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).
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?
wouldnt it be (x-3)^2
with the rest
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
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)
dhat peace :) im hitting the bed
good night :)
I hear you, nite nite