# Thread: Polynomial Long Division

1. ## Polynomial Long Division

Ok, let me start by saying that I know how to do polynomial long divison. At least I thought I did. But, I've encountered one that's giving me trouble.

$\displaystyle (-7x^5-9x^4+5x^3-9x^2-8x+7)/(3x^2+8x+1)$

I'm new to formatting this stuff on the web, but so far, I've gotten $\displaystyle -2x^3$ as my quotient, with $\displaystyle x^5+7x^4+7x^3-9x^2-8x+7$ still left to be divided.

The problem is that I don't see how you can divide $\displaystyle 3x^2$ into $\displaystyle x^5$.

Am I missing something?

I can attempt to format the work I've done thus far, but I'm hoping that won't be necessary. If it is, can anyone give me an example of how to format long division on here?

Thanks.

2. Originally Posted by franzpulk

The problem is that I don't see how you can divide $\displaystyle 3x^2$ into $\displaystyle x^5$.

$\displaystyle x^5\div 3x^2 = \frac{x^3}{3}$

Originally Posted by franzpulk

I can attempt to format the work I've done thus far, but I'm hoping that won't be necessary. If it is, can anyone give me an example of how to format long division on here?
I've never tried this myself. Could be a world of pain. Until then check this link. Its a goody.

Polynomial Long Division

3. Thanks or the reply, but that doesn't make a bit of sense to me.
That isn't something that we've covered at all.

Also, I've been to that site, as well as purplemath. Every example they give, the factor divides evenly every time.
I can do those, no problem. It's this one that's giving me grief.

4. Well I think this example is a good one. In your case you need to deal with some fractions and maybe that's what is giving you a hard time.

First compare the leading terms of each expression $\displaystyle -7x^5\div 3x^2 = \frac{-7x^3}{3}$ Take that result and multiply it through the divisor.

$\displaystyle \frac{-7x^3}{3}\times (3x^2+8x+1) = \dots$ and subtract this from the original dividend.

What do you get?

Spoiler:
$\displaystyle \displaystyle \frac{29x^4}{3}+\frac{22x^3}{3}-9x^2$

5. First, thanks for the help.
Second, that confuses me even more.

If I were to do that ($\displaystyle \frac{-7x^3}{3}\times (3x^2+8x+1) = \frac{29x^4}{3}+\frac{22x^3}{3}-9x^2$), I'd end up with $\displaystyle \frac{-21x^5}{3}+\frac{-56x^4}{3}+\frac{-7x^3}{3}$

I just don't see how you came up with: $\displaystyle \displaystyle \frac{29x^4}{3}+\frac{22x^3}{3}-9x^2$

I posted a pic of what I have so far on the problem I posted about, as well as a similar problem that's also giving me trouble.

6. Hi

I know nothing but have a look here

7. Thanks for the link, but it still doesn't tell me how to get the answer.
I could just just plug the answer in for a correct answer (this is from my online homework), but that's not what I'm after.
If I don't understand the concepts, the answers aren't going to do squat for me.

8. Hi

No I appreciate that, but I thought that this may help

9. Originally Posted by franzpulk

I just don't see how you came up with: $\displaystyle \displaystyle \frac{29x^4}{3}+\frac{22x^3}{3}-9x^2$
This was the answer after the subtraction! not the multiplication part

In your working you need to eliminate the $\displaystyle x^5$ term.

I would suggest you reload with some easier examaples to get a better understanding before tackling a problem this tricky.

10. The problem is, this is on my homework. It's nothing we've covered, but it's due tomorrow night. And, with it being fall break on campus, there is no one here to talk to about it.
I've already emaild my prof regarding the lack of instruction on this type of problem, but haven't yet received a response.
So, without a response, and with it being due tomorrow night, it's up to me to understand it and do it by then.

11. Originally Posted by pickslides
$\displaystyle x^5\div 3x^2 = \frac{x^3}{3}$
I'm still trying to process this. How did you get this?

12. @RH -

I plugged that into my homework and got the answer correct.

Now, I just need to understand it.

13. Can you follow this?

$\displaystyle x^5\div 3x^2 = \frac{x^5}{ 3x^2} = \frac{x\times x \times x \times x \times x }{3 \times x \times x }= \frac{x^3}{3}$

14. I should've been more clear. I understand cancelling out exponents/variables within a fraction.
I guess what I'm not so clear on is when do you know you'll have to do a step like that?
Would you need to do something like this when you're left with a leading coefficient on the variable of the dividend that's smaller than the leading coefficient in the divisor?

For example:

$\displaystyle 3x^2$ cannot divide evenly into $\displaystyle x^5$, so would this ALWAYS tell me to do the problem as discussed? Or is there some other circumstance that should tell me to divide only the leading dividend variable by the leading divisor variable?

I hope all that made sense. I was having trouble making sense of it myself as I was typing it.

BTW, thanks for all the help thus far.

15. When you look at a problem like this, you have to ask yourself: "What when multiplied by $\displaystyle 3x^2$ will give me $\displaystyle 7x^5$. Since 3 is not a divisor of seven, you know that there will be a fraction involved. So we have:

$\displaystyle 3x^2*\frac{7x^3}{3}=7x^5$

Thus, we come to the conclusion that the first term of the coefficient is:

$\displaystyle \frac{7x^3}{3}$

Continue the problem as such, and you will arrive at the answer.

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