Question

Simplify:
x^2-9 (over) / x^2+6x+9

Answer 1

\[\frac{ x ^{2}-9 }{x ^{2}+6x+9 }\]

Answer 2

factor both polynomials

Answer 3

I'm not good at factoring is my problem .. @Peter14

Answer 4

what multiplies to -9 and adds to zero?

Answer 5

-3 and 3? @Peter14

Answer 6

yes,
so x^2 -9 = (x+3)(x-3)
got this ?

Answer 7

Oh okay, yeah that makes sense. So then what about the bottom?

Answer 8

or you can use the formula \(a^2-b^2= (a+b)(a-b)\)
a= x and b= 3 here

Answer 9

ok, for x^2+6x+9
you need to find 2 numbers whose product is 9 and sum is 6

Answer 10

3 & 3?

Answer 11

yes.
so, i split 6x as 3x+3x
x^2+6x+9 = x^2+3x+3x+ 9
now what can you factor out from 1st 2 terms x^2+3x .... ?
now what can you factor out from last 2 terms 3x+9 ..... ?

Answer 12

You can factor out the x & then the 3 ?

Answer 13

X in the first, 3 in the second.

Answer 14

Wait so would the bottom look like \[x^{2}(x+3)(x+3)+9\]

Answer 15

more like
\(\large x^2+3x+3x+9= x(x+3)+3(x+3)\)

Answer 16

Okay

Answer 17

then you can factor out x+3 entirely
\(\large x^2+3x+3x+9= x(x+3)+3(x+3)= (x+3)(x+3)\)
go this?
what gets cancelled ?

Answer 18

So I have this so far correct?:

\[\frac{(x+3)(x-3) }{x(x+3)+3(x+3)}\]

Answer 19

denominator = (x+3)(x+3)
by factoring out x+3 entirely

Answer 20

and then something gets cancelled!

Answer 21

x+3 gets cancelled then right?

Answer 22

yes! :)

Answer 23

what remains is your answer...

Answer 24

So then I'm left with: \[\frac{ x-3 }{ x+3 }\]?

Answer 25

correct :)
thats exactly your answer.

Answer 26

Thank you so much!

Answer 27

you are always welcome ^_^