x^2 - 2x - 5/x - 3 ÷ x - 5/x^2 - 9

- anonymous

x^2 - 2x - 5/x - 3 ÷ x - 5/x^2 - 9

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- schrodinger

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- Nnesha

factor the quadratic equation

- Nnesha

\[\huge\rm \frac{ \frac{ x^2-2x-5 }{ x-3 } }{ \frac{ x-5 }{ x^2-9 }}\]
and change division to multiplication
to do that multiply top fraction with the RECIPROCAL of the bottom fraction
example \[\huge\rm \frac{ \frac{ a }{ b } }{ \frac{ c }{ d } }=\frac{ a }{ b } \times \frac{ d }{ c}\]

- Astrophysics

\[\frac{ x^2-2x-5 }{ x-3 } \div \frac{ x-5 }{ x^2-9 }\] you need to factor the numerator \[(x^2-2x-5)\] and remember when you divide by fractions you flip the second fraction and multiply, \[\frac{ a }{ b } \div \frac{ c }{ d } \implies \frac{ a }{ b } \times \frac{ d }{ c }\]

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## More answers

- Nnesha

x^2-9
apply difference of squares rule \[\huge\rm a^2-b^2 =(a+b)(a-b)\]

- Astrophysics

You will have to complete the square or use quadratic formula for x^2-2x-5

- anonymous

I'm trying to keep up! Could you run me through the factoring process really quickly? I think I must be doing something wrong because my results don't match any of the possible answers.

- Nnesha

are you sure its x^2-2x-5 ??

- Astrophysics

Yeah I don't think that's right either haha

- anonymous

I'm so sorry! It's x^2 - 2x - 15, not 5. But the rest should be right!

- Nnesha

o^_^o

- Astrophysics

Oh ok so now just find two numbers that add up to -2 and multiply together to give -15, can you think of two?

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Juliette2120
I'm trying to keep up! Could you run me through the factoring process really quickly? I think I must be doing something wrong because my results don't match any of the possible answers.
\(\color{blue}{\text{End of Quote}}\)
show your work plz :)

- anonymous

3 and - 5?

- Astrophysics

Perfect!

- myininaya

did you factor x^2-9 yet @Juliette2120

- Astrophysics

So you have \[\frac{ (x-5)(x+3) }{ x-3 } \times \frac{ x^2-9 }{ x-5 }\] now you can notice we can cancel out some terms, but lets first factor \[x^2-9\] as nnesha mentioned here \(\color{blue}{\text{Originally Posted by}}\) @Nnesha
x^2-9
apply difference of squares rule \[\huge\rm a^2-b^2 =(a+b)(a-b)\]
\(\color{blue}{\text{End of Quote}}\)

- myininaya

an example
\[x^2-25=(x-5)(x+5) \\ \text{ notice: I just replaced } a \text{ with } x \text{ and } b \text{ with 5 } \\ \text{ since } x^2-25=x^2-5^2\]

- myininaya

can you write 9 as a some number squared?

- anonymous

@myininaya 3^2?

- myininaya

right
so a is x
and b is 3
in this case

- myininaya

\[x^2-9=x^2-3^2 \\ x^2-3^2=(x-3)(x+3)\]

- anonymous

@Astrophysics I'm having a difficult time understanding the difference of squares rule! Could you explain that to me? Or show me how to factor this specific example?

- anonymous

@myininaya Okay! I understand this. What's the next step?

- Astrophysics

Well @myininaya just showed it but, it can be a bit tricky as it's not very intuitive to go from \[a^2-b^2 \implies (a+b)(a-b)\] unless you distribute \[(a+b)(a-b)\] itself

- myininaya

\[\frac{ x^2-2x-15 }{ x-3 } \div \frac{ x-5 }{ x^2-9 } \\ \frac{x^2-2x-15}{x-3} \times \frac{x^2-9}{x-5}
\]

- Astrophysics

I guess for now just remember it, \[a^2-b^2 \implies x^2-3^2 \]

- myininaya

notice to change it to multiplication we just flip the second fraction

- myininaya

now let's put in all of our factored forms

- myininaya

\[\frac{(x-5)(x+3)}{x-3} \cdot \frac{(x-3)(x+3)}{x-5} \\ \frac{(x-5)(x+3)(x-3)(x+3)}{(x-3)(x-5)}\]
do you see anything that cancels?

- anonymous

Do they have to be on the same level to cancel?

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Juliette2120
@Astrophysics I'm having a difficult time understanding the difference of squares rule! Could you explain that to me? Or show me how to factor this specific example?
\(\color{blue}{\text{End of Quote}}\)
take square root of both terms
(sqrt of 1st term + sqrt of 2nd term)(sqrt of 1st term - sqrt of 2nd term)
easy to remember.

- myininaya

you have to have a factor on top that matches a factor on bottom to cancel that common factor

- anonymous

So it would be (x - 5) and (x - 3)?

- Astrophysics

Nope you can always factor them out, also remember for example\[\frac{ (x+1) }{ (x+1) } = 1\]

- Astrophysics

I guess I'll let @myininaya help you too much information all at once lol

- anonymous

@Astrophysics I'm sorry about that! You've both been very helpful - it's all making a bit more sense!

- myininaya

as you see you have an (x-5) on top and bottom
so as @Astrophysics says (x-5)/(x-5)=1

- myininaya

or in other words you can cancel the (x-5) on top with the one on bottom

- myininaya

do you see anything else that can cancel?

- anonymous

Is (x - 3) one?

- myininaya

yes
\[\frac{(x-5)(x+3)}{x-3} \cdot \frac{(x-3)(x+3)}{x-5} \\ \frac{\cancel{(x-5)}(x+3)\cancel{(x-3)}(x+3)}{\cancel{(x-3)}\cancel{(x-5)}} \]

- myininaya

\[\frac{(x+3)(x+3)}{1} \text{ or } (x+3)(x+3)\]
do you know how to multiply (x+3)(x+3) out?

- anonymous

So... It would be (x + 3)^2?

- myininaya

yes that is right

- myininaya

can we leave it as (x+3)^2?

- myininaya

or do they want it in standard form?

- anonymous

Yes! That's an answer!

- myininaya

k!
do you want to talk more about the difference of squares formula?

- anonymous

Trust me - I will be back shortly with more questions! I've been doing really well with math, but for some reason this module has me stumped!

- myininaya

just in case you need something to look back on in the future:
\[(a-b)(a+b) =a(a+b)-b(a+b) \\ (a-b)(a+b)=a(a)+a(b)-b(a)-b(b) \\ (a-b)(a+b)=a^2+ab-ab-b^2 \\ (a-b)(a+b)=a^2-b^2 \\ \text{ Examples: } \\ x^2-1=(x-1)(x+1) \\ x^2-4=(x-2)(x+2) \\ x^2-9=(x-3)(x+3) \\ x^2-16=(x-4)(x+4) \\ x^2-25=(x-5)(x+5) \\ \text{ More Examples: } \\ 4x^2-25=(2x-5)(2x+5) \text{ note: I hope you see that } 4x^2=(2x)^2 \text{ and } 25=5^2\]

- anonymous

Thank you so much! I'll keep that!

- anonymous

Thanks to all of you for your help! @myininaya @Astrophysics @Nnesha

- myininaya

@Astrophysics and @Nnesha are the most awesomest! :)
Yes I know that isn't a word.
And @Juliette2120 you did great too.

- Astrophysics

Haha, thanks and no problem! Everyone was great! XD

- Nnesha

my pleasure.

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