k is a polynomial of degree 6. K has a root of multiplicity 3 at t=1, a root of multiplicity 2 at t=-5, k(-2) = -247.86, and k(-8)=0. find the algebraic equation for k

Question

k is a polynomial of degree 6. K has a root of multiplicity 3 at t=1, a root of multiplicity 2 at t=-5, k(-2) = -247.86, and k(-8)=0. find the algebraic equation for k

Answer 1

(t-1)^3(t+5)^2(t-x)

k(-8) = (-9)^3(-3)^2(-8-x) = -243(9)(-8-x) ]=0

= 243(9(8))= -243(9)x

x = -243)(9)(8)/243(9) = -8

then the 6th degree polynomial is k(t) = (t-3)^3(t+5)^2(t+8)

but if you used the last condition

k(-2) = (-3)^3(3)^2(-2-x) = -247.86

-243(-2) +243x = -247.86

243x = -247.86 -486 = -733.86

x = -733.86/243 = -3.02

then the 6th degree polynomial would be

k(t) = (t-1)^3(t+5)^2(t+3.02)