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you know this problem?

f(x) = a*x(x+3)(x-1)^2
solve for a using given point
16 = a*(-1)(-1+3)(-1-1)^2
solve for a

wait,
a*x?

sorry, just means multiplying

a*x?

where you get a and x anyway?

a is like a scale factor, it determines how skinny or wide the graph is kind of like the slope

so it should be x infrotn of (-1) etc?

then -1(-1) etc, ? i got confused with ax , but i get it the you replace it.

i understand what you have to do next, multy,

yes multiply and get a by itself

yeah i see it,
so is going to be ,
16=-1(etc?

yeah

ok thx, going go to sleep, at what time you usually on dude?

depends, usually at night im on pacific time

ok, so at this time?

yeah and definitely later too

ok, bye

hey, ok, you subtitude, then what?

was stoke for awhile, while working on it

what i got right now
16=a(-1)(-1+3)(-1-1)

ok use PEMDAS
do the parenthesis first
(-1-1)^2 = (-2)^2 = 4

ok now you got, 16=a(-1)(4)

what happened to the (-1+3) ??

forogt..

yep good
16 = a(-1)(8)

16=-8a
-2=a

we found a, now we must replace a with -2

correct
now use that and put it back in the equation

to what equipon. 16=-2x(x+3)(x-1)^2?

the very first one
now we put the x's and y's back in

f(x) = ax(x+3)(x-1)^2

just replace a with -2

-2a(1)(9)
-18a

how you find the fourt root?

woah
not x, replace a with -2
thats what we just solved for

???
fourth root of what

you looking for the fourt root

no its a fourth degree polynomial
it just means our function has a x^4 term

ok, how you start then?

i dont know what you mean
you just did it
answer is
f(x) = -2x(x+3)(x-1)^2

but the question is, find the fourt root

oh NOT

is find the fourt dree polynomial function

i thought all along,to find the fourt root

there are 3 real roots
0,-3,1

no really, their is four, 1 has multicplicity of 2

correct

ok thank, yoiu being great help to me,good night

:)