If h(x)=2x^4+x^3-3x^2+x-1 find h(2) - h(-2)

Question

So from what I understand I replace the x's with (2) for the problem then once I get the number I then replace the x's with (-2) to get a different answer then the first one. So do I leave the two numbers on the paper or do I take the first answer minus the second and that is my overall answer? I'm confused on how to do this.

Answer 1

y=2x^4+x^3 -3x^2 +x-1 . . . evaluate y at x=2...2*16 +8 -3*4 +2-1 ....32+8-12+2-1...=29 . . . at x=-2...2*16 -8 -3*4 -2 -1....32-8 -12-3 ....=9 . . . 29-9=20

Answer 2

"Solution: Given h(x)=2x^4+x^3-3x^2+x-1

h(2)=2x^4+x^3-3x^2+x-1

=2*2^4+2^3-3*2^2+2-1

=29

h(-2)=2x^4+x^3-3x^2+x-1

=2(-2)^4+(-2)^3-3*(-2)^2+(-2)-1

= 9

h(2)-h(-2)=29-9

=20"