Question

Let $P\left( x \right)={{x}^{6}}+a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f $ be a polynomial such that $P\left( 1 \right)=1$,$P\left( 2 \right)=2$,$P\left( 3 \right)=3$,$P\left( 4 \right)=1$,$P\left( 5 \right)=5$,$P\left( 6 \right)=6$, then find the value of $P\left( 7 \right)$.\[\]

Answer 1

Hint: We see that the polynomial $P\left( x \right)=x$ for the values$x=1,2,3,4,5,6$. We then use factor theorem to get $P\left( x \right)-x=C\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\left( x-4 \right)\left( x-5 \right)\left( x-6 \right)$. We assign $f=P\left( x \right)-x$ and find the value of C by comparison of leading coefficients and then $f$. We add and subtract 7 to find $P\left( 7 \right)$ using $P\left( x \right)-x$.\[\]

Complete step-by-step answer:
We know from the factor theorem that a polynomial $p\left( x \right)$ has a factor $\left( x-a \right)$ if and only if $p\left( a \right)=0$ in other words $a$ is a zero of $p\left( x \right)$ . We also know that the factorial of a whole number $n$ is product of first $n$ numbers, $n!=1\times 2\times 3...\times n$\[\]

The given polynomial is
\[P\left( x \right)={{x}^{6}}+a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f\]
We see that it is a polynomial of degree 6 with leading or the coefficient of highest power is 1. It is also given that $P\left( 1 \right)=1$,$P\left( 2 \right)=2$,$P\left( 3 \right)=3$,$P\left( 4 \right)=1$,$P\left( 5 \right)=5$,$P\left( 6 \right)=6$. So we have $P\left( x \right)=x$ for $x=1,2,3,4,5,6$ and $P\left( x \right)-x=0$ for $x=1,2,3,4,5,6$. \[\]

We can conclude that $P\left( x \right)=x$ is a polynomial of degree 6 and $x=1,2,3,4,5,6$ are the zeros of the polynomial$P\left( x \right)=x$. So using factor theorem we can say $\left( x-1 \right)$,$\left( x-2 \right)$,$\left( x-3 \right)$,$\left( x-4 \right)$,$\left( x-5 \right)$,$\left( x-6 \right)$ are the factors of the given polynomial $P\left( x \right)$. So we can write $P\left( x \right)-x$ (denoted as $f$) and have
\[f=P\left( x \right)-x=C\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\left( x-4 \right)\left( x-5 \right)\left( x-6 \right) \]
where C is a constant. As the coefficient of the leading term ${{x}^{6}}$ is 1 then $C=1$. Let us put $x=0$ in the above equation
\[\begin{align}
  & f=P\left( 0 \right)-0=C\left( -1 \right)\left( -2 \right)\left( -3 \right)\left( -4 \right)\left( -5 \right)\left( -6 \right)=C\cdot 6! \\
 & \Rightarrow C=\dfrac{f}{6!} \\
 & \Rightarrow f=1\cdot 6!=6! \\
\end{align} \]

Let us put $x=7$ in the expression for $f$ and we get
\[P\left( 7 \right)-7=C\left( 7-1 \right)\left( 7-2 \right)\left( 7-3 \right)\left( 7-4 \right)\left( 7-5 \right)\left( 7-6 \right)\]
Now we add and subtract 7 to use above expression,
\[\begin{align}
  & P\left( 7 \right)=7+\left[ P\left( 7 \right)-7 \right] \\
 & =7+C\left( 7-1 \right)\left( 7-2 \right)\left( 7-3 \right)\left( 7-4 \right)\left( 7-5 \right)\left( 7-6 \right) \\
 & =7+\dfrac{f}{6!}\left( 6! \right)\left( \because C=\dfrac{f}{6!} \right) \\
 & =7+f \\
\end{align}\]
We have already obtained that $f=6!$. So $ P\left( 7 \right) = 7 + 6! = 7 + 720 = 727.$\[\]

Note: The addition and subtraction of 7 is the most important step in this problem. The factor theorem is a special case of remainder theorem which states that if linear polynomial $x-a$ divides the polynomial $p\left( x \right)$ then it will leave the remainder $p\left( a \right)$. When $p\left( a \right)=0$ , the remainder theorem is specialized as a factor theorem.