Algorithm

Question

A polynomial p(x) is such that p(0)=5 ,p(1)=4 ,p(2)=9 and p(3)=20  The minimum degree it can have is..

Comments

https://gateoverflow.in/651/gate2000_2-4    Answer 0 for your question .

---

@Vinay: A polynomial with no (real) roots can exist. There is no crossing the x-axis doesn't mean that the min degree required is 0.

A polynomial with degree 0 is a constant.

Answer 1

The degree can't be $0$ since that will give us a constant polynomial, but we have $4$ different points to satisfy.

The points that we need to satisfy aren't co-linear either. For example, $\frac{p(1)-p(0)}{1-0} = -1 \quad\neq\quad  5 = \frac{p(2)-p(1)}{2-1}$

Let's try to fit a parabola.

Let $p(x) = ax^2 + bx + c$

Since $p(0) = 5$, we have that:

$$\begin{align}
a\cdot 0^2 + b\cdot 0 + c &= 5\\
c &= 5\\[1em]
\hline
\implies p(x) &= ax^2 + bx + 5
\end{align}$$

Also, $p(1) = 4$. Thus,

$$\begin{align}
a\cdot 1^2 + b\cdot 1 + 5 &= 4\\[1em]
a + b &= -1\tag{1}\label{1}
\end{align}$$

Finally, $p(2) = 9$. This gives us:

$$\begin{align}
a\cdot 2^2 + b\cdot 2 + 5 &= 9\\[1em]
4a + 2b &= 4\\[1em]
2a + b &= 2\tag{2}\label{2}
\end{align}$$

Using $\eqref{1} \,\&\, \eqref{2}$ we get:

$$\begin{array}{rrll}
+ (&2a + b &= 2 &)\\
- (&a + b &= -1 &)\\[1em]
\hline
\implies&a&=3\\
&b&=-4
\end{array}$$

So we have $p(x) = 3x^2 - 4x + 5$.

Testing it on our final point, we have that $p(3) = 3 \cdot 3^2 -4 \cdot 3 + 5 = 27 - 12 + 5 = 20$, which is equal to the given value of $p(3)$.

Hence, the minimum degree is $2$ and the polynomial is $p(x) = 3x^2 - 4x + 5$.