已知多项式mx^5+nx^3+px-4，当x=2时，此多项式的值是5，求当x=-2时，多项式的值是多少？

x=2时,
mx^5+nx^3+px-4=32m+8n+2p-4=5
32m+8n+2p=9
-(32m+8n+2p)=-9
x=-2时
mx^2+nx^3+px-4=-32m-8n-2p-4=-(32m+8n+2p)-4
把-(32m+8n+2p)=-9代入
mx^2+nx^3+px-4=-32m-8n-2p-4=-(32m+8n+2p)-4=-9-4=-13

m2^5+n2^3+2p-4=5
m2^5+n2^3+2p=9

m(-2)^5+n(-2)^3+p(-2)-4
=-m2^5-n2^3-2p-4
=-(m2^5+n2^3+2p)-4
=-9-4
=-13
很详细，给分哦。谢谢

32m+8n+2p-4=5
-32m-8n-2p-4=-(5+4)-4=-13