Question

Let U be a root of the polynomial x3+3x+3. In Q(u), express (7-2u+u2)-1 in the form a+bu+cu2

Answer 1

Q(u)=(7-2u+u²)-1=u²-2u+6.

x³+3x+3 has only one real root. To find it we can use Newton's iterative Method.

Let f(x)=x³+3x+3, then f'(x)=3x²+3, x_n+1=x_n-f(x_n)/f'(x_n). Let x₀=0.

x₁=-3/3=-1, x₂=-1-(-1)/6=-⅚, x₂=-449/549=-0.81785 approx, ..., x=u=-0.8177316739 approx (stable value) after a few more iterations.

Q(u)=Q(-0.8177316739)=8.304148438 approx.