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Mathematics 21 Online
OpenStudy (anonymous):

How would you solve (x^2-4) ÷ (x-8) ÷ (x - 2)

OpenStudy (anonymous):

\[\large a \div b \div c = \frac{a}{b} \div c \implies \frac{\frac{a}{b}}{c} \implies \frac{a}{b \cdot c}\]

OpenStudy (anonymous):

Before going forward, tell me can you solve for \(x^2-4\) ??

OpenStudy (anonymous):

Use here : \[a^2 - b^2 = (a +b)(a-b)\]

OpenStudy (anonymous):

No but you can simplify... right

OpenStudy (anonymous):

????

OpenStudy (anonymous):

I mean to simplify it sorry.. :)

OpenStudy (anonymous):

Use that formula and convert that \(x^2-4\) into factors first..

OpenStudy (anonymous):

Can you do this or not?? You can write \(4\) as \(2^2\) there and then apply the formula.

OpenStudy (anonymous):

(Writing it on paper,might take a sec)

OpenStudy (anonymous):

Just try and let me know if in doubt.

OpenStudy (anonymous):

Take your time and go slow..

OpenStudy (anonymous):

What do u do after getting (x+2)(x -2)?

OpenStudy (anonymous):

Okay, this becomes simple now..

OpenStudy (anonymous):

Here, as I mentioned above: \(a = (x+2)(x-2)\) \(b = (x - 8)\) \(c = (x-2)\) So can you plug in above?

OpenStudy (anonymous):

Yeah I think so (1 sec again) :D

OpenStudy (anonymous):

Take your time, I am in no hurry..

OpenStudy (anonymous):

X+2 over x-8?

OpenStudy (anonymous):

Hopefully yes... :)

OpenStudy (anonymous):

Could you walk me through another one? Not sure if the same formulas apply...

OpenStudy (anonymous):

Sure..

OpenStudy (anonymous):

X^2+2x+1 x -2 x^2 -1 x^2 - 4

OpenStudy (anonymous):

Thank you so much 4 your help on the first 1 btw!!!

OpenStudy (anonymous):

What is the question?

OpenStudy (anonymous):

Those numbers over eachother. Each line is a different set over the ones above

OpenStudy (anonymous):

You mean like this?? \[\huge \frac{\frac{a}{b}}{\frac{c}{d}} \]

OpenStudy (anonymous):

Yes!

OpenStudy (anonymous):

Sorry I cant do that on my kindle

OpenStudy (anonymous):

So if you have this then this simplifies as : \[\large \frac{\frac{a}{b}}{\frac{c}{d}} \implies \frac{a \cdot c}{b \cdot d}\]

OpenStudy (anonymous):

Now, can you simplify this ? \(x^2 + 2x + 1\) = ??

OpenStudy (anonymous):

Just make factors of it..

OpenStudy (anonymous):

X•x+ x+x+1?

OpenStudy (anonymous):

I mean to factorize it.. Like this: \[x^2 + 2x + 1 = x^2 + x + x + 1 \implies x(x + 1) + 1(x+1) \implies (x+1)(x+1)\]

OpenStudy (anonymous):

Now can you factorize \(x^2-1\) and \(x^2-4\) as you did earlier in previous question?

OpenStudy (anonymous):

Yes?

OpenStudy (anonymous):

Can you simplify them? If yes then do.

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