Question

Multiply Rational Expressions

I have my notes the teacher gave me but when I attempt to do these problems I'm doin something wrong. I'm supposed to find what x cannot equal.

21x-9/15x^2 • x^2-x-56/3x+21

Answer 1

\[\dfrac{21x -9}{x^2}*\dfrac{x^2-x-56}{3x+21}\]
Right?

Answer 2

Hey, don't throw it here and let others do your homework.

Answer 3

I didn't dump it here for you to do. I was hoping that someone would do it step by step and explain it so I could do my other ones.

Answer 4

Can you please help me with this? @loser66

Answer 5

Hey Arielle :)
\[\Large\rm \frac{21x -9}{x^2}\cdot\frac{x^2-x-56}{3x+21}\]We need to find restrictions on x?

Answer 6

Yes... like what x cannot equal.

Answer 7

In the land of math, there are some really dirty dingey places that you want to avoid...\[\Large\rm \sqrt{-4} \quad\leftarrow\quad the~ghetto\]bad bad place, don't go there.

Answer 8

the 21x-9 should be over 15x^2 -35x^3
I guess i must have messed that up when i typed it

Answer 9

\[\Large\rm \frac{4}{0}\quad\leftarrow\quad docks~at~night\]Another bad place. Don't go here. We can't divide by 0 in the land of math, ya?
That's what leading to restrictions in our problem.

Certain values of x will lead to a 0 showing up in the bottom.

Answer 10

\[\Large\rm \frac{12x-9}{15x^2 -35x^3}\cdot\frac{x^2-x-56}{3x+21}\]Like this? :) Oh ok

Answer 11

Yes :)

Answer 12

If you coukd explain what youre doing and why thatd be great!!!

Answer 13

We have a problem when our equation looks like this:\[\Large\rm \frac{12x-9}{0}\cdot\frac{x^2-x-56}{3x+21}\]Or when our equation looks like this:\[\Large\rm \frac{12x-9}{15x^2 -35x^3}\cdot\frac{x^2-x-56}{0}\]Zeros in the bottom, bad bad bad.


So this is a little nerve wracking i know :O
We're like the police, we're going to go to that scary place...
and then we'll say, "no! you can't do that!"

Answer 14

So we'll let one of our factors in the bottom equal 0,
\(\Large\rm 15x^2-35x^3=0\)
and now we want to solve for x, we want to figure out which x values make this happen.

Answer 15

So to solve for x here.. ummm
looks like we need to do some factoring...
each term has something in common.
What can we factor out of them?

Answer 16

What do you think? :)
Gimme

Answer 17

15x?

Answer 18

15x^2?

Answer 19

Mmm that's pretty close! :)
Yes, we can pull 2 of the x's out.
But we can't take a 15 out of 35, I guess we'll have to settle with a 5, ya?

Answer 20

that makes sense

Answer 21

\[\Large\rm 15x^2-35x^3=0\]\[\Large\rm 5x^2(3-7x)=0\]Do you understand how to solve for x in this setup?
Do you remember your Zero Factor Property?

Answer 22

I have to move my 3 and -7x over, square both sides, then move my 7x back to combine like terms and then divide that by both sides?

Answer 23

No no, too fancy.
Zero Factor Property tells us that when we multiply two things together and get 0, at least one of them is zero.
\[\Large\rm a\cdot b=0, \qquad\to\qquad a=0,\quad\text{and/or}\quad b=0\]When we deal with x like this, we say that both will be zero at different times.

So from here,\[\Large\rm 5x^2(3-7x)=0\]We can jump to here,\[\Large\rm 5x^2=0\qquad\text{and}\qquad (3-7x)=0\]

Answer 24

And we'll look for solutions separately.
Process look familiar I hope? XD

Answer 25

Yes i remember that

Answer 26

5x^2=0,
divide by 5,
x^2=0
square root,
x=0

Ok that gives us a solution, ya?
How bout the other equation. Can you find the x value? :)

Answer 27

Yes. for the other one I subtracted my 3 and divided by -7 and got 3/7

Answer 28

Ok good good good.
So we've found two values for x that are bad!
We went to that bad place, and found em.

So we can confidently say that \(\Large\rm x\ne0,\quad x\ne\frac{3}{7}\).

Answer 29

We've found two restrictions for x.
Our other denominator will give us the other ones.

Answer 30

okayy

Answer 31

\[\Large\rm 3x+21=0\]So repeat the process with the other denominator.
Turns out this one is much simpler though hehe

Answer 32

Factor? 3(x+7)

Answer 33

and then equal it to zero and solve?

Answer 34

You could, yes.
But notice that both terms don't have x's in them.
So we don't have to be as picky with this one.

We could simply subtract 21 to the other side,
divide by 3,
ya?

Answer 35

Yes so the answer i s-7?

Answer 36

Ok great.
So our restrictions will be:\[\Large\rm x\ne0,\quad x\ne\frac{3}{7},\quad x\ne-7\]

Answer 37

Those are all different values that would cause a 0 to show up in the denominator.

Answer 38

Make a little more sense I hope? 0_o
Math is weird, i know i know :D

Answer 39

Okayy. So my teacher had taught us a way to just factor and then cross out like terms and x could not equal the opposite of those... So if i factor out each part of the equation and do it that way would this be correct for an answer left from the euation? 3/5x^2 times x-8/3 with x not equaling 0, 3/7, -7

Answer 40

So your instructions are to, not only find restrictions, but to also simplify?
Ok then ya I guess you have to factor out the top and bottom and do a bunch of fun stuff.

We already factored the bottoms earlier, so our equation is:\[\Large\rm \frac{12x-9}{5x^2(3 -7x)}\cdot\frac{x^2-x-56}{3(x+7)}\]Yah you would cross out like terms :)
So ummm the top...

Answer 41

well she had said factor those too and cross out each two terms alike so i crossed out x+7 and the 3-7x on both leaving 3/5x^2 times x-8/3

Answer 42

No no no silly.
You cross out something from the top, with something from the bottom.
Those are both in the bottom, they don't cross out.
And further more, they're not the same, so they wouldn't cross out anyway.

I don't understand where this x-8/3 is coming from 0_o

Answer 43

Oh I think I see what you're saying maybe.. ummm

Answer 44

Oh it was a 21 lolol my bad

Answer 45

well when i factored i got 3(7x-3) and if you solve that and the bottom denominator the equal the same so i can cross them out...

Answer 46

\[\Large\rm \frac{21x-9}{5x^2(3 -7x)}\cdot\frac{x^2-x-56}{3(x+7)}\]Becomes,\[\Large\rm \frac{-3(3-7x)}{5x^2(3 -7x)}\cdot\frac{(x-8)(x+7)}{3(x+7)}\]You can't cross out a (7x-3) with a (3-7x).
Do you understand why I factored a -3 out instead of a 3 in the upper left? :)

Answer 47

yes i see now.

Answer 48

Could you walk me through another problem, but division this time?

Answer 49

yaaa maybe :)
open a new question though >.< this one is getting really laggy with all of the LaTeX

Answer 50

okayy.