Polynomial Long Division

**franzpulk** · October 11th 2010, 12:43 PM

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.

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

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

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

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.

**pickslides** · October 11th 2010, 01:22 PM

Originally Posted by franzpulk

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

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

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

**franzpulk** · October 11th 2010, 01:27 PM

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.

**pickslides** · October 11th 2010, 01:40 PM

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 $-7x^5\div 3x^2 = \frac{-7x^3}{3}$ Take that result and multiply it through the divisor.

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

What do you get?

Spoiler:

**franzpulk** · October 11th 2010, 01:57 PM

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

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

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

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.

**RHandford** · October 11th 2010, 02:14 PM

Hi

I know nothing but have a look here

**franzpulk** · October 11th 2010, 02:20 PM

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.

**RHandford** · October 11th 2010, 02:23 PM

Hi

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

**pickslides** · October 11th 2010, 02:27 PM

Originally Posted by franzpulk

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

This was the answer after the subtraction! not the multiplication part

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

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

**franzpulk** · October 11th 2010, 02:39 PM

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.

**franzpulk** · October 11th 2010, 02:41 PM

Originally Posted by pickslides

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

I'm still trying to process this. How did you get this?

**franzpulk** · October 11th 2010, 02:45 PM

@RH -

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

Now, I just need to understand it.

**pickslides** · October 11th 2010, 04:26 PM

Can you follow this?

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

**franzpulk** · October 11th 2010, 04:56 PM

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:

$3x^2$ cannot divide evenly into $x^5$ , 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.

**rtblue** · October 12th 2010, 07:13 AM

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

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

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

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

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

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