# Homework Help: How do I factor this trinomial

1. Sep 7, 2012

### priceofcarrot

So I can't find an example in my book that shows how to factor a trinomial like this one:

2(x)^3 + 3(x)^2 - 1

I tried finding a number that multiplied to -2 and added to 3, but that didn't work. I then tried just factoring x out of the equation, but I didn't know what to do with the -1 that was left over.

Is there another method? Thanks

2. Sep 7, 2012

### CAF123

Have you tried synthetic division?

3. Sep 7, 2012

### LCKurtz

Do you know the rational roots theorem?

4. Sep 7, 2012

### SammyS

Staff Emeritus

Maybe try graphing it.

5. Sep 7, 2012

### HallsofIvy

The rational roots theorem that LCKurtz mentioned is good. Or just "try" simple numbers are roots, or graphing it as SammyS suggested. Those are all ways of finding roots. And the "remainder theorem" that SammyS also suggested says that if x= a satisfies makes a polynomial equal to 0, then x- a is a factor. And could then use "synthetic division" as CAF123 suggested to find another factor.

6. Sep 8, 2012

### priceofcarrot

I tried doing the find a value of x that makes the polynomial equal 0, and I got 0.5 by trial and error, but when I then went to divide the trinomial in the original post by x - 0.5, I got a remainder. Doesn't that mean that x - 0.5 isn't a factor of the trinomial?

But how is that possible if x-0.5 makes the trinomial equal 0? Did I make a mistake in my division?

Thanks

7. Sep 8, 2012

### LCKurtz

Yes. Try again. And use $1/2$ instead of .5, maybe that will help your arithmetic. Or show us your work.

8. Sep 8, 2012

### priceofcarrot

Last edited by a moderator: May 6, 2017
9. Sep 8, 2012

### SammyS

Staff Emeritus
Last edited by a moderator: May 6, 2017
10. Sep 8, 2012

### LCKurtz

And you left out the $0x$ term in the dividend. If you have trouble with long division, why don't you just try plugging $x=1/2$ into the equation and see if it works?

11. Sep 8, 2012

### priceofcarrot

I put 0.5 in the original trinomial, and it equalled 0. But the question wants me to completely factor it. Doesn't that mean I need to use another tool to get the 2nd factor?

12. Sep 8, 2012

### Bohrok

Just try doing the long division again and you'll get a quadratic that you can factor; no need for any other tool to get the other factors.

13. Sep 9, 2012

### priceofcarrot

The first one is 3x^2 - 1, and the second is 5x^2 - 1.

I just did the long division again, and this time I got a remainder of -1 + 2x. I don't understand what I'm doing wrong. I've successfully done long division of polynomials before this. I also read all the posts in this topic. Although I don't understand why I should include 0x in the dividend. Isn't that always 0?

14. Sep 9, 2012

### SammyS

Staff Emeritus
No matter whether you take
2x3 + 4x2 - 1 - (2x3 - x2)​
or
2x3 + 4x2 - 1 - 2x3 + x2
the result is the same:
5x2 -1 .​

Now, divide (5x2 -1) by (x - 1/2) .

Last edited: Sep 9, 2012
15. Sep 9, 2012

### LCKurtz

Corrections in red below:

No matter whether you take
2x3 + 3x2 - 1 - (2x3 - x2)​
or
2x3 + 3x2 - 1 - 2x3 + x2
the result is the same:
4x2 -1 .​

Now, divide (4x2 -1) by (x - 1/2) .

16. Sep 9, 2012

### vela

Staff Emeritus
You just didn't finish. You can still divide 2x-1 by x-1/2. You keep going until the degree of the remainder is less than the degree of the divisor.

17. Sep 9, 2012

### ehild

Why don't you try a negative root, -1 for example?

ehild

18. Sep 10, 2012

### priceofcarrot

Thanks a lot for this post. I finally got it to work. I ended up using -1 though after ehild raised it as a possibility. I see that it would have worked with 0.5 too though.

I also corrected the mistake I made in my original work.

Thanks a lot guys!

19. Sep 10, 2012

### ehild

Now, that the problem has been solved by the OP, I can show an other way of factoring the expression.
2x3+3x2-1=(2x3+2x2)+(x2-1)=2x2(x+1)+(x+1)(x-1)=(x+1)(2x2+x-1)=(x+1)(x+1)(2x-1)

ehild

20. Sep 10, 2012

### SammyS

Staff Emeritus
Nice !