# Thread: (-x)^8

1. ## (-x)^8

$y=(-x)^8-3(-x)^4+2(-x)^2+2$

lets look at the first term...

this is the same as $(-x)(-x)(-x)(-x)(-x)(-x)(-x)(-x)=x^8$

but if you wrote it as $-x^8$ it is the same as...

$(-1)\cdot(x)^8=(-1)(x)^8=-x^8$

different answers.

now lets look at the second term...

this is the same as
$(-3)\cdot[(-x)(-x)(-x)(-x)] = (-3)\cdot(x)^4=-3x^4$

but i though you could do this...

$(-3)\cdot(-x)^4=3x^4$

these are different answers.

which is correct?

i really hope this isn't another missing dollar riddle type situation because then i would feel silly.

2. Right. The exponent only applies to the thing immediately preceding it. That's either a variable, or a function, or something in parentheses. Remember PEMDAS?

So, for the expression

$(-3)\cdot(-x)^4=3x^4,$

which is, incidentally, incorrect, your problem is that you didn't to the operations in the correct order. You need to do the parentheses first, then the exponentiation, yielding

$-3x^{4}.$

Make sense?

3. yes this does make sense.

so odd functions keep negatives negative and even functions make negatives positive.

4. Originally Posted by jonnygill
yes this does make sense.

so odd functions keep negatives negative and even functions make negatives positive.
Better to say that odd exponents keep negatives negative, and even exponents make negatives positive (if the minus sign is inside the parentheses). However, even here you have to be careful with expressions like

$(x-y)^{3}.$

To find out what happens here, you'd need to use the binomial theorem.