dividing polynomials

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kimmy2892000

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Nov 12, 2005
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Can you tell me if I have done these 2 problems right?

(4h^5k-6h^2k^2)/(-2h^2k)
=(4h^5k/-2h^2k)-(6h^2k^2/-2h^2k
= 2h^3+3k

2u^3-13u^2-8u+7/u-7
2u^3-13u^2/u-7 = 2u^2
u^2-8u/u-7 = u
-u+7/u-7 = -1

answer = 2u^2+u-1

Thanks for your help.
Kimmy
 
the first one is right...but the second equation, you cant do the same because the denomonator is put together different...

u-7 isnt multiplied the like -2h^2k so you cant do this one like the first equation
what i mean is instead of multipling u and 7, your subtracting them

my best guess is to put the original equation equal to zero and multiply both sides by u-7 then go from there....


If your still confused repost and ill see if i can explain it better :)
 
did you understand how to do it now or...?

keep in mind its just a guess...its been a couple of months since i've done this.
 
Are you familiar with synthetic division? If you are having trouble long dividing in the second problem, try doing it synthetically.
 
I am not familiar with synthetic division...but I did get a problem solver to check it and the answer is correct. Thanks.
 
You can do some factoring.
(4h^5k-6h^2k^2)/(-2h^2k) =
(2h^2k)(2h^3-3k)/(-2h^2k) =
-(2h^3-3k)
You were close but the minus sign gotcha.

Long division works fine. You can show it as with the code button. Click twice on it then type between the boxes.
2u^3-13u^2-8u+7/u-7
Code: 
        2u^2 +  u    -1
      _______________________
  u-7 | 2u^3 -13u^2  -8u  +7
        2u^3 -14u^2
        -----------
                u^2  -8u
                u^2  -7u
                ----------
                      -u  +7
                      -u  +7

Sometime when you have nothing to do, look at
http://www.purplemath.com/modules/synthdiv.htm
You might like it.
 
Hello, Kimmy!

The recommended approach is Factor-and-Cancel.

\(\displaystyle \L\frac{4h^5k\,-\,6h^2k^2}{-2h^2k}\)
Click to expand...
Factor: .\(\displaystyle \L\frac{2h^2k(2h^3-3k)}{-2h^2k}\)

Cancel: .\(\displaystyle \L\frac{2h^3\,-\,3k}{-1}\:=\:-2h^3\,+\,3k\)


\(\displaystyle \L\frac{2u^3\,-\,13u^2\,-\,8u\,+\,7}{u\,-\,7}\)
Click to expand...
The <u>only</u> chance we have to simplify is if the denominator divides exactly into the numerator.
. . Well, does it divide exactly? . . . We can use long division to find out.

And we find that: .\(\displaystyle (2u^3\,-\,13u^2\,-\,8u\,+\,7)\,\div\,(u\,-\,7)\:=\:2u^2\,+\,u\,-\,1\)

Therefore: .\(\displaystyle \L\frac{(u\,-\,7)(2u^2\,+\,u\,-\,1)}{u\,-\,7}\:=\:2u^2\,+\,u\,-\,1\)

[Edit: Beat me again, Gene!]
 
kimmy2892000 said:
Can you tell me if I have done these 2 problems right?

(4h^5k-6h^2k^2)/(-2h^2k)
=(4h^5k/-2h^2k)-(6h^2k^2/-2h^2k
= 2h^3+3k : close! should be: 3k - 2h^3

2u^3-13u^2-8u+7/u-7
2u^3-13u^2/u-7 = 2u^2
u^2-8u/u-7 = u
-u+7/u-7 = -1
answer = 2u^2+u-1 : CORRECT; good work 8-)

Thanks for your help.
Kimmy
Click to expand...
 
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