1.5 = 1.5
3.2x = 3.24.71 = 15.072 → 15.1
6.1x² = 6.14.714.71 = 135.32301 →135
x³ = 4.714.714.71 = 104.48711 → 104
Therefore fl(x) = 104 - 135 + 15.1 + 1.50 = -14.4
Is this correct. Especially when I round offing x³ is it correct?
No. You are forgetting to round after each individual arithmetic operation. While $$(4.71)^3=104.487111$$ rounds to $104$, this is not the value your machine will produce. It will initially compute $$(4.71)^2 = 22.1841$$ which is rounded to $22.2$. It then produces $$22.2 \times 4.71 = 104.562$$ which is rounded to $105$, rather than $104$.
In order to avoid such problem in the future it is vital that you identify all the operations that your machine will perform and write down the algorithm applied. A polynomial $$p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3$$ is often evaluated using Horner's method which exploits that $$p(x) = ( (a_3 x + a_2) x + a_1)x + a_0$$ and is executed as \begin{alignat}{2} p_0 &= a_3 & \\ t_1 & = x p_0, \quad p_1 &= t_1 + a_2, \\ t_2 & = x p_1, \quad p_2 &= t_2 + a_1, \\ t_3 & = x p_2, \quad p_3 &= t_3 + a_0. \end{alignat} It is straight forward to verify that $p_3 = p(x)$. This exposes the 6 operations required. You must remember to round the result of each operation before you continue to the next.