Am I cubic?

Let us assume that the given polynomial $f(x)$ is cubic. Then, we can construct a difference table for $f(x)$ using the given function values. The $k^{\textrm{th}}$ difference column will be constant. The table comes out as follows:

$\begin{array} {c c c c c c c} n & f(n) & D_1(n) & D_2(n) & D_3(n)\\ 1 & 4 & -1 & 2 & 0\\ 2 & 3 & 1 & 2 & 0\\ 3 & 4 & 3 & & 0\\4 & 7 & & & 0\end{array}$

We see that the elements of the column $D_3(n)$ are all $0$ and we know that the $k^{\textrm{th}}$ difference for the assumed cubic polynomial will be $3!$ times the leading coefficient, i.e., coefficient of $x^3$ in the polynomial. But the $k^{\textrm{th}}$ difference is $0$ that we found out earlier.

Thus, the polynomial, if assumed cubic comes out as the form:

$f(x)=0\cdot x^3 + ax^2 + bx +c$

Since coefficient of $x^3$ is $0$ , the polynomial $f(x)$ can never be cubic.

Hence, the answer is No.

Let $g(x)=(x-2)^2+3-f(x)$ . If $f$ is a polynomial, then so is $g$ ; furthermore, $g(x)=0$ for $x=1,2,3,4$ . A cubic polynomial (with non-zero leading coefficient) can't have four distinct roots; therefore $g$ is of degree $\geq 4$ and hence so is $f$ . The answer, therefore is $\boxed{\text{no}}$ (although batman is close to the truth as well).

Assuming that $f(x)$ is a cubic polynomial, we can write in matrix form:

$XA=B\quad \Rightarrow \begin{bmatrix} 1^3 & 1^2 & 1^1 & 1^0 \\ 2^3 & 2^2 & 2^1 & 2^0 \\ 3^3 & 3^2 & 3^1 & 3^0 \\ 4^3 & 4^2 & 4^1 & 4^0 \end{bmatrix} \begin{bmatrix} a_3 \\ a_2 \\ a_1 \\ a_0 \end{bmatrix} = \begin{bmatrix} 4 \\ 3 \\ 4 \\ 7 \end{bmatrix}$

We can find $A$ as follows:

$A = X^{-1}B = \begin{bmatrix} -\frac{1}{6} & \frac {1}{2} &-\frac {1}{2} & \frac {1}{6} \\ \frac {3}{2} & -4 & \frac {7}{2} &-1 \\ -\frac {13}{3} & \frac {21}{2} & -7 & \frac{11}{6} \\ 4 & -6 & 4 & -1 \end{bmatrix} \begin{bmatrix} 4 \\ 3 \\ 4 \\ 7 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ -4 \\ 7 \end{bmatrix}$

It is shown that: $f(x) = x^2-4x+7$ showing that: $\boxed{No}$ , it is not a cubic polynomial.

1 1 1 4

1 2 4 3

1 3 9 4

1 4 16 7

Since this 4 x 4 determinant is zero

1 1 1 1

1 2 4 8

1 3 9 27

1 4 16 64

and this 4 x 4 determinant is non zero of 12

f(x) = 7 - 4 x + x^2 + 0 x^3

Two points for a straight line, three points for a parabola and four points for a cubic curve and etc on a plane, where d = 0 tells that they cannot form a cubic curve but one among the four points is redundant or stays on parabola formed by 3 of them.

Hence, the answer is No.

This is pretty cumbersome for this question, since we're given nice values, but it's a good general solution:

Lemma: There is at most one polynomial with degree $\le n$ that contains any $n+1$ specific points

Suppose $p(x)$ and $q(x)$ have degree $\le n$ and contain $(x_1,y_1)...(x_{n+1},y_{n+1})$

Then let $r(x)=p(x)-q(x)$ .

$r(x)$ has degree $\le n$ and at least $n+1$ zeros at $x_1,...,x_{n+1}$ . Thus $r(x)=0$ by the factor theorem. $\square$

You might recognize this polynomial as $f(x) = (x-2)^2 +3$ but again a systematic method:

We can brainlessly write

$f(x)= 4 \dfrac{(x-2)(x-3)(x-4)}{(1-2)(1-3)(1-4)} + 3 \dfrac{(x-1)(x-3)(x-4)}{(2-1)(2-3)(2-4)} \\ + 4 \dfrac{(x-1)(x-2)(x-4)}{(3-1)(3-2)(3-4)} + 7 \dfrac{(x-1)(x-2)(x-3)}{(4-1)(4-2)(4-3)}$

We don't actually need to evaluate this; we just need to check the $x^3$ coefficient:

$\dfrac{4}{(1-2)(1-3)(1-4)} + \dfrac{3}{(2-1)(2-3)(2-4)} \\+ \dfrac{4}{(3-1)(3-2)(3-4)} + \dfrac{7}{(4-1)(4-2)(4-3)}$

$=-\dfrac{4}{6}+\dfrac{3}{2}-\dfrac{4}{2}+\dfrac{7}{6}=0$

So this is not a cubic.