rootbeer003:
hElp
satellite73:
ok
satellite73:
you got \(x^6\) with a minus sign in front \(\color{red}-x^6\) forget about the two out at the end
satellite73:
looks like this |dw:1490147218007:dw|
rootbeer003:
k
satellite73:
a big sad face as x goes to the right, you are going down so as x goes to negative infinity \(-x^6\) goes to negative infinity as x goes to negative infinity \(-x^6\) also goes to negative infinity it is headed down and both ends
rootbeer003:
b?
satellite73:
ok i made a typo there, i meant as x goes to infinity \(-x^6\) goes to negative infinity
satellite73:
yes B
satellite73:
to find this, replace \(x\) by \(-3\) and compute \[(-3)^2-5\times (-2)-22\]
rootbeer003:
ohh ok
satellite73:
either that or you can divide but that way is probably easier
rootbeer003:
um i got -3
satellite73:
yeah me too hmm i must have made a mistake
rootbeer003:
replace the 5X with -3?
rootbeer003:
like 5(-3)
satellite73:
lol yeah because i am an idiot i said replace x by -3 so that is what we should do \[(-3)^2-5\times (-3)-22\]
satellite73:
\[9+15-22=24-22=2\]
rootbeer003:
dddddddddd
satellite73:
yeah that one
satellite73:
i can't show you how to do long division here, but i can show you how to figure out the correct answer easily
satellite73:
it is going to be \[(x+3)( something )=2x^3-3x^2-17x+\color{red}{30}\]
satellite73:
what would you multiply 3 by to get 30?
rootbeer003:
10
rootbeer003:
c
rootbeer003:
?
satellite73:
yes, and only one choice has ten as a constant
satellite73:
yes, it has to be that one, you don't need to check anything else in this example
satellite73:
this one you actually have to do the division, which i can't show you i would use technology
satellite73:
look here, then we will get it http://www.wolframalpha.com/input/?i=(x%5E3-6)%2F(x%2B2)
satellite73:
ok that didn't work, the answer is D
satellite73:
in the first one replace x by 1, which is real easy
satellite73:
\[4+3+2-10=?\]
rootbeer003:
-1
satellite73:
done
satellite73:
for the next one we have to do some work, but it can't be A or B because 3 doesn't go in to 8 or 16
rootbeer003:
i think c
satellite73:
i would have thought so too, but maybe i made a mistake somewhere
satellite73:
yes you are right, it is C
rootbeer003:
bAM
satellite73:
because 4 goes in to 40. mpt 3
satellite73:
ok whatever i was trying to say how we doing?
rootbeer003:
good
satellite73:
we got more?
rootbeer003:
duh
satellite73:
ok the coefficients of \(3x^2-x-2\) are \[\large 3,-1,-2\] but that is in every answer
satellite73:
the because you are dividing by \(x+4\) you need to put \(\large -4\) in on the side that eliminates A and D
satellite73:
bring down the 3, what is \(3\times -4\)?
rootbeer003:
i think b
rootbeer003:
-12
satellite73:
right, so you have the -12 on that line, then you ADD what is \(-1+(-12)\) ?
rootbeer003:
13
satellite73:
um no actually it is \(-13\)
rootbeer003:
yyayaya
satellite73:
so C
satellite73:
\[8n^2-26n+20\] replace \(n\) by \(2\)
satellite73:
what is \[8\times 2^2-26\times 2+20\]?
rootbeer003:
o
satellite73:
yes, that means there is no remainder, so B and D are out
satellite73:
and since \(20\div -2=-10\) i would go with A
satellite73:
the coefficients are \(\large 3, 0,-2\) so \[3x^2+0x-2\] i.e. just \[3x^2-2\]
rootbeer003:
d
satellite73:
put in a 4 where you see a g, that will give you the remainder
rootbeer003:
c?
satellite73:
\[2\times 4^3-2\times 4-8\]
satellite73:
no
rootbeer003:
112
satellite73:
at least that is not what i got i got \(112\)
satellite73:
so C is the one with that as a remainder
rootbeer003:
I SAID C BRO
rootbeer003:
lol
satellite73:
then you were right
satellite73:
replace k by -1, that will give you the remainder
rootbeer003:
d?
satellite73:
oh damn we still have two choices
rootbeer003:
b or c
satellite73:
B or D
rootbeer003:
right
rootbeer003:
d
satellite73:
D
rootbeer003:
im always right like the good ol days
rootbeer003:
give me a medal i deserve it
satellite73:
you deserve two medals
rootbeer003:
ik
rootbeer003:
ok 5 more
satellite73:
the remainder the last entry 7
rootbeer003:
b
satellite73:
yup
satellite73:
\[4\times 5^2-5\times 5+2\]
satellite73:
i know what i got ...
rootbeer003:
77
satellite73:
bingo
satellite73:
\[2\times 2^4-5\times 2^2-3\times 2+2\]
rootbeer003:
8
satellite73:
you are right again
satellite73:
you are matching up the first column and the last column
satellite73:
\[(-2-16), (-1,-8), (1,-4), (2,-20)\] are all of them
satellite73:
evidently \((-2,-2)\) is not there
satellite73:
too much work you have to plug in each of those numbers
rootbeer003:
a
satellite73:
yes
rootbeer003:
ok Im done for now But do u wanna do more?
satellite73:
it is getting late for me, school day tomorrow we can do more later
rootbeer003:
Ok Thankyou
satellite73:
your welcome, hope you learned something see you soon
rootbeer003:
Bye
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