Question

Simplify

Answer 1

\[x ^{2}-9/x ^{2}+5x+6 * 1/x+3 \]

Answer 2

\[\frac{ x^2 - 9 }{ x^2 + 5x + \frac{ 6*1 }{ x+3 } } \]
is this right?

Answer 3

or is it more like this?

\[\frac{ x^2−9 }{ x^2+5x+6 } \times \frac{ 1 }{ x+3 }\]

?? @charlieberzak

Answer 4

ummm

Answer 5

i got three answers :D

Answer 6

1

1/x+2

x-3/(x+2)(x+3)

Answer 7

yeah but what's the question?

Answer 8

above

Answer 9

\[x^2 - \frac{ 9 }{ x^2 } + 5x + 6\times \frac{ 1 }{ x+3 }\]
???

Answer 10

which one?

Answer 11

here: http://curriculum.kcdistancelearning.com/courses/ALG1x-HS-A06/b/Exams/9Quiz2/9Quiz2_q6.gif

Answer 12

ok so as it's multiply the 2 fractions, simply multiply straight across ie numerator x numerator and denominator by denominator
ie 1/6 * 4 / 10 = (1*4) / (6 * 10) = 4 / 60

Answer 13

then simplify the single fraction by removing factors from top and bottom

Answer 14

Nooooo. DO NOT multiply first! Factor the num'r and den'r of the first rational expression, first.

ALWAYS better to factor and reduce first, THEN multiply. This goes for fractions as well as rational expressions, but expecially here, taking that product first will be one HOT MESS. Factor first and things will cancel. You'll be left with something very simple when you multiply.

Answer 15

\[\Large \dfrac{ x^2−9 }{ x^2+5x+6 } \times \dfrac{ 1 }{ x+3 }\]

That first numerator is a difference of squares, so it factors:
\(\large a^2-b^2=(a-b)(a+b)\)

The first den'r also factors into the product of 2 binomials ("undo" the FOIL).

Answer 16

And in fact, in your answer choices everything is in factored form, or reduces completely, so you won't even really "multiply" when you multiply.

Answer 17

\[\frac{ x^2-9 }{ (x+3) (x^2+5x+6) } = \frac{ x^2-9 }{ (x+3) (x+2) (x+3) }\]
... wasn;t that hard and it still dosent simplify ...?

Answer 18

x^2 - 9 = (x+3) (x-3) yeah?

Answer 19

so u can remove one of those as a factor if u like @charlieberzak

Answer 20

lol ok, you didn't really "multiply" in that den'r, you wrote it all as a product in one rational expression, but when you told him to "multiply across" I thought you were telling him to actually TAKE THE PRODUCT of that trinomial with the binomial in the den'r (that would, indeed, be a hot mess).

What you did in the den'r is exactly what I was telling him to do - factor.

And yes, it is important that the num'r also factors. And you don't "remove one of those as a factor if u like". The problem is to "SIMPLIFY" the expression, so you MUST reduce all common factors.

What you are left with is one of the answer choices.

Answer 21

true

Answer 22

\[(x ^{2}-9) \div (x^2+5x+6)(x+3)=(x+3)(x-3)\div(x+2)(x+3)(x+3) = (x-3)\div((x+2)(x+3)\]

Answer 23

is it 1?

Answer 24

Read what's above, @charlieberzak. @jack1 did pretty much the whole problem for you, just reduce the common factors. What's left??

\[\frac{ x^2-9 }{ (x+3) (x^2+5x+6) }= \frac{(x+3)(x-3) }{ (x+3) (x+2) (x+3) }\]

Answer 25

THe only way it would be =1 is if ALL the factors in num'r and den'r cancelled.

Answer 26

(x+3) (x-3)??

Answer 27

???

For what? That isn't even one of your answer choices.

That's the num'r, before you reduce.

Answer 28

What does this reduce to??

\(\Large \dfrac{(x+3)(x-3) }{ (x+3) (x+2) (x+3) }\)

Answer 29

o so its would have to be the third choice

Answer 30

Yes.

but you really need to understand how to factor these, yourself.

Answer 31

x-3
--------
(x+2)(x+3)

Answer 32

\[a^2-b^2=(a+b)(a-b) \] so by applying it to \[x^2-9=x^2-3^2\]

Answer 33

we get (x+3)(x-3)

Answer 34

dude, appreciate the sentiment but maybe @DebbieG is a better choice for that, considering who stayed around and helped you vs who went a looked at pictures of cats instead (sorry they're just hilarious sometime tho) ;D

Answer 35

@charlieberzak ^^

Answer 36

:D

Answer 37

No worries @jack1.... :) And who doesn't like looking at pictures of cats?? :)