+...) or you could have something of the form (6x+...)(x+...) as either will generate a 6x^2 when expanded.

The way to do this is to write out the factors of 6:

2,3
1,6

And write out the factors of 2:

1,2

And find combinations that make 13 by multiplying pairs of numbers and adding them. It's basically trial and error here.

Starting with '2,3' and '1,2':

2 x 1 + 3 x 2 = 8
2 x 2 + 3 x 1 = 7

Neither combination is going to work. Note that 2 x 3 + 1 x 2 is not a combination to consider.

So it must be from 1,6 and 1,2
1 x 6 + 1 x 2 = 8
1 x 1 + 6 x 2 = 13

So there's the combination we need. Then you need to think about the correct arrangements. You need to multiply 6 and 2, so they need to be in opposing brackets. This means the only possible result is:
(x+2)(6x+1)

3. So it would be x(x+2)(6x+1) ?

4. Originally Posted by Arise01 So it would be x(x+2)(6x+1) ?
Remember that it's an equation you need, not an expression. You must write x(x+2)(6x+1)=0.
...So what are your three solutions this time?

5. x=0, x+2=0, 6x+1=0

So 0, -2, 1/6 are the x-intercepts?

6. Originally Posted by Arise01 x=0, x+2=0, 6x+1=0

So 0, -2, 1/6 are the x-intercepts?
Nearly, there's a sign error:

If 6x+1=0
then 6x=-1
So x=-1/6

The other two are fine.

7. Thanks again quacky.

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