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Question: Let \(P\left( x \right)={{x}^{6}}+a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+ex+f \) be a polynomia...

Let P(x)=x6+ax5+bx4+cx3+dx2+ex+fP\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(1)=1P\left( 1 \right)=1,P(2)=2P\left( 2 \right)=2,P(3)=3P\left( 3 \right)=3,P(4)=1P\left( 4 \right)=1,P(5)=5P\left( 5 \right)=5,P(6)=6P\left( 6 \right)=6, then find the value of P(7)P\left( 7 \right).$$$$

Explanation

Solution

We see that the polynomial P(x)=xP\left( x \right)=x for the valuesx=1,2,3,4,5,6x=1,2,3,4,5,6. We then use factor theorem to get P(x)−x=C(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)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(x)−xf=P\left( x \right)-x and find the value of C by comparison of leading coefficients and then ff. We add and subtract 7 to find P(7)P\left( 7 \right) using P(x)−xP\left( x \right)-x.$$$$

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

The given polynomial is
P(x)=x6+ax5+bx4+cx3+dx2+ex+fP\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(1)=1P\left( 1 \right)=1,P(2)=2P\left( 2 \right)=2,P(3)=3P\left( 3 \right)=3,P(4)=1P\left( 4 \right)=1,P(5)=5P\left( 5 \right)=5,P(6)=6P\left( 6 \right)=6. So we have P(x)=xP\left( x \right)=x for x=1,2,3,4,5,6x=1,2,3,4,5,6 and P(x)−x=0P\left( x \right)-x=0 for x=1,2,3,4,5,6x=1,2,3,4,5,6. $$$$

We can conclude that P(x)=xP\left( x \right)=x is a polynomial of degree 6 and x=1,2,3,4,5,6x=1,2,3,4,5,6 are the zeros of the polynomialP(x)=xP\left( x \right)=x. So using factor theorem we can say (x−1)\left( x-1 \right),(x−2)\left( x-2 \right),(x−3)\left( x-3 \right),(x−4)\left( x-4 \right),(x−5)\left( x-5 \right),(x−6)\left( x-6 \right) are the factors of the given polynomial P(x)P\left( x \right). So we can write P(x)−xP\left( x \right)-x (denoted as ff) and have
f=P(x)−x=C(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)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 x6{{x}^{6}} is 1 then C=1C=1. Let us put x=0x=0 in the above equation

& 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{aligned} $$ Let us put $x=7$ in the expression for $f$ and we get $$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{aligned} & 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{aligned}$$ **We have already obtained that $f=6!$. So $ P\left( 7 \right) = 7 + 6! = 7 + 720 = 727.$$$$$** **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.