設多項式ax5+bx3+cx+d=M,已知當x=0時,M=-5;當x=-3時,M=9,則當x=3時,M=________.
-19
分析:將數(shù)值x=0,M=-5代入多項式ax5+bx3+cx+d=M,得d=-5.由ax5+bx3+cx+d=M得:ax5+bx3+cx=M-d,再將x=-3與x=3分別代入ax5+bx3+cx=M-d中,觀察式子的特點,從而找到解答問題的突破口.
解答:將x=-3代入ax5+bx3+cx=M-d,
得a(-3)5+b(-3)3+(-3)c=9-(-5)
∴-a(3)5-b(3)3-3c=14
再將x=3代入ax5+bx3+cx=M-d
得a(3)5+b(3)3+(3)c=M-(-5)
∴M=a(3)5+b(3)3+(3)c-5
=-14-5
=-19
故填-19.
點評:將x=-3與x=3分別代入ax5+bx3+cx=M-d中,觀察出-a(3)5-b(3)3-3c與a(3)5+b(3)3+3c是一對相反數(shù),是解答本題的關鍵.