Question

Fan and medal
Multiply. Show all work. Make sure to simplify. Assume no denominator equals zero.

Answer 1

Like i always tell my Algebra students, don't let the length of the problem intimidate you; take it one step at a time. Look at each fraction individually, numerator then denominator. In the first fraction, you have 8m - 8. Is there any way you can see to simplify that one at all or is it as simple as it can be?

Answer 2

Its as simple as it could be

Answer 3

@IMStuck

Answer 4

Ok, but look closer at it. There is an 8 with the m, and an 8 after the - sign. Can we factor an 8 out of that, do you think?

Answer 5

Then it would just be m?

Answer 6

Here, let's do this. If you divide 8 by 8 what do you get?

Answer 7

U get 1, sorry my computer is really slow right now

Answer 8

That's right. Also, of course if you divide the 8 after the -sign, you also get 1. The m cannot be divided out by an 8, so it stays put. Here is what it looks like to factor 8m -8\[8(m-1)\]You took an 8 out of 8m and you are left with 1m or just m. You take an 8 out 8 and you are left with 1. If you FOIL that outside that is outside the parenthesis back into the m - 1, you would get 8m - 8 right back. FOILing is the opposite of factoring. Do you see what I did with that? Does it make sense?

Answer 9

I kind of bumbled up that post. It's supposed to say if you FOILed the 8 that is outside the parenthesis back into the m - 1, you would 8m - 8 back.

Answer 10

yeah it makes sense

Answer 11

Ok, so what we have so far is this. And keep in mind that I am doing this one thing at a time, so it might take a while to get through it, but it may help tremendously.\[\frac{ 8(m-1) }{ m ^{2}+8m+7 }\times \frac{ m ^{2}+10m+21 }{ -2m+2 }\] Now we will work on the denominator of the first fraction.

Answer 12

Are you familiar with how to factor polynomials into two sets of parenthesis?

Answer 13

Yep

Answer 14

Ok, then try your hand at \[m ^{2}+8m+7\]

Answer 15

I don't think that can be factored, don't I have to use the quadratic formula?

Answer 16

What two numbers multiply to be 7 and add to be 8? There is only one set of numbers that multiply to get 7. What are they?

Answer 17

Ohhhh 1 and 7, wow I didn't think of that

Answer 18

There you go! so what do your two sets of parenthesis look like then?

Answer 19

Ok, I got (m+7)(m+1)

Answer 20

That's right. Now so far we have\[\frac{ 8(m-1) }{ (m+7)(m+1) }\times \frac{ m ^{2}+10m+21 }{ -2m+2 }\]Now let's look at the second numerator there,\[m ^{2}+10m+21\]Can you factor that into two sets of parenthesis as well?

Answer 21

(m+7)(m+3)

Answer 22

right again! So NOW we have \[\frac{ 8(m-1) }{ (m+7)(m+1) }\times \frac{ (m+3)(m+7) }{ -2m+2 }\]Lat thing you have to do is to factor that last denominator. You can't get that factored into two sets of parenthesis, though, can you? But what is common to both terms (the -2m and the 2 after the plus sign)? Something common that you can take out, just like you did with the 8m - 8

Answer 23

Would the common number be 2?

Answer 24

Or 1?

Answer 25

The common number would be 2, but take it out as a negative 2 like this\[-2(m-1)\]because when you FOIL the -2 back in you get -2m + 1. Remember your rules for multiplying negative and positive numbers. Ok? see that? NOW we have almost come to the end here. What we have left is\[\frac{ 8(m-1) }{ (m+7)(m+1) }\times \frac{ (m+3)(m+7) }{ -2(m-1) }\]Do you know how to simplify from here?

Answer 26

No im not sure, should it be multiplied by the reciprocal or..?

Answer 27

You only have to multiply by a reciprocal when you are dividing a rational expression by another rational expression. Here, now that we are multiplying, we can cross cancel out like terms. Getting these simplified showed us that we have several things in common now that we can cancel. BUT we can only cancel from top to bottom and bottom to top...NEVER top and top and bottom and bottom. Here