Multiplying/Dividing Rational Expressions

**Shrinkwrap** · October 12th 2008, 09:18 PM

Hey, this is actually a College class but im not very good at math so Im in an Intermediate Algebra, I still think it most likely goes here but, anyway... I missed a lecture on some of this and while I managed to learn a majority of the chapter myself im stuck on a few problems. I simply dont even understand how to start them.

Find the product - Rational Expressions/Equations
(Sofar I canceled ad, and 3b but thats about all I got)

ac + 3a +2c +6......ad - 5a + 2d - 10
--------------- X -----------------
ad + a + 2d + 2......bc + 3b - 4c - 12

Find the quotient - Rational Expressions/Equations
(I flipped the second fraction in both already, they were originally division but thats about all I got)

4a³ - 8a².....5a^4
--------- X -------
.....15a......3a² - 6a

y² + y - 20...........y
----------- X --------------
......7x²........3y² + 19y + 20

Those three I am really having trouble with. Sorry if I posted them wrong, I couldnt get the math code to work. I tried making them as readable as possible.

Thanks

**earboth** · October 12th 2008, 10:03 PM

Originally Posted by Shrinkwrap

...

Find the quotient - Rational Expressions/Equations[/B]
(I flipped the second fraction in both already, they were originally division but thats about all I got)
Those three I am really having trouble with. Sorry if I posted them wrong, I couldnt get the math code to work. I tried making them as readable as possible.[/INDENT]
Thanks

$\dfrac{ac + 3a +2c +6}{ad + a + 2d + 2} \cdot \dfrac{ad - 5a + 2d - 10}{bc + 3b - 4c - 12 } = \dfrac{a(c + 3) +2(c +3)}{a(d + 1) + 2(d + 1)} \cdot \dfrac{a(d - 5) + 2(d - 5)}{b(c + 3) - 4(c +3) } =$ $\dfrac{(a+2)(c + 3) }{(a+2)(d + 1) } \cdot \dfrac{(a+2)(d - 5)}{(b-4)(c + 3)} = \dfrac{(a+2)(d-5)}{(d+1)(b-4)}$

**earboth** · October 12th 2008, 10:20 PM

Originally Posted by Shrinkwrap

...

Find the quotient - Rational Expressions/Equations[/B]
(I flipped the second fraction in both already, they were originally division but thats about all I got)
....

$\dfrac{4a^3 - 8a^2}{15a} \cdot \dfrac{5a^4}{3a^2 - 6a }= \dfrac{4a^2(a-2)}{15a} \cdot \dfrac{5a^4}{3a(a-2)}= \dfrac{20a^6}{45a^2} = \dfrac49a^4$

I get you started with the last one: (I changed the $7x^2$ into $7y^2$ )

$\dfrac{y^2 + y - 20}{7y^2} \cdot \dfrac{y}{3y^2 + 19y + 20 } = \dfrac{(y-4)(y+5)}{7y^2} \cdot \dfrac{y}{(3y+4)(y+5)} =$

**Shrinkwrap** · October 13th 2008, 10:55 AM

oh, very nice... thank you very much. Much easier now that you've shown me how its done.

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