Question 1209938
Let's solve this functional equation problem step by step.

**1. Simplify the equation:**

f(x) + f(2x + y) + 5xy = f(4x - y) - x² + 5xy - 8x + 17y + 1

Subtract 5xy from both sides:

f(x) + f(2x + y) = f(4x - y) - x² - 8x + 17y + 1

**2. Strategically substitute values for x and y:**

* **Let x = 0:**

f(0) + f(y) = f(-y) + 17y + 1

* **Let y = 0:**

f(x) + f(2x) = f(4x) - x² - 8x + 1

* **Let y = 2x:**

f(x) + f(4x) = f(0) - x² + 34x + 1

* **Let y = 4x:**

f(x) + f(6x) = f(0) - x² + 68x + 1

* **Let y = -2x:**

f(x) + f(0) = f(6x) - x² - 8x - 34x + 1
f(x) + f(0) = f(6x) - x² - 42x + 1

**3. Combine equations:**

From the equations with y=4x and y=-2x:

f(x) + f(6x) = f(0) - x² + 68x + 1
f(x) + f(0) = f(6x) - x² - 42x + 1

Subtract the second equation from the first:

f(6x) - f(0) = f(0) - x² + 68x + 1
f(6x) = 2f(0) - x² + 68x + 1

Subtract f(x) from f(x) + f(2x) = f(4x) - x² - 8x + 1 and f(x) + f(4x) = f(0) - x² + 34x + 1

f(2x) - f(4x) = f(4x) - f(0) - 42x

f(2x) -2f(4x)+f(0)=-42x

**4. Test for a quadratic form:**

Let f(x) = ax² + bx + c.

* f(0) = c

Substitute f(x) into the original equation:

ax² + bx + c + a(2x+y)² + b(2x+y) + c + 5xy = a(4x-y)² + b(4x-y) + c - x² - 8x + 17y + 1

ax² + bx + c + a(4x² + 4xy + y²) + 2bx + by + c + 5xy = a(16x² - 8xy + y²) + 4bx - by + c - x² - 8x + 17y + 1

ax² + bx + c + 4ax² + 4axy + ay² + 2bx + by + c + 5xy = 16ax² - 8axy + ay² + 4bx - by + c - x² - 8x + 17y + 1

(5a)x² + (3b)x + 2c + (4a+5)xy + ay² +by = (16a-1)x² + (4b-8)x + c + (-8a)xy + ay² + (-b+17)y + 1

Equating coefficients:

* x²: 5a = 16a - 1 => 11a = 1 => a = 1/11
* x: 3b = 4b - 8 => b = 8
* c: 2c = c + 1 => c = 1
* xy: 4a + 5 = -8a => 12a = -5 => a = -5/12
* y²: a = a
* y: b = -b + 17 => 2b = 17 => b = 17/2

Since the coefficients don't agree, the form f(x) = ax² + bx + c does not work.

**5. Try f(x) = ax + b:**

ax + b + a(2x+y) + b + 5xy = a(4x-y) + b - x² - 8x + 17y + 1

ax + b + 2ax + ay + b + 5xy = 4ax - ay + b - x² - 8x + 17y + 1

3ax + 2b + ay + 5xy = 4ax - ay + b - x² - 8x + 17y + 1

* x² = 0 => 0 = -1 (contradiction)

**6. Try f(x) = ax^2 + bx + c, y=0**

f(x) + f(2x) = f(4x) - x^2 - 8x + 1
ax^2+bx+c+a(2x)^2+b(2x)+c = a(4x)^2+b(4x)+c-x^2-8x+1
5ax^2+3bx+2c = 16ax^2+4bx+c-x^2-8x+1
0=11ax^2+bx-c-x^2-8x+1
0=(11a-1)x^2+(b-8)x+(1-c)

11a-1=0 => a=1/11
b-8=0 => b=8
1-c=0 => c=1

f(x) = x^2/11 + 8x + 1

f(10) = 100/11 + 80 + 1 = 100/11 + 81 = 100/11 + 891/11 = 991/11