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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