Multiplying and Dividing Polynomials

demha · 2012-12-05 23:56:10

I am having a hard time understanding how to do these correctly. Could someone please help explain how to do them step by step?

1. (3x + 6x)/3x

2. (21x3 + 14x) / 7

3. (4x5 + 8x2) / 4x

4. [2(x 1) + 3(x 1)] / (x 1)

5. [3(2x 3) x(2x 3)] / (2x - 3)

6. [x2(5x + 6) 3(5x + 6)] / (5x + 6)

7. [3x2(2x - 3) + 27x3(2x - 3)] / [3x(3 2x)]

Bob · 2012-12-06 01:29:40

hi demha

OK let's start with Q1

Now you could simplify the top first but, looking at the later ones, it is probably better to split into two fractions.

Can you finish this one from here?

Bob

demha · 2012-12-08 18:51:18

Sorry I haven't answered sooner. I have been busy with other matters.

To finish from there. If I'm correct, we have to reduce it.
So:

3x/3x + 6x/3x

Reducing:
1x/1x + 3x/1x

For first fraction:
1x goes into 1x one time so it will become just x.

For second fraction:
1x goes into 3x three times so it will become 3x.

So the problem becomes x + 3x. So the answer is 4x?

Agnishom · 2012-12-08 18:58:33

For first fraction:
1x goes into 1x one time so it will become just x.

No it will become 1

Similarly, 6x/3x = 2

demha · 2012-12-08 22:03:56

So you don't reduce them?

So then:
3x/3x + 6x/3x

For first fraction:
3x goes into 3x 1 time.

For second fraction:
3x goes into 6x 2 times.

Then it becomes 1 + 2 and of course the answer is 3.

Bob · 2012-12-08 23:53:19

hi demha

That's correct. Shall we try Q2 ?

Try to split this one into two fractions and then simplfy each one.

Bob

demha · 2012-12-09 21:06:42

So that would be set as:

21x^3/7 + 14x/7

For the first fraction:
7 goes into 21x 3x times I suppose. This will leave it with 3x^3?

For the second fraction:
7 goes into 14x 2x times.

The problem then becomes:

3x^3 + 2x

They are not entirely alike so you can't add them. Therefor that is the answer I believe?

Bob · 2012-12-09 21:16:33

hi demha,

Yes, that's correct, well done.

The others are similar, although the expressions get more complicated.

How about trying Q3 and Q4 and then I'll check them.

Bob

demha · 2012-12-09 23:57:09

3.
(4x^5 + 8x^2) / 4x

4x^5/4x + 8x^2/4x

For first equation:
4x goes into 4x 1 time so it will be 1^5

For second equation:
4x goes into 8x 2 times so it will be 2^2.

It will then be:
1^5 + 2^2

I hope I did this one correct.

Number 4 seems to be pretty complicated. I'm not too sure how to start it!

Bob · 2012-12-10 01:36:02

hi demha,

You shouldn't have lost the xs. Have a look at this:

Cancel the 4 top and bottom and one x top and bottom to leave

Bob

demha · 2012-12-10 14:10:14

Oh alright! And the: 8x^2/4x would become 2x?

bobbym · 2012-12-10 14:14:19

That is correct!

demha · 2012-12-11 20:48:03

So how would you do number 4?

Bob · 2012-12-11 21:07:20

hi demha

Q4. Same method:

Bob

demha · 2012-12-11 22:17:39

2(x 1) + 3(x 1) / (x 1)

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

For the first fraction:
so (x - 1) goes into (x - 1) one time making it 2(1). Now I want to stop right here and make sure that I'm doing this right.

Bob · 2012-12-12 04:47:06

yes. first part is 2.

Bob

demha · 2012-12-12 17:46:41

Glad to know I'm on the right track.
Second part of the equation:
3(x 1)/( x - 1)

Same way, (x - 1) goes into (x - 1) 1 time making it 3(1).

So now the equation is:
2 + 3 and the final answer will be 5.

Am I correct?

bobbym · 2012-12-12 19:20:12

Hi;

Correct! Very good!

Bob · 2012-12-12 20:19:59

Well done.

5, 6 and 7 are very similar to 4, so have a go and post your answers.

Bob

demha · 2012-12-12 20:59:09

#5
[3(2x 3) x(2x 3)] / (2x - 3)

First fraction:
3(2x - 3)/(2x - 3)
(2x - 3) goes into (2x - 3) 1 time.
3(1)

Second fraction:
x(2x - 3)/(2x - 3)
(2x - 3) goes into (2x - 3) 1 time.
x(1)

Answer:
3 - x

#6
[x2(5x + 6) 3(5x + 6)] / (5x + 6)

First fraction:
x^2(5x + 6)/(5x + 6)
(5x + 6) goes into (5x + 6) 1 time.
x^2(1)

Second fraction:
3(5x + 6)/(5x + 6)
(5x + 6) goes into (5x + 6) 1 time.
3(1)

x^2 - 3

#7
[3x^2(2x - 3) + 27x^3(2x - 3)] / [3x(3 2x)]

This one seems a little confusing.

---

I also have two more having some trouble with:

8. 15x^7 45x^5 / 3x4

19. (3x 2)(x - 4) (x - 4)(6 5x) / (4 x)(8x 1)

Bob · 2012-12-12 21:17:18

#5
[3(2x 3) x(2x 3)] / (2x - 3)

First fraction:
3(2x - 3)/(2x - 3)
(2x - 3) goes into (2x - 3) 1 time.
3(1)

Second fraction:
x(2x - 3)/(2x - 3)
(2x - 3) goes into (2x - 3) 1 time.
x(1)

Answer:
3 - x
correct!
#6
[x2(5x + 6) 3(5x + 6)] / (5x + 6)

First fraction:
x^2(5x + 6)/(5x + 6)
(5x + 6) goes into (5x + 6) 1 time.
x^2(1)

Second fraction:
3(5x + 6)/(5x + 6)
(5x + 6) goes into (5x + 6) 1 time.
3(1)

x^2 - 3
correct!
#7
[3x^2(2x - 3) + 27x^3(2x - 3)] / [3x(3 2x)]

This one seems a little confusing.

At this stage you can divide each part and then simplify.

(3-2x) goes into (2x-3) exactly minus 1 times

Let's get this one finished and then we can look at 8 and 9.

Bob

---

I also have two more having some trouble with:

8. 15x^7 45x^5 / 3x4

19. (3x 2)(x - 4) (x - 4)(6 5x) / (4 x)(8x 1)

demha · 2012-12-12 22:38:21

[3x^2(2x - 3) + 27x^3(2x - 3)] / [3x(3 2x)]

First one:
3x^2(2x - 3)/3x(3 2x)]

So since it is (2x - 3) and (3 - 2x), it goies into it but it becomes a -1? And as for the 3x^2 and 3x, would this becomes x^2?

demha · 2012-12-12 23:05:42

Wait, I think it just becomes x(-1)

27x^3(2x - 3) / 3x(3 2x)

(2x - 3) goes into (3 - 2x) -1 time. 3x goes into 27 9 times and takes down the ^3 to ^2 which makes it 9^2(-1)

-x + -9^2

Is that correct?

Bob · 2012-12-12 23:54:41

Nearly

takes down the ^3 to ^2 which makes it 9^2(-1)

should be 9x^2(-1)

Bob

demha · 2012-12-14 04:25:43

So the equation is

-x + -9x^2 then?

Math Is Fun Forum

#1 2012-12-05 23:56:10

Multiplying and Dividing Polynomials

#2 2012-12-06 01:29:40

Re: Multiplying and Dividing Polynomials

#3 2012-12-08 18:51:18

Re: Multiplying and Dividing Polynomials

#4 2012-12-08 18:58:33

Re: Multiplying and Dividing Polynomials

#5 2012-12-08 22:03:56

Re: Multiplying and Dividing Polynomials

#6 2012-12-08 23:53:19

Re: Multiplying and Dividing Polynomials

#7 2012-12-09 21:06:42

Re: Multiplying and Dividing Polynomials

#8 2012-12-09 21:16:33

Re: Multiplying and Dividing Polynomials

#9 2012-12-09 23:57:09

Re: Multiplying and Dividing Polynomials

#10 2012-12-10 01:36:02

Re: Multiplying and Dividing Polynomials

#11 2012-12-10 14:10:14

Re: Multiplying and Dividing Polynomials

#12 2012-12-10 14:14:19

Re: Multiplying and Dividing Polynomials

#13 2012-12-11 20:48:03

Re: Multiplying and Dividing Polynomials

#14 2012-12-11 21:07:20

Re: Multiplying and Dividing Polynomials

#15 2012-12-11 22:17:39

Re: Multiplying and Dividing Polynomials

#16 2012-12-12 04:47:06

Re: Multiplying and Dividing Polynomials

#17 2012-12-12 17:46:41

Re: Multiplying and Dividing Polynomials

#18 2012-12-12 19:20:12

Re: Multiplying and Dividing Polynomials

#19 2012-12-12 20:19:59

Re: Multiplying and Dividing Polynomials

#20 2012-12-12 20:59:09

Re: Multiplying and Dividing Polynomials

#21 2012-12-12 21:17:18

Re: Multiplying and Dividing Polynomials

#22 2012-12-12 22:38:21

Re: Multiplying and Dividing Polynomials

#23 2012-12-12 23:05:42

Re: Multiplying and Dividing Polynomials

#24 2012-12-12 23:54:41

Re: Multiplying and Dividing Polynomials

#25 2012-12-14 04:25:43

Re: Multiplying and Dividing Polynomials

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