Determining if function odd or even This exercise on the Khan Academy requires you to determine whether the following function is odd or even
f(x) = $-5x^5 - 2x - 2x^3$
To answer the question, the instructor goes through the following process


*

*what is f(-x)

*f(-x) = $-5(-x)^5 - 2(-x) - 2(-x)^3$

*f(-x) = $5(x)^5 +2(x) + 2(x)^3$

*f(-x) = $-(-5x^5 - 2x - 2x^3)$

*f(-x) = -f(x)


Therefore f(x) is odd.
Question. Why is everything converted to a negative both inside and outside the parenthesis at step 4?
At step 2, he's making x negative. At step 3, we get the result of making the x values negative. So why add the negatives again at step 4?
 A: If you have troubles with this approach, what you can do instead is the following:
Step 1 Calculate $f(-x)$. This is 1.-3. in your computation.
Step 2. Calculate $-f(x)$. In this case
$$-f(x)=-(-5x^5 - 2x - 2x^3)=5x^5+2x+2x^3$$
Step 3 Compare. If they are equal, the function is odd, if not, it is not odd.
[ Step 4 Compare $f(-x)$ to $f(x)$. If they are equal, the function is even, if not, not.]
A: we have $$f(-x)=-5(-x)^5-2(-x)-2(-x)^3=5x^5+2x+2x^3=-f(x)$$
since $$(-x)^5=-x^5$$
$$-x=-x$$
and $$(-x)^3=-x^3$$
A: Let's follow the simplest process
Notice, $$f(x)=-5x^5-2x-2x^3$$ $$f(-x)=-5(-x)^5-2(-x)-2(-x)^3$$ $$ =5x^5+2x+2x^3$$ $$\implies f(x)+f(-x)=-5x^5-2x-2x^3+5x^5+2x+2x^3=0$$
Hence, $f(x)$ is odd. 
A: In step 4, the instructor is factoring out a $-1$.
Just like $2a + 4b + 6c$ can be factored into $2(a + 2b + 3c)$, so can $5x^5 +2x + 2x^3$ be factored into $-1(-5x^5 - 2x - 2x^3)$.  He just omitted the $1$ when writing it.
The reason he factored out a $-1$ is because that makes the inner expression the same as $f(x)$, and he's attempting to prove that $f(-x) = -f(x)$.
