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okay first question
heres the first 3
ok can you hold a minute I'm answering another question
yes I can
so GCF is the greatest common factor.. what doe the equation have in common
so what does the equation \[6y^2+12xy^2 \] have in common? hint: there's a variable and a number in common
umm this is for 10?
they both have 6 and Y^2
yes they do
so when there is something in common we can place that common part to the side.. so 6y^2 ( whatever there is left)
so since a 1 and 2x is left behind we have 6y^2(1+2x) we can check easily by distributing them back 6y^2+12xy^2
so it A?
for #11.. there is only a number in common because there are 2 terms with a variable but the last one is just a number
3 is a common factor
so 3 is on the side and what do we have left ( ) ?
no because originally we have 3 terms.. B only has two terms.
there's a way to make this a bit faster... since 3 is the common factor we just divide 12/3 18/3 and 3/3
Got you.. so its D?
for twelve, i have to right the answer out.
hmm there is just a number is common.. it's similar to the previous problem
so what number can be used to divide 3, 9, and 27 ?
is my answer ...
wow you're picking this up fast that's right \[3(x^2-3x+9) \rightarrow 3x^2-9x+27\]
okay next 3
is 13 d?
just a quick guess :/
wait no I mean b
yes it is B for 13
14 b also?
okay 15 now
2 is common
yes 2 is common
but I have to do the +C -C
hmmm that would mean that x^2+9x+8 is factorable
so there's more than just factoring out the 2 .... SOOO FAST! :D
did I get it right?
yay! next 3
well 2 lol
we need long division for this xD
what does that mean?
I don't understand :/
I'm getting burned out.. it's like .. oh mai brain T_T
lol geeeeez louiseeee
wait. see now I'm not thinking right... we just use factoring and take out the like terms
the choices gave me a hint... if it was long division the answer would look different.. ok I can do this now xD!
for the first problem, there is a perfect square binomial
Yay we are back on the road lol and okay.
so x^2-16... that's a perfect square in the form of (x^2-y^2) = (x+y)(x-y)... so what is the square root of 16?
yes so we have (x+4)(x-4) in the denominator.
that makes it A?
I think we got lucky on the numerator part of this since we need to find a combination of numbers that can produce an 8 and a 16 4 x 4 = 16 4+4 is also 8 and since the pattern is all plus signs, our end result should be something like (x+4)(x+4) and wow it is A
because there is at least one pair of x+4 (both in numerator and denominator) in common so that gets cancelled.
okay 17 now
numerator is that there is one number in common denominator is that it can be split up since 12 is 6 x 2 = 12 and 6-2 = 4 the problem is should the 2 be negative or the 6 be negative in order to achieve the result of x^2+4x-12
yes an for the numerator?
yes 5 is common 5(x-2)/ (x+6)(x-2) who gets canceled out?
yes so the answer is 5/(x+6)
I have to go for a bit... xD need a break xD!
I have to go eat dinner. I'll be back