Let f(x)=2x2+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 use the Remainder Theorem.
The Remainder Theorem states that if a polynomial f(x) is divided by x - c, the remainder is equal to f(c). In other words, if f(x) leaves the same remainder when divided by x - a as when divided by x + 2a, it means that f(a) = f(-2a).
Let's find f(a):
f(a) = 2a^2 + 40a + 25
Now, let's find f(-2a):
f(-2a) = 2(-2a)^2 + 40(-2a) + 25
= 8a^2 - 80a + 25
Since f(a) = f(-2a), we can set the two equations equal to each other:
2a^2 + 40a + 25 = 8a^2 - 80a + 25
Now, we can simplify and solve for a:
6a^2 - 120a = 0
6a(a - 20) = 0
From this equation, we have two possible solutions:
a = 0 or a = 20
However, we are given that a is a positive integer. Therefore, the value of a is 20.