# Why won't 2x^2 change when doing f(-x)?

School is trying to teach me about Descartes' rule of signs, in that how many times the signs (+, -) change in an equation shows how many possible positive zeros there are. To do this, they are showing an equation like this:

f(x) = 2x^5 - 3x^4 + 2x^2 + 5x - 1

and then when they perform f(-x), which, to my knowledge is just flipping the signs on each number, they show an equation like this:

f(-x) = -2x^5 - 3x^4 + 2x^2 - 5x - 1

I cannot wrap my head around why in the world, for some reason, both "2x^2" remains positive after the operation, as well as "-1" remaining negative. Can someone please explain?

### 2 Answers

- PuzzlingLv 71 month agoFavourite answer
If the power is odd (5, 3, 1) the sign is flipped. That's because (-1)^n = -1 when n is odd.

If the power is even (4, 2, 0) the sign is NOT flipped. That's because (-1)^n = 1 when n is even.

Example:

(-1)³ → (-1)(-1)(-1) = -1 ← sign changes when you multiply by -1

(-1)² → (-1)(-1) = 1 ← sign doesn't change when you multiply by 1.

Now back to your specific polynomial:

f(x) = 2x^5 - 3x^4 + 2x^2 + 5x - 1

........ odd ... even .. even . odd . even

........ flip .... keep .. keep .. flip . keep

f(-x) = -2x^5 - 3x^4 + 2x^2 - 5x - 1

Update:

There is no x in the constant -1 at the end so it doesn't change. Or as you can think of it as having an implicit x⁰ since x⁰ = 1.

Also noticed there isn't an x³ term so it messes up the alternating pattern. That's why it is best to just look at the exponents and flip the signs on the odd powers.

- AshLv 71 month ago
f(x) = 2x⁵ - 3x⁴ + 2x² + 5x - 1

For f(x) → f(-x), every x is replaced by -x

f(-x) = 2(-x)⁵ - 3(-x)⁴ + 2(-x)² + 5(-x) - 1

f(-x) = -2x⁵ - 3x⁴ + 2x² - 5x - 1