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Mathematics 12 Online
OpenStudy (iwanttogotostanford):

HELP NEE

OpenStudy (sadlalkdja12111312):

do you want the graph

OpenStudy (iwanttogotostanford):

no i just need to know how to solve this one i have desmos if i want to graph but i need to know how

OpenStudy (mjdennis):

A zero is someplace the graph wold cross the x-axis (in other words, where y is zero). Multiplicity says how many times that solution appears in the eequation.

OpenStudy (iwanttogotostanford):

one post of how to do it would be great, because this is how I learn these are my examples so if you could tell me the steps that would be awesome

OpenStudy (iwanttogotostanford):

what could I do with that next ??

OpenStudy (mjdennis):

Suppose you had y = (x - 3)^3(x - 1)^2(x+1)

OpenStudy (iwanttogotostanford):

so I would make them into 0's or equaling 0's right

OpenStudy (mjdennis):

Do you agree that for y to be 0, that either y = (x - 3)^3 = or (x - 1)^2 = 0 or (x+1) = 0 right? the only way for 2*x = 0 to be true is if x = 0. And the only way fro x*y = 0 is for x=0 or y=0.

OpenStudy (mjdennis):

So if you mean "set each term in parentheses equal to zero", yes, that is right.

OpenStudy (mjdennis):

So don't worry about multiplicity for a moment, just tell me what the zeros are for your equation.

OpenStudy (iwanttogotostanford):

how to i find the zeros... sorry!

OpenStudy (mjdennis):

The zeros for my equation, y = (x - 3)^3(x - 1)^2(x+1) are 3, 1, and -1 because 3-3 is zero, so when x=3, the equation becomes 0*12*4, which is zero.

OpenStudy (iwanttogotostanford):

oh ok so mine would be -5,-64, and 512?

OpenStudy (mjdennis):

Take a term, set it to zero, find x: Your first term is (x + 8)^2 Set it to zero (x + 8)^2 = 0 x+8 = 0 x+8 -8 = 0-8 x=-8 Got it?

OpenStudy (iwanttogotostanford):

yes i got that

OpenStudy (iwanttogotostanford):

what next? @mjdennis

OpenStudy (mjdennis):

There is another zero. What is that value?

OpenStudy (mjdennis):

I'll go back to my example to explain multiplicity. y = (x - 3)^3(x - 1)^2(x+1) The three zeros are 3, 1, -1 But because we have (x-3)^3 , which is (x-3)(x-3)(x-3), and (x-1)^2, my problem really has six answers 3,3,3,1,1,-1 even though some are duplicates of each other, there are soem parts of math where you will need to know how many times a dupe answer appears. so the example has zeros (3 multiplicity 3), (1multiplicity 2), and (1 multiplicity 1). That last one we can just shorten to "1"

OpenStudy (iwanttogotostanford):

x=3

OpenStudy (iwanttogotostanford):

my other "0"

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