Question

*help please! ^^ will give a medal and fan!!!*


[(x+3)/x] times [(x+4)/(x^2+7x+12)]

Answer 1

\[\frac{ x+3 }{ x} * \frac{ x+4 }{ x^{2}+7x+12 }\]
Thank you so much for helping!!

Answer 2

What do you get when you factor x^2 + 7x + 12

Answer 3

factoring x^2 + 7x + 12 is -4, and -3.

Answer 4

two numbers that multiply to 12 and add to 7 are 4 and 3

so that means x^2 + 7x + 12 factors to (x+4)(x+3)

Answer 5

\[\large \frac{ x+3 }{ x} * \frac{ x+4 }{ x^{2}+7x+12 }\]
\[\large \frac{ x+3 }{ x} * \frac{ x+4 }{ (x+3)(x+4) }\]

which terms cancel?

Answer 6

i'm not really getting this, could you try to explain it further please? thank you:)

Answer 7

Do you see how x^2 + 7x + 12 factors to (x+4)(x+3) ?

Answer 8

yes, i understand that!

Answer 9

so we factor and then we cancel like so...

\[\large \frac{ x+3 }{ x} * \frac{ x+4 }{ x^{2}+7x+12 }\]
\[\large \frac{ x+3 }{ x} * \frac{ x+4 }{ (x+3)(x+4) }\]
\[\large \frac{ x+3 }{ x} * \frac{ \cancel{x+4} }{ (x+3)\cancel{(x+4)} }\]
\[\large \frac{ x+3 }{ x} * \frac{ 1 }{ x+3 }\]
\[\large \frac{ \cancel{x+3} }{ x} * \frac{ 1 }{ \cancel{x+3} }\]
\[\large \frac{ 1 }{ x} * \frac{ 1 }{ 1 }\]
\[\large \frac{ 1 }{ x} * 1\]
\[\large \frac{ 1 }{ x}\]
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This means
\[\large \frac{ x+3 }{ x} * \frac{ x+4 }{ x^{2}+7x+12 }\]
simplifies to
\[\large \frac{ 1 }{ x}\]

Answer 10

The cancellations occur because when you divide any number by itself (excluding 0) you get 1

examples

2/2 = 1
7/7 = 1
22/22 = 1

in general

x/x = 1

where x is not zero

Answer 11

and the 1s go away because 1*x = x

examples

1*9 = 9
1*17 = 17
etc etc

Answer 12

I'm still not quite understanding where you are going with this? @jim_thompson5910

Answer 13

do you see the steps I listed out?

Answer 14

that went from

\[\large \frac{ x+3 }{ x} * \frac{ x+4 }{ x^{2}+7x+12 }\]

to

\[\large \frac{ 1 }{ x}\]

Answer 15

yes, how would i continue to solve this problem?

Answer 16

once you get to \[\large \frac{ 1 }{ x}\] you are done

Answer 17

because it's fully simplified

Answer 18

ahh, i see!! thank you so much for explaining it so thoroughly. :)

Answer 19

you're welcome