Question

t^2-49/t^2+4t-21 * t^2+8t+15/t^2-2t-35

multiply or divide

Answer 1

\[\frac{ t^2-49 }{ t^2+4t-21 } * \frac{ t^2+8t+15 }{ t^2-2t-35 }\]

Answer 2

first you need to factor everything

Answer 3

You know how to do that or do you need help factoring? ^^

Answer 4

im ok at factoring

Answer 5

see if you can factor:
t^2-49
t^2+4t-21
t^2+8t+15
t^2-2t-35

Answer 6

(t+7)(t-7)/(t+3)(t-7)

Answer 7

Close.
(t+3)(t-7) would be t^2-7t+3t-21 which would be t^2-4t-21 instead of +4t

Answer 8

so the first fraction factors into\[\frac{ (t+7)(t-7) }{ (t+7)(t-3) }\]

Answer 9

See if you can factor the other fraction

Answer 10

(t+3)(t+5)/ i dont know im having trouble with the bottom

Answer 11

Well, it has to be factors that multiply together to get -35, and add together to get -2
7 and 5 make 35, and because the 2 is negative the bigger factor also has to be negative.
(t-7)(t+5)

Answer 12

o okay i see .

Answer 13

so you have:
\[\frac{ (t+7)(t−7) }{ (t+7)(t−3) } * \frac{ (t+3)(t+5) }{ (t-7)(t+5) }\]

Answer 14

Now you cross out any like terms

Answer 15

would it be 0 or im doing t way wrong

Answer 16

\[\frac{ (t-7)(t+3) }{ (t-3)(t-7) }\]

Answer 17

So you're left with:
\[\frac{ t+3 }{ t-3 }\]

Answer 18

what happen to the (t-7) and (t+7)

Answer 19

They were both negative so they cancled out

Answer 20

canceled *

Answer 21

o okay

Answer 22

would t+3/t−3 be the answer or there another stoep

Answer 23

step

Answer 24

Thats it ^^

Answer 25

okay thank you . i appericate the help

Answer 26

Mhm^^