Question

(t+9)^2/t+8 * (t+8)^2/t+9

mulitply and simplify

Answer 1

\[\frac{(t+9)^2}{(t+8)}*\frac{(t+8)^2}{(t+9)}\]

See anything you can cancel to simplify before you multiply everything out?

Answer 2

not really im so lost

Answer 3

What if I rewrite it like this:
\[\frac{(t+9)^2}{(t+9)}*\frac{(t+8)^2}{(t+8)}\]

Answer 4

would those be 1?

Answer 5

Here, let me rewrite it yet again :-)

\[\frac{(t+9)(t+9)}{(t+9)}*\frac{(t+8)(t+8)}{(t+8)}\]

Answer 6

idk that just made me a little more lost lol sorry

Answer 7

I haven't done math since high school

Answer 8

Okay, do you agree that \[(t+9)^2 = (t+9)(t+9)\]?

Answer 9

yes i get that one

Answer 10

Good. So the last thing I did was I just replaced \( (t+9)^2\) with \((t+9)(t+9)\) and \( (t+8)^2\) with \((t+8)(t+8)\) — are we cool with that?

Answer 11

that wont change the answer??

Answer 12

no, they are equal! (something)^2 is just (something)(something)

Answer 13

ok whats next

Answer 14

And in the first step, all I did was switch the denominators around — that's okay because they are getting multiplied together, and a*b = b*a, right?

Answer 15

ok

Answer 16

so, no reason to be confused :-)

Now, what can we do with a fraction if the same thing appears in the numerator and the denominator?

Answer 17

Take for example
\[\frac{2*3}{3*3}\]

Answer 18

Do you remember canceling matching things in fractions?

Answer 19

\[\frac{3}{3} = 1\] right?

Answer 20

not at all =(

Answer 21

i remember that

Answer 22

Okay, so our fraction could be written as\[\frac{2*3}{3*3} = \frac{2}{3}*\frac{3}{3} = \frac{2}{3}*1 = \frac{2}{3}\]Okay with that?

Answer 23

ok

Answer 24

Great. That's all we need to make this problem much easier :-)

\[\frac{(t+9)(t+9)}{(t+9)} * \frac{(t+8)(t+8)}{(t+8)} = \frac{(t+9)}{(t+9)}*(t+9)*\frac{(t+8)}{(t+8)}*(t+8)\]

Answer 25

What does \[\frac{(t+9)}{(t+9)}\] equal?

Answer 26

1

Answer 27

Excellent! And the same goes for (t+8)/(t+8). What do we have left after we turn those two fractions in 1s?

Answer 28

into 1s, that is

Answer 29

Doesn't it become
\[1*(t+9)*1*(t+8) = (t+9)(t+8)\]?

Answer 30

Do you remember how to expand that product?

Answer 31

not at all ugh

Answer 32

Okay, it's straightforward. If you've got (a+b)(c+d), that is equal by the distributive property to:
\[(a+b)(c+d) = a(c+d) + b(c+d) = ac + ad + bc + bd\]
Basically, each term of the first thing gets multiplied by each term of the second thing

Your problem is\[(t+9)(t+8) = \] I'll do a similar one as an example, in case the symbols don't make sense to you:

\[(x+3)(x+4) = x(x+4) + 3(x+4) = x*x + 4*x + 3*x + 3*4 = \]\[x^2 + 4x + 3x + 12 = x^2 + 7x + 12\]

Can you do the same procedure with yours? I'll check your answer...

Answer 33

t(t+8)+9(t+8)

Answer 34

Okay, that's the first step. What do you get after you expand that?

Answer 35

i get losttttt

Answer 36

\[t(t+8)+9(t+8) = t*t + t*8 + 9(t+8)\]Can you do the other half?

Answer 37

t^2+9t+8t...

Answer 38

+ 9*8
\[t(t+8) + 9(t+8) = t*t + t*8 + 9*t + 9*8 = t^2 + 17t + 72\]

That's your final answer.

Here's a nice video on multiplying binomials:
https://www.khanacademy.org/math/algebra/polynomials/multiplying_polynomials/v/multiplying-binomials

There are many good videos on that site, and it would probably be worth your time to watch some and brush up on your skills.

I have to leave now, but I'll leave you with a different way of thinking of this multiplication problem:

Another way to think of it is as if you are doing multiplication of multiple-digit numbers:

t + 9
X t + 8
-------
8t +72 <- this is 8 * (t+9)
t^2 + 9t <- this is t * (t+9)
-------------
t^2+17t+72

Answer 39

is that in lowest terms??

Answer 40

By the way, aren't you glad we canceled things out first? Otherwise, we would have had
\[\frac{t^4+34t^3+433t^2+2448t+5184}{t^2+17t+72}\] :-)

Yes, \[t^2+17t+72\] is in lowest terms, and cannot be further simplified. We don't have any fractions, we don't have any parentheses, and we only have one term with each power of t.

Answer 41

its telling me that is wrong

Answer 42

Well, maybe you copied the problem wrong...

Answer 43

... no i did it exactly

Answer 44

i appreciate you sticking with me though

Answer 45

is that factored form??

Answer 46

http://www.wolframalpha.com/input/?i=%28t%2B9%29%5E2%2F%28t%2B8%29+*+%28t%2B8%29%5E2%2F%28t%2B9%29

Answer 47

See the first choice under alternate forms.

Answer 48

Anyhow, I have to go now. Have a look at those videos, the lecturer is usually pretty clear. Feel free to shoot me a message if you get stuck on something and I'll try to help you out the next time I'm online.

Answer 49

ok thanks