Let f(x)=2x^2+40x+25. Given that f(x) leaves the same remainder when divided by x−a as when divided by x+2a for a positive integer a, what is the value of a?

To find the value of a, we need to equate the remainders when f(x) is divided by x - a and x + 2a.

When f(x) is divided by x - a, the remainder can be found by substituting x = a into the function f(x). Similarly, when f(x) is divided by x + 2a, the remainder can be found by substituting x = -2a into the function f(x).

Let's calculate the remainders for both cases:

1. Dividing f(x) by x - a:
Substituting x = a into f(x):

f(a) = 2a^2 + 40a + 25

2. Dividing f(x) by x + 2a:
Substituting x = -2a into f(x):

f(-2a) = 2(-2a)^2 + 40(-2a) + 25
= 8a^2 - 80a + 25

Since we are given that these remainders are the same, we can set them equal to each other:

2a^2 + 40a + 25 = 8a^2 - 80a + 25

Now, we can simplify and solve for a:

2a^2 + 40a = 8a^2 - 80a

Subtracting 2a^2 and 40a from both sides:

0 = 6a^2 - 120a

Dividing both sides by 6a:

0 = a(a - 20)

Now, we have two possible solutions: a = 0 or a = 20. Since a needs to be a positive integer, the value of a is 20.

Therefore, the value of a is 20.