How would you solve (x^2-4) ÷ (x-8) ÷ (x - 2)
\[\large a \div b \div c = \frac{a}{b} \div c \implies \frac{\frac{a}{b}}{c} \implies \frac{a}{b \cdot c}\]
Before going forward, tell me can you solve for \(x^2-4\) ??
Use here : \[a^2 - b^2 = (a +b)(a-b)\]
No but you can simplify... right
????
I mean to simplify it sorry.. :)
Use that formula and convert that \(x^2-4\) into factors first..
Can you do this or not?? You can write \(4\) as \(2^2\) there and then apply the formula.
(Writing it on paper,might take a sec)
Just try and let me know if in doubt.
Take your time and go slow..
What do u do after getting (x+2)(x -2)?
Okay, this becomes simple now..
Here, as I mentioned above: \(a = (x+2)(x-2)\) \(b = (x - 8)\) \(c = (x-2)\) So can you plug in above?
Yeah I think so (1 sec again) :D
Take your time, I am in no hurry..
X+2 over x-8?
Hopefully yes... :)
Could you walk me through another one? Not sure if the same formulas apply...
Sure..
X^2+2x+1 x -2 x^2 -1 x^2 - 4
Thank you so much 4 your help on the first 1 btw!!!
What is the question?
Those numbers over eachother. Each line is a different set over the ones above
You mean like this?? \[\huge \frac{\frac{a}{b}}{\frac{c}{d}} \]
Yes!
Sorry I cant do that on my kindle
So if you have this then this simplifies as : \[\large \frac{\frac{a}{b}}{\frac{c}{d}} \implies \frac{a \cdot c}{b \cdot d}\]
Now, can you simplify this ? \(x^2 + 2x + 1\) = ??
Just make factors of it..
X•x+ x+x+1?
I mean to factorize it.. Like this: \[x^2 + 2x + 1 = x^2 + x + x + 1 \implies x(x + 1) + 1(x+1) \implies (x+1)(x+1)\]
Now can you factorize \(x^2-1\) and \(x^2-4\) as you did earlier in previous question?
Yes?
Can you simplify them? If yes then do.
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