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).

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

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

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

7 years agook, how do you find that?

7 years agothe -1/2?

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

7 years agowith the rest

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

7 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

7 years agoconstant =-2

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

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

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

7 years agour formula was wrong i believe

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

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

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

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

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

7 years agoinstead of multiplicity 2 at x =3

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

7 years agoOh, got it

7 years agoplug a 4 in and find multiplier

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

7 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)

7 years agothanks :)

7 years agonp

7 years agodhat peace :) im hitting the bed

7 years agogood night :)

7 years agoI hear you, nite nite

7 years ago