General

[quote=bloodIsShed]nope. a is just a constant[/quote]

Ah, thanks I got it now.

nope. a is just a constant

[quote=bloodIsShed]
f(x) = f(-x)
p(x - a) = p(-x -a)
[/quote]

But shouldn't it be p(x-a) = p(-x+a) ?
For example:
f(b) = f(-b) || b = a+1
f(a+1) = f(-a-1)

?

f(x) = p(x-a) is an even function.
since f(x) is an even function, f(x) = f(-x)

p(x) is a 2. degree polynomial and its roots are x = 1 and x = 5.
since the roots are given, it's clear that p(x) = (x - 1) * (x - 5)
or p(x) = x^2 - 6x + 5
so p(x - a) = (x-a)^2 - 6(x-a) + 5

f(x) = f(-x)
p(x - a) = p(-x -a)
(x-a)^2 - 6(x-a) + 5 = (-x-a)^2 - 6(-x-a) + 5
x^2 -2ax + a^2 - 6x +6a + 5 = x^2 + 2ax + a^2 + 6x + 6a + 5
-2ax - 6x = 2ax + 6x
-2a - 6 = 2a + 6
4a = -12
a = -3

[quote=dadenong]it's been awhile since i've done this kind of math but i'll give it a shot 1 sec
yeah nope sorry can't remember the method[/quote]

Thanks for trying though.

The answer is supposed to be a = -3. However, I can't even get to eliminate all the x.

[quote=isiah13]don't ask basil...[/quote]

basil is the best place for math questions, its full of nerds asians

don't ask basil...