SOLUTION: Let p(x) be a cubic polynomial. If p(0) = 0, p(1) = 1, p(2) = 2, and p(3) = 3, then compute p(4).

Question 1209699: Let p(x) be a cubic polynomial. If p(0) = 0, p(1) = 1, p(2) = 2, and p(3) = 3, then compute p(4).

Found 3 solutions by CPhill, ikleyn, math_tutor2020:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Let $p(x)$ be a cubic polynomial. We are given that $p(0) = 0$, $p(1) = 1$, $p(2) = 2$, and $p(3) = 3$.
Consider the polynomial $q(x) = p(x) - x$. Then $q(0) = p(0) - 0 = 0 - 0 = 0$, $q(1) = p(1) - 1 = 1 - 1 = 0$, $q(2) = p(2) - 2 = 2 - 2 = 0$, and $q(3) = p(3) - 3 = 3 - 3 = 0$.
Since $q(x)$ has roots at $x = 0, 1, 2, 3$, and $p(x)$ is a cubic polynomial, $q(x)$ must be a quartic polynomial.
However, since $p(x)$ is a cubic polynomial, $p(x) - x$ can be at most a cubic polynomial. Thus, we must have $q(x) = ax(x-1)(x-2)(x-3)$ for some constant $a$.
Then $p(x) = q(x) + x = ax(x-1)(x-2)(x-3) + x$.
Since $p(x)$ is a cubic polynomial, the coefficient of $x^4$ must be zero. The $x^4$ term in $ax(x-1)(x-2)(x-3)$ is $ax^4$. Since $p(x)$ is a cubic, $a$ must be 0.
Therefore, $q(x) = 0$ and $p(x) = x$.
Then $p(4) = 4$.
However, if $p(x)$ is a cubic, then $q(x)$ is also a cubic.
Since $q(0)=q(1)=q(2)=q(3)=0$, $q(x)$ must be identically 0, which means $p(x)=x$.
Then $p(4) = 4$.
Consider the finite differences.
x | p(x) | Δp | Δ²p | Δ³p
---|---|---|---|---
0 | 0 | 1 | 0 | 0
1 | 1 | 1 | 0 |
2 | 2 | 1 | |
3 | 3 | | |
Since the third differences are zero, the polynomial is at most quadratic. Since the first differences are all 1, the polynomial is $p(x) = x$. Then $p(4) = 4$.
Final Answer: The final answer is $\boxed{4}$

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Let p(x) be a cubic polynomial. If p(0) = 0, p(1) = 1, p(2) = 2, and p(3) = 3, then compute p(4).
~~~~~~~~~~~~~~~~~~~~~~~~

        It can be solved/worded/presented differently, in much shorter way.

Consider the polynomial p(x) - x. It is easy to check that the polynomial p(x)-x takes the value of 0 at the points x= 0, 1, 2, 3. In other words, the polynomial p(x)-x has 4 (four) different roots. Since the polynomial P(x)-x has the degree not greater than 3, from Algebra, it implies that the polynomial is identically equal to zero. So, we proved that p(x) - x = 0. It implies p(x) = x. Then p(4) = 4. ANSWER. We proved that under given condition, the polynomial p(x) is equal to x: p(x) = x, and p(4) = 4.

Solved.

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Strictly speaking, the problem formulation, as it is presented in the post, is INCORRECT.

A correct formulation should be in THIS form:

Polynomial p(x) has a degree less than 4. If p(0) = 0, p(1) = 1, p(2) = 2, and p(3) = 3, then compute p(4).

As it is given in the post, it demonstrates
to all people around that the problem's creator does not know basics of Algebra
and can not express his thoughts in mathematically correct form.

Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

p(4) = 4

The function is p(x) = x which is the same as y = x
This is a straight line through the points (0,0) and (1,1) and (2,2) etc

p(x) is not a cubic polynomial and instead it's linear. Your teacher made a typo.

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