Find the polynomial ${P(z)}$ of degree ${3}$ such that ....

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

Write $\ P(x) = 8 + (x\!-\!6)(a+(x\!-\!4)(b+(x\!-\!2)c)).\, $ Evaluating successively at $\,x = 4,2,-1\,$ yields that $\ a=5,\ b=15/8,\ c =253/840.$