Find the polynomial ${P(z)}$ of degree ${3}$ such that .... I meant that if we have ${P(z)}$ of degree ${3}$ such that .....
$${P(-1)=7} , {P(2)=3}  ,{P(4)=-2} ,{P(6)=8}$$ 
Find the polynomial 
 A: Here's an idea: suppose we had four polynomials $P_1$, $P_2$, $P_3$, and $P_4$ with the following properties:
\begin{align}
P_1(-1)&=7, & P_1(2)&=0, & P_1(4)&=0, & P_1(6)&=0, \\
P_2(-1)&=0, & P_2(2)&=3, & P_2(4)&=0, & P_2(6)&=0, \\
P_3(-1)&=0, & P_3(2)&=0, & P_3(4)&=-2, & P_3(6)&=0, \\
P_4(-1)&=0, & P_4(2)&=0, & P_4(4)&=0, & P_4(6)&=8, \\
\end{align}
If this were the case, the polynomial $P=P_1+P_2+P_3+P_4$ would have the properties we are looking for. So let's make these four polynomials.
It's easy to find a polynomial which vanishes at $2$, $4$, and $6$: take
$Q_1(x)=(x-2)(x-4)(x-6)$ (in fact, all polynomials that vanish at those points are a multiple of this one). Then the polynomial
$$
P_1(x) = 7\cdot\frac{Q_1(x)}{Q_1(-1)} = -\frac{1}{15}(x-2)(x-4)(x-6)
$$
has all the necessary properties. Now do the same for the other three polynomials and add them up. The idea we used is called Lagrange interpolation, as Git Gud mentioned.
A: Hint:
let 
$$P(x)=ax^3+bx^2+cx+d$$
the desired polynomial hence by the hypothesis:
$$\left\{\begin{array}\\P(-1)=7&\Rightarrow& -a+b-c+d=7\\
P(2)=3&\Rightarrow&????\\
P(4)=-2&\Rightarrow&????\\
P(6)=8&\Rightarrow&????
\end{array}\right.$$
and finaly solve this system of equations.
A: Write $\ P(x) = 8 + (x\!-\!6)(a+(x\!-\!4)(b+(x\!-\!2)c)).\, $ Evaluating successively at $\,x = 4,2,-1\,$ yields that $\ a=5,\ b=15/8,\ c =253/840.$ 
