Simplify: ((2x^n)^2 - 1)/(2x^n-1)
where x is an integer and n is a positive interger.
for this one igot the answer the same as 2n^x+1, but will the value of the given expression be even, odd, or either? Please explain. :) Thank Yoo
factor the numerator.
(2x^n -1)(2x^n+1)
dividing, leaving you with 2x^n+1
Well, 2(any number) is even, add one, it is always odd.
How do you get from x^n to n^x?
2x^n - 1
will be even if n is even, but will never be odd, because 1 = x^0 is an even power.
f(-x) = 2(-x)^n - 1 = 2x^n-1 = f(x) if n is even
f(-x) = 2(-x)^n - 1 = -2x^n-1 ≠ -f(x) if n is odd
Oops. bobpursley got it right. I was reading it wrong, think about even or odd functions, not the value of the expression. :-(
Thank you guys so so so so much! I have been looking at this forever trying to see how to solve it, when it was right in front of me!! :D
But Steve, you used the equation 2x^n-1 rather than 2x^n+1, does that make a difference??
my bad - typo.
it makes no difference to the even/odd results.
To simplify the expression ((2x^n)^2 - 1)/(2x^n-1), let's break it down step by step.
Step 1: Simplify the numerator.
((2x^n)^2 - 1) can be expanded using the exponent rule (a^2 = a * a):
(2x^n * 2x^n) - 1
4x^2n - 1
Step 2: Simplify the denominator.
2x^n - 1
Now, let's rewrite the expression with the simplified forms:
(4x^2n - 1)/(2x^n - 1)
To determine whether the value of the expression will be even, odd, or either, we need to examine whether the numerator and denominator have any common factors.
If we factor out (2x^n - 1) from both numerator and denominator, we get:
(2x^n - 1)(2x^n + 1)/(2x^n - 1)
Now, we can cancel out the common factor from the numerator and denominator:
2x^n + 1
So, the simplified expression is 2x^n + 1.
Now let's analyze this expression. Since x is an integer and n is a positive integer, both x^n and 1 are always non-negative.
If x is even, x^n will also be even.
If x is odd, x^n will also be odd.
Adding 1 to any even or odd value will always yield an odd number.
Therefore, the expression 2x^n + 1 will always be an odd number regardless of the values of x and n.
So, the value of the given expression will always be odd.