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).

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

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

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

ok, how do you find that?

the -1/2?

wouldnt it be (x-3)^2

with the rest

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

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

constant =-2

OOOH!!! gotcha. I will try that.

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

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

ur formula was wrong i believe

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

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

oh right, I misread the question. mathmajor is right

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

lol i know what u did u saw the 2

instead of multiplicity 2 at x =3

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

Oh, got it

plug a 4 in and find multiplier

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

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)

thanks :)

np

dhat peace :) im hitting the bed

good night :)

I hear you, nite nite