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).

8 years ago(x+1)(x-2)^2

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

8 years agosorry it is -1/2(x+1)(x-2)^2

8 years agook, how do you find that?

8 years agothe -1/2?

8 years agowouldnt it be (x-3)^2

8 years agowith the rest

8 years ago(x-3)^2(x+1)

8 years agosubstitute 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

8 years agoconstant =-2

8 years agoOOOH!!! gotcha. I will try that.

8 years agono mathmajor, you are incorrect. constant is -1/2

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

8 years agour formula was wrong i believe

8 years agoif you substitute 4 in that you get 5 times 1^2 = 5

8 years agoI understand the -1/2 part, not thereal zeroes part.

8 years agooh right, I misread the question. mathmajor is right

8 years agoi thought it had a multiplicity of 2 at x =2

8 years agolol i know what u did u saw the 2

8 years agoinstead of multiplicity 2 at x =3

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

8 years agoOh, got it

8 years agoplug a 4 in and find multiplier

8 years agoit is -2*(x-3)^2(x+1)

8 years agobecause 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)

8 years agothanks :)

8 years agonp

8 years agodhat peace :) im hitting the bed

8 years agogood night :)

8 years agoI hear you, nite nite

8 years ago