\(\sf Simplify ~~\huge x^2-7x-\frac{8}{x^2}-4x-32\)
\[\huge\rm \frac{ x^2 -7x-8 }{ x^2 -4x-32 }\] factor both numerator and denominator
do you know how to factor quadratic equations ??
FACTOR IT
nope
for first one ?? then nope
you need to use quadatic formula to find root so you can factor
okay lets factor first one \[\huge\rm x^2 -7x -8\]let me know what you get :)
find two number if you multiply them you should get -8 and if you add or subtract them you should get -7 :)
-8 and 1
yes that is right :) :)
\[\huge\rm(x-8)(x+1)\] because leading coefficient is one so you can write like this ;)
\[\huge\rm x^2-4x -32\] what about this one ??
-8 and 4?
yes right -8 times 4 = -32 and -8 +4 = -4
Ah, thank you so much! Would you mind helping me with one more problem? :)
\[\frac{ (x-8)(x+1) }{ (x-8)(x+4) }\]
got it!! ??
my pleasure and SURE!!
Yes, you're a great help!! And the other problem says to multiply, x^2/x+7 * x^2+11x+28/x^2+4x
\[ \huge\rm \frac{ x^2 }{ x+7 } \times \frac{ x^2 +11x +28 }{ x^2 +4x }\] O_O this is your question right ?
Correct!
again factor both den. and numerator on right side
Okay, how do you factor if there is only 2 numbers?
\[\frac{ x^2 +11x +28 }{ x^2 +4x }\] do you mean den. ? x^2 +4x ? this one ??
Yes .
oaky so what is common in this both term x^2 and 4x ??
I believe for the top it is 7 and 4? And the x.
yes right 7 times 4 = 28 7 +4 = 11 so yes
yes for the bottom part just x so what you left
So would it be (x+7)(x+4) / (x+4)?
yes right but don't forget x which suppose to be at the bottom
Okay, so then the two x+4 would cancel out.
\[\frac{ (x+7) (x+4)}{ x(x+4) }\] like this
yes right
Oh I see, so I canceled them out now I'm left with x^2/x+7 * x+7/x
yes right
Okay, can you please help me with wht the next step is(:
what's the final answer ??
ohh okay i thought you said with next question
Oh haha, I'm sorry!!
|dw:1424579104183:dw| you can cancel out x+7 ;)
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