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\[\frac{ x + 1 }{ x^2 + x -6 } \div \frac{ x^2 + 5x + 4 }{ x + 4 }\]
@satellite73

http://www.wolframalpha.com/input/?i=%28x%2B1%29%2F%28x^2%2Bx-6%29*%28x%2B4%29%2F%28x^2%2B5x%2B4%29

you go that right?

nice use of the equation tool btw

i can show you another trick in using wolfram

okay and heres the question

simplify (12z^2 - 25z + 12)/ (3z^2 + 2z - 8)

using the link i got (4z - 3)/(x + 2)

ok i so take what you wrote, copy and paste it directly in to wolfram

yeah i did that and got (4z - 3)/(x + 2)

yeah i get that too, but of course with a z

i could also show you how to do it without wolfram, it is "factor and cancel"

oh yeah typo lol sorry okay i think it'll be better without wolframe alpha in case my teacher asks

ok then here is the deal
do you know how to factor?

sometimes but not really it'll be helpful if you showed me though

i don't actually, believe it or not if i wanted to factor i would cheat

what i mean is that i have no good way to show you how

i usually use a site for factoring

that works, of course wolfram will do that too

so in case the teacher asks, you can say something like this ;

\[ (12z^2 - 25z + 12)=(4 z-3) (3 z-4)\] and
\[(3z^2 + 2z - 8)=
(3 z-4) (z+2)\]

so
\[\frac{12z^2+25z+12}{3z^2+2z-8}=\frac{4z-3)(3z-4)}{(3z-4)(z+2)}\]

oh and then you cross out (3z - 4)?

exactly what you said

that is the idea of all of these
factor and cancel

okay cool i get it now

as to how to factor, you are pretty much on your own, but you can use wolfram to do it for sure

okay thnx

ill also ask someone else on os

want to try another one ?

yeah

ok i get a cup of coffee, you post

ok What polynomial identity should be used to prove that 21 = 25 - 4?

wow do you have any choices? \(21=25-4\) is simple arithmetic

ooh ok

25 and 4 are two squares

then B?

\[25=5^2\] and \[2^2\] so yeah B

what a dumb retricequestion
next?

thnx

when you subtract inside the function it move is 3 units to the RIGHT

thnx

what is the corresponding y value when x = 1?

ok ok

\[\frac{-1-8}{4-1}\] is whatyou have to compuote

i get
\[\frac{-9}{3}=-3\]

thnx

yw

how many more?

the conjugate of \(a+bi\) is \(a-bi\)

so b?

the conjugate of \(2+3i\) is \(2-3i\) that was an easy one

yeah b

okay thnx

yw next?

got that ?

so 18/9 = 2

yes

see the trick? it is easy

yup

\[g(x) = 4x + 18\]
\[h(x)=6x+13\]
\[g(x)+h(x)=4x+18+6x+13\]

combine like terms,what do you get?

10x + 31

ok good so C or D

yup but i think its with team work so C?

not sure
if bob works 3 hours alone he makes
\[f(3)=6\times 3+13=18+13=31\]

oh so D

don't jump the gun

if they both work 3 hours they make a total of
\pg(3)=10\times 3+31=61\]

oh

\[h(3)=10\times 3+31=61\]\]

if he has to split it, then yes he is better off working alone

so is its C then?

i think D
not really clear, but i think D

ok

next?

What is the equation of the quadratic graph with a focus of (3, 6) and a directrix of y = 4?

this takes a second

half way between 6 and 4 is 5, so the vertex is \((3,5)\)

therefore it will look like
\[4p(y-5)=(x-3)^2\] we need \(p\)

\[4(y-5)=(x-3)^2\\
y-5=\frac{1}{4}(x-3)^2\\
y=\frac{1}{4}(x-3)^2+5\]

x^2 + 2x +8

you want do step by step? or wolfram it?

step by step so i can learn

ok
a) subtract 8 from both sides, what do you get?

if that is not clear say so i will tell you

oh note that you are starting with
\[x^2+2x+8=0\]

x^2 + 2x = 8

i meaN -8

ok good

SO X^2 + 2X = -8

\[x^2+2x=8\] now what is half of 2?`

oops
\[x^2+2x=-8\] what is half of two?

so x^2 + x = -8/2?

no slow

oh sorry

half of 2 is 1, so we go right to
\[(x+1)^2=-8+1^2\] or \[(x+1)^2=-7\]

ok

take the square root, get
\[x+1=\pm\sqrt{-7}\] or
\[x+1=\pm\sqrt{7}i\]

subtract 1, get
\[x=-1\pm\sqrt{7}i\]

thnx

ok we do this, then i gotta split

\[f f(x) = -3x^4 - 5x^3 - x^2 - 8x + 4\]

it is a polynomial of degree 4, so it has 4 zeros for sure

\[f(x) = -3x^4 - 5x^3 - x^2 - 8x + 4\]
\[f(-x) = -3x^4 + 5x^3 - x^2 + 8x + 4\]

now i really have to bolt
ask your sister how to do the rest
good luck!

thnx and ok bye

bye !