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=532m+8n+2p=9-(32m+8n+2p)=-9x=-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