Can someone help me with finding the formula for P(x)?
The polynomial of degree 3, P(x), has a root of multiplicity 2 at x=3 and a root of multiplicity of 1 at x=-1. Another point on the polynomial is (4,-10).

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OpenStudy (anonymous):

(x+1)(x-2)^2

OpenStudy (anonymous):

I know that part, it's going through the points I dont get.

OpenStudy (anonymous):

sorry it is -1/2(x+1)(x-2)^2

OpenStudy (anonymous):

ok, how do you find that?

OpenStudy (anonymous):

the -1/2?

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OpenStudy (anonymous):

wouldnt it be (x-3)^2

OpenStudy (anonymous):

with the rest

OpenStudy (anonymous):

(x-3)^2(x+1)

OpenStudy (anonymous):

substitute x =4 in (x+1)(x-2)^2 you get 20
but -10 is what you should get for x =4
so the constant multiplier has to be -1/2

OpenStudy (anonymous):

constant =-2

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OpenStudy (anonymous):

OOOH!!! gotcha. I will try that.

OpenStudy (anonymous):

no mathmajor, you are incorrect. constant is -1/2

OpenStudy (anonymous):

hey, wait a minute. Why does (x+1)(x-3)^2 not work?

OpenStudy (anonymous):

ur formula was wrong i believe

OpenStudy (anonymous):

if you substitute 4 in that you get 5 times 1^2 = 5

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OpenStudy (anonymous):

I understand the -1/2 part, not thereal zeroes part.

OpenStudy (anonymous):

oh right, I misread the question. mathmajor is right

OpenStudy (anonymous):

i thought it had a multiplicity of 2 at x =2

OpenStudy (anonymous):

lol i know what u did u saw the 2

OpenStudy (anonymous):

instead of multiplicity 2 at x =3

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OpenStudy (anonymous):

Yes, butI put in -1/2*(x-3)^2(x+1) and It was incorrect. whats wrong with it?

OpenStudy (anonymous):

Oh, got it

OpenStudy (anonymous):

plug a 4 in and find multiplier

OpenStudy (anonymous):

it is -2*(x-3)^2(x+1)

OpenStudy (anonymous):

because the formula I supplied was wrong
f(x) = k(x-3)^2(x+1)
-10 = k (1^2) * 5
5k =-10
k = -2
so the correct answer is -2(x-3)^2(x+1)

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