Curious Remainders

First, notice that $deg (P(x)-x)=5$ . Since $P(x)-x=0$ for $x=1, 2, 3, 4, 5$ , it follows that $P(x)-x=c(x-1)(x-2)(x-3)(x-4)(x-5)$ for some constant $c$ .

Since we are given $P(0)=-1$ , we can try plugging 0 in: $P(0)-0=c(-1)(-2)(-3)(-4)(-5)=-1 \Rightarrow c=\frac{1}{120}$ .

Therefore, we have $P(x)-x=\frac{1}{120} (x-1)(x-2)(x-3)(x-4)(x-5)$ .

Now, plugging in $x=12$ , we have $P(12)-12=\frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{120}=11 \cdot 7 \cdot 6=462$ . It follows that $P(12)=462+12=\boxed{474}$ *

*This can be computed in many different ways. See http://faculty.atu.edu/mfinan/2033/section12.pdf

From Remainder-Factor Theorem we have $P(a)=a$ $,$ for $a=1,2,3,4,5$ .

$\Rightarrow$ $P(x)=x$ $,$ for $x=1,2,3,4,5$ .

Since $P(x)$ is a degree $5$ polynomial, so we let $P(x)-x=A\cdot(x-1)\cdot(x-2)\cdot(x-3)\cdot(x-4)\cdot(x-5)\:,$ for some $A\neq0$ .

From given conditions i.e. $P(0)=\,-1$ we have

$-1-0=A\cdot(-1)\cdot(-2)\cdot(-3)\cdot(-4)\cdot(-5)$

$\Rightarrow$ $A=\large\frac{1}{1\cdot2\cdot3\cdot4\cdot5}$

Now $P(12)=12+A\cdot(12-1)\cdot(12-2)\cdot(12-3)\cdot(12-4)\cdot(12-5)$

$\Rightarrow$ $P(12)=\boxed{474}$ .

I recently learned how to solve problems like these, so here goes:

From the remainder theorem, we know that $P(a) = a$ for $a = 1, 2, 3, 4, 5$ . Let us define a new function $Q(x) = P(x) - x$ . Now, we know that $x-a$ is a factor of $Q(x)$ for $a = 1, 2, 3, 4, 5$ . Note that $Q(x)$ is also a degree 5 polynomial. Hence:

$Q(x) = P(x) -x = a(x-1)(x-2)(x-3)(x-4)(x-5)$ , where $a$ is some arbitrary constant.

But we know that $P(0) = -1$ . Substituting this into the above function:

$\Longrightarrow P(0) - 0 = a(0-1)(0-2)(0-3)(0-4)(0-5)$

$\Longrightarrow -1 = a(-1)(-2)(-3)(-4)(-5)$

$\Longrightarrow a = \dfrac{1}{120}$

Hence, $Q(x) = \dfrac{1}{120}(x-1)(x-2)(x-3)(x-4)(x-5)$ , which means that $P(x) = \dfrac{1}{120}(x-1)(x-2)(x-3)(x-4)(x-5) +x$ .

Substituting $x=12$ into the function, we get:

$\Longrightarrow P(12) = \dfrac{1}{120}(12-1)(12-2)(12-3)(12-4)(12-5) + 12$

$\Longrightarrow P(12) = \boxed{474}$

The given polynomial can be written as

$P(x)=k(x-1)(x-2)(x-3)(x-4)(x-5)+x$

where k is a constant.

The remainder when $P(x)$ is divided by $(x-a)$ is $P(a)$ which holds true for 1 to 5 according to the given statement.

Since $P(0)$ is $-1$ , we get $P(0)=k(-1)(-2)(-3)(-4)(-5)=-1 \implies k= \frac{1}{5!}$ .

Thus, we get $P(12)= \boxed{474}$

By the Factor Theorem , we have:

$P(x)-x=n\left(\prod_{a=1}^5 (x-a)\right)$

Here $n$ is a constant. Since we have $P(0)=-1$ , plugging in $x=0$ , we have:

$P(0)-0=n\left(\prod_{a=1}^5 -a\right)~\Longrightarrow -1=n\cdot (-120) ~\Longrightarrow n=\dfrac{1}{120}$

$\therefore ~~~ P(x)=x+\dfrac{1}{120} \left(\prod_{a=1}^5 (x-a)\right)$

Plugging in $x=12$ , we get our desired result:

$P(12)=12+\dfrac{1}{120} \left(\prod_{a=1}^5 (12-a)\right)=12+\dfrac{11\times 10\times 9\times 8\times 7}{120}=12+462=\fbox{474}$

consider..... $g(x)=P(x)-x$ . then for $x=1\rightarrow5$ $g(x)=K(x-1)(x-2)(x-3)(x-4)(x-5)$ where K is a constant

Given $P(0)=-1$ we put x=0 $\Rightarrow c(-1)(-2)(-3)(-4)(-5)=-1$ $\Rightarrow c=\frac {1}{120}$ $P(x)=g(x)+x=x+\frac {1}{120}(x-1)(x-2)(x-3)(x-4)(x-5)$ $Put x=12...we get P(x)=\boxed{474}$

I did it the long way. There should be a better way. Solution: constant term = -1 use 5 eq and 5 variables to find the equation P(x) then find P(6)=474

This is basically an application of remainder Theorem and linear equation solution. You need to know the theorem that when f(x) is divided by (x-a), the remainder is always f(a). Thus we get the following linear equtions f(0)=-1; f(1)=1; f(2)=2; f(3)=3; f(4)=4; f(5)=5; Observe that these equations are linear with the coefficients being the unknowns. Now all you need to know is how to solve a linear equation of type Ax=b. the solution is inverse(A).b there . is the usual matrix multiplication.

P.S. I don't have enough patience to invert a 5*5 matrix, so I wrote i small piece of R-code for everything. Here is the code:

A=matrix(1,5,5); for(i in 1 :5) { A[i,]=i^c(1:5) }; a=c(2:6); beta=ginv(A)% %a; beta=c(-1,beta); ans=sum(12^c(0:5) beta); print(ans); This app has some problem with newlines. place a newline at every ";".

By the remainder theorem we have $P(a)=a$ , so we define $Q(x)=P(x)-x$ , consequently $Q(x)$ has roots at $1,2,3,4,5$ and we can write it as

$Q(x)=(x-1)(x-2)(x-3)(x-4)(x-5)C \longrightarrow P(x)=(x-1)(x-2)(x-3)(x-4)(x-5)C+x$ Plugging in $0$ we can evalute $C$ , so

$\displaystyle P(0)=-1=-5!C \longrightarrow C=\frac{1}{120}$

Finally

$\displaystyle P(12)=11\cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot \frac{1}{120} + 12=474$ .

Since P(x), when divided by (x-a) gives P(a) as the result, P(1)=1, P(2)=2, P(3)=3, etc. and P(0)=-1. Using these points, one can create a 6x7 matrix. Then use rref (reduced-row-echelon-form) on the matrix to get the coefficient values of the 5th degree polynomial and then plug in 12 into the resultant polynomial.

Let $P(x)-x=Q(x)$ for some polynomial $Q(x)$ . Since $P(x)$ is of degree 5, so is $Q(x)$ . Note that $Q(x)$ has roots $1, 2, 3, 4, 5$ , so we can write $Q(x)=c(x-1)(x-2)(x-3)(x-4)(x-5)$ . We are also given that $P(0)=-1 \implies -120c=-1 \implies c=\frac{1}{120}$ .

Finally, plugging in 12 gives $P(12)-12=\frac{1}{120}(11)(10)(9)(8)(7) \implies P(12)=474$ .

We can easily see that $P(x) - x$ has roots $1,2,3,4,5$ . Since $P(x) - x$ is also a quintic polynomial, we have that $P(x) - x = a(x-1)(x-2)(x-3)(x-4)(x-5)$ for some real $a$ . Substituting $(0,-1)$ gives us $P(0) = a(-1)(-2)(-3)(-4)(-5) = -1 \implies a = \frac{1}{120}$ So $P(12) - 12 = \frac{1}{120} (11)(10)(9)(8)(7) = 462 \implies P(12) = 474 .$