Question

x^2 - 2x - 5/x - 3 ÷ x - 5/x^2 - 9

Answer 1

factor the quadratic equation

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

o^_^o

Answer 11

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

Answer 12

\(\color{blue}{\text{Originally Posted by}}\) @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 :)

Answer 13

3 and - 5?

Answer 14

Perfect!

Answer 15

did you factor x^2-9 yet @Juliette2120

Answer 16

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 \[\huge\rm a^2-b^2 =(a+b)(a-b)\]
\(\color{blue}{\text{End of Quote}}\)

Answer 17

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

Answer 18

can you write 9 as a some number squared?

Answer 19

@myininaya 3^2?

Answer 20

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

Answer 21

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

Answer 22

@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?

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

now let's put in all of our factored forms

Answer 29

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

Answer 30

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

Answer 31

\(\color{blue}{\text{Originally Posted by}}\) @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.

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

do you see anything else that can cancel?

Answer 40

Is (x - 3) one?

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

yes that is right

Answer 45

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

Answer 46

or do they want it in standard form?

Answer 47

Yes! That's an answer!

Answer 48

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

Answer 49

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!

Answer 50

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

Answer 51

Thank you so much! I'll keep that!

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

my pleasure.