If (3x+1)(Ax-3)=Px2 +Qx+R,find the constants P,Qand R
*PS* the Px2 that 2 is in the upper side of x
~中二數 F.2 Maths~
2011-09-03 8:40 pm
回答 (2)
2011-09-03 9:55 pm
✔ 最佳答案
(3x+1)(Ax-3)3Ax2-9x+Ax-3
3Ax2+(-9+A)x-3
SO,
P=3A..Q=-9+A..R=-3
2011-09-03 10:42 pm
(3x+1)(Ax-3)=Px^2+Qx+R
3Ax^2-9x+Ax-3=Px^2+Qx+R
3Ax^2+(A-9)x-3=Px^2+Qx+R
P=3A
Q=A-9
R=-3
3Ax^2-9x+Ax-3=Px^2+Qx+R
3Ax^2+(A-9)x-3=Px^2+Qx+R
P=3A
Q=A-9
R=-3
參考: my answer
收錄日期: 2021-04-13 18:12:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110903000051KK00323