Polynomial Equation

The key is to recognize that since $f(x)$ is a polynomial of some arbitrary degree $n$ , then $2f(x+1)$ is also a polynomial of the same degree. Therefore, $2f(x+1) - f(x) = (x+1)^3 = x^3+3x^2+3x+1.$ Furthermore, two polynomials that are equal to each other must have the same degree and the same coefficient on each term of the same degree.

It is thus clear that $f(x)$ must be a cubic polynomial and have the general equation $f(x)=ax^3+bx^2+cx+d$ for $a, b, c, d \in \mathbb{R}.$ We can thus find that $f(x+1)=a(x^3+3x^2+3x+1)+b(x^2+2x+1)+c(x+1)+d$ and $2f(x+1)-f(x)=ax^3+x^2(6a+b)+x(6a+4b+c)+2a+2b+2c+d.$

Since this equation must be equal to $(x+1)^3=x^3+3x^2+3x+1$ , we can match coefficients and deduce that $a=1, 6a+b=3, 6a+4b+c=3,$ and $2a+2b+2c+d=1.$

Solving through, we find that $a=1, b=-3, c=9,$ and $d=-13.$ . Therefore, we have found the explicit form of $f(x)$ to be $f(x)=x^3-3x^2+9x-13$ and thus, $f(10)=777.$

If the degree of $f$ is less than $3,$ then the left hand side has higher degree than the right hand side, which cannot occur. If the degree of $f$ is greater than $3$ , then the degrees of the left hand side the right hand side are the same, but the coefficient for the highest power on the right is twice that on the left, which is impossible.

This shows the degree of $f$ is $3$ . If $f(x)=kx^3+...,$ then $k+1=2k,$ so $k=1$ . Suppose $f(x)=x^3+ax^2+bx+c$ . Then multiplying everything out and simplifying, we get the following: $2x^3+(a+3)x^2+(b+3)x+(c+1)=2x^3+(2a+6)x^2+(4a+2b+6)x+(2a+2b+2c+2)$ This gives the following system of equations: $\begin{cases} a+3=2a+6\\ b+3=4a+2b+6\\ c+1=2a+2b+2c+2 \end{cases}$ From the first equation $a=-3$ . Then from the second equation $b=9$ . Then from the third equation $c=-13.$ So $f(x)=x^3-3x^2+9x-13,$ therefore $f(10)=777$ .

Suppose a polynomial f(x) has a form :

$f(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} +... +a_0$ for $n$ is an integer number.

Then the equation will be :

$a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} +... +a_0 + x^3 + 3x^2 + 3x + 1 = 2 ( a_n (x+1)^n + a_{n-1} (x+1)^{n-1} + a_{n-2} (x+1)^{n-2} +... +a_0 )$

$a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} +... +a_0 + x^3 + 3x^2 + 3x + 1 = 2 (a_n x^n + (a_n {n \choose 1} + a_{n-1} ) x^{n-1} + (a_n {n \choose 2} + a_{n-1} {n \choose 1} + a_{n-2} ) x^{n-2} + ... + a_0 )$

suppose $n>3$ , then by comparing the coefficient on the same degree of $x$ , we will get $a_n$ = 0. This means the highest degree of polynomial now is $n-1$ .

But suppose $n-1 > 3$ , then by doing the same thing we will get $a_{n-1} = 0$ . Hence, this sequence is continuing until $x^3$ . Thus the possible highest degree is $x=3$ .

with $n=3$ , consequently we can get $a_3 = 1$ , $a_2 = -3$ , $a_1 = 9$ , and $a_0 = -13$ . Thus :

$f(x) = x^3 -3x^2 +9x -13$

$f(10) = 1000 - 300 +90-13 = 777$

Because $(x + 1)^3$ is a third degree polynomial, the expression $2 f(x + 1)$ must also be a third degree polynomial for the equation to hold. It follows that $f(x)$ is a third degree polynomial.

Let $f(x) = ax^3 +bx^2 + cx + d$ . Then we have $\begin{aligned} f(x) + (x + 1)^3 & = & (ax^3 + bx^2 + cx + d) + (x + 1)^3 \\ & = & (ax^3 + bx^2 + cx + d) + (x^3 + 3x^2 + 3x + 1) \\ & = & (a + 1) x^3 + (b + 3) x^2 + (c + 3) x + (d + 1) \end{aligned}$

Similarly, we expand $2 f(x+1)$ as follows. $\begin{aligned} 2 f(x+1) & = & 2 (a (x+1)^3 + b(x+1)^2 + c(x+ 1) + d) \\ & = & 2(ax^3 + (3a + b)x^2 + (3a +2b + c)x + (a + b + c + d)) \\ & = & 2ax^3 + (6a + 2b)x^2 +(6a + 4b + 2c)x + (2a + 2b + 2c + 2d) \end{aligned}$

Because these two expressions are equal, we have $\begin{aligned} a + 1 & = & 2a \\ b + 3 & = & 6a + 2b \\ c + 3 & = & 6a + 4b + 2c \\ d + 1 & = & 2a + 2b + 2c + 2d \end{aligned}$

This system of equations solves to $a = 1$ , $b = -3$ , $c = 9$ , and $d = -13$ , so $f(x) = x^3 - 3x^2 + 9x - 13$ . Thus, $f(10) = 1000 - 300 + 900 - 13 = \boxed{777}$ .

Note that, in order for the LHS and the RHS to have the same degrees, f(x) must be cubic since if the degree is higher, f(x) and 2f(x+1) would have different leading coefficients. Thus, let f(x)=ax^3+bx^2+cx+d.

The LHS becomes ax^3+bx^2+cx+d+x^3+3x^2+3x+1, or (a+1)x^3+(b+3)x^2+(c+3)x+(d+1).

The RHS becomes, by the Binomial Theorem, 2[ax^3+(3a+b)x^2+(3a+2b+c)x+(a+b+c+d)], or (2a)x^3+(6a+2b)x^2+(6a+4b+2c)x+2(a+b+c+d).

Now, we may match coefficients based on the degree of the terms. 1. a+1=2a 2. b+3=6a+2b 3. c+3=6a+4b+2c 4. d+1=2a+2b+2c+2d. Clearly, from equation 1, a=1. Substituting into equation 2, b=-3. Similarly, we get c=9 and d=-13, so f(x)=x^3-3x^2+9x-13. Finally, f(10)=777.

Let $f(x) = a x ^ 3 + b x ^ 2 + c x +d$ .

Then,

$f(x + 1) = a (x + 1) ^ 3 + b (x + 1) ^ 2 + c (x + 1) +d$ .

We get this in a series of basic transformations:

$f(x)+(x+1)^3=2f(x+1)$

$a x ^ 3 + b x ^ 2 + c x +d+(x+1)^3=2(a (x + 1) ^ 3 + b (x + 1) ^ 2 + c (x + 1) +d)$

$(1-a)x^3+(3-6a-b)x^2+(3-6a-4b-c)x+1-2a-2b-2c-d=0$

Now, we get the equations:

$1-a=0$

$3-6a-b=0$

$3-6a-4b-c=0$

$1-2a-2b-2c-d=0$

The solution of these equations is: $a=1, b=-3, c=9, d=-13$ , which gives as the function:

$f(x)=x^3-3x^2+9x-13$

Now, it's easy to calculate that:

$f(10)=777$

Let $deg f(x) = n$ and the coefficient of $x^n$ in $f(x)$ = $a$ . If $n>3$ , coefficient of $x^n$ in $L.H.S.\ = a \neq 2a =$ coefficient of $x^n$ in $R.H.S.$ . If $n<3$ , degree of $x$ in $L.H.S.$ = 3 $neq n$ = degree of $x$ in $R.H.S.$ . Thus $n=3$ . Let $f(x) = ax^3+bx^2+cx+d$ .

$f(x)+(x+1)^3=2f(x+1)\Rightarrow f(x-1)+x^3=2f(x)$ $a(x-1)^3+b(x-1)^2+c(x-1)+d+x^3=2ax^3+2bx^2+2cx+2d$ .

By comparing the coefficients of $x^3$ , $x^2$ and $x$ and constant terms respectively,

$\begin{cases} a+1+2a \\ -3a+b=2b \\ 3a-2b+c=2c \\ -a+b-c+d=2d \end{cases}$

$a=1, b=-3,c=9,d=-13\Rightarrow f(x)=x^3-3x^2+9x-13$

$f(10)=(10)^3-3(10)^2+9(10)-13=\boxed{777}$

Let f(x) = a x^3 + b x^2 + c x + d

We know that f(x) is cubic by counting degrees. (if f(x) has degree 4, it cannot be equal to 2f(x+1) as only a polynomial of degree 3 is added to it)

Thus RHS = 2 [ a (x+1)^3 + b (x+1)^2 + c (x+1) + d]

Expanding and simplifying both sides,

x^3 + 3x^2 + 3x + 1 = a x^3 + (6a + b) x^2 + (6a + 4b+ c)x + (2a+2b+2c+d)

Comparing coefficients, a = 1, b = -3, c = 9, d = -13

Thus f(x) = x^3 -3x^2 + 9x + 3

f(10) = 10^3 - 3(10^2) + 9(10) + 3 = 777

From a simple analysis, we know that the degree of $f(x)$ is 3 and the coefficient of $x^3$ is 1. Then suppose $f(x)=x^3+g(x)$ the degree of $g(x)$ is 2 and we have

$x^3+g(x)+(x+1)^3=2g(x+1)+2(x+1)^3,\\ 2g(x+1)-g(x)=-3x^2-3x-1$

so the coefficient of $x^2$ is -3. Then suppose

$g(x)=-3x^2+h(x)$ the degree of $h(x)$ is 1 and we have

$2h(x+1)-6(x+1)^2+3x^2-h(x)=-3x^2-3x-1, \\2h(x+1)-h(x)=9x+5$

so the coefficient of $x$ is 9. Then suppose

$h(x)=9x+b$

we have

$18(x+1)+2b-9x-b=9x+5, \\b=-13$

So

$f(x)=x^3-3x^2+9x-13,f(10)=1000-300+90-13=777$

We can clearly see that fx is a cubic function. Let us assume that fx=x3+ax2+bx+c where x3 represents x raised to power 3,so we can now put x+1 in place of above x then we will get f(x+1) so by keeping it in equation given in the question we will get values of a,b and c by comparing hence we found fx now put x=10 and get the ans......IT IS A BIT LONGER APPROACH BUT IT IS VERY EASY.....😊😊

Supposing you can calculate f(0)=d=-13, then f(0)/2¹⁰+∑_{i=1,10}(i^3)/(2^(10+1-i)) = -13/2¹⁰+1/2¹⁰+2³/2⁹+3³/2⁸+...+9³/2²+10³/2¹=777.

Substantially the same as Mr Zuniga's approach using Python sympy module to do symbolic algebra.

Make an initial guess that f is cubic (knowing that it would have been easy to rework the solution for higher degree polynomials). Define the indeterminate in the polynomial, then the coefficients, then the polynomial itself. Expand the lefthand side of given functional, then the righthand side, collecting similar terms in the indeterminate in each case. Solve for the values of the coefficients then evaluate the polynomial for these values of the coefficients and the given value of the indeterminate.

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>>> from sympy import *
>>> var('x')
x
>>> var('a0 a1 a2 a3 a4')
(a0, a1, a2, a3, a4)
>>> f=a0+a1*x+a2*x**2+a3*x**3
>>> (2*f.subs(x,x+1)).expand().collect(x)
2*a0 + 2*a1 + 2*a2 + 2*a3*x**3 + 2*a3 + x**2*(2*a2 + 6*a3) + x*(2*a1 + 4*a2 + 6*a3)
>>> (f+(x+1)**3).expand().collect(x)
a0 + x**3*(a3 + 1) + x**2*(a2 + 3) + x*(a1 + 3) + 1
>>> solve([a3 + 1-2*a3,a2 + 3-(2*a2 + 6*a3),a1 + 3-(2*a1 + 4*a2 + 6*a3),a0+1-(2*a0 + 2*a1 + 2*a2 + +2*a3)])
{a3: 1, a0: -13, a2: -3, a1: 9}
>>> f.subs(a3,1).subs(a0,-13).subs(a2,-3).subs(a1,9).subs(x,10)
777

This is only possible if degree of polynomial is 3.Let's see how.If I consider polynomial less than 3 then degree of f(x) +(x+1)^3 will be three whereas of 2f(x+1) will be less than 3 so both sides of equality sign can never be equal in this case.If we take degree greater than 4 say n and assume that coefficent of x^n in f(x) is A then in LHS of equation coefficent of x^n will be A and on RHS coefficent of x^n will be 2A.Again a contradiction. Now take a general 3 degree equation and equate coeffiients on both sides. Cubic comes out to be x^3-3x^2+9x-13 Put x=10

We can say that f(x) is a third degree polynomial because all of the coefficients increase when you add (x+1)^3. Let the polynomial be f(x)=ax^3+bx^2+cx+d. f(x)+(x+1)^3=(a+1)x^3+(b+3)x^2+(c+3)x+(d+1). 2f(x+1)=(2a)x^3+(2b+6)x^2+(4b+2c+6)x+(2b+2c+2d+2) In each coefficient, we get that a+1=2a b+3=2b+6 c+3=4b+2c+6 d+1=2b+2c+2d+2 From these equations, we get a=1, b=-3, c=9, d=-13. Plugging these values in the original polynomial, f(x)=x^3-3x^2+9x-13. So f(10)=777

1st analyse the equation then you will get the general form of the function which is f(x) = ax^3 + bx^2 + cx + d then just find the value a, b, c, d then you will find f(x) = ax^3 + bx^2 + cx + d substitute x=10 then f(x)=777