Find the values of A, B and C in the following identitiy.
(2y+3)(y-4)+(y-2)(y+5)=Ay^2+By+C
Please show your working steps and explain your answer in details.
A F. 2 maths question (最佳解答全取10分)
2007-01-03 2:41 am
回答 (4)
2007-01-03 2:50 am
✔ 最佳答案
L.H.S(Left Hand Side)=(2y+3)(y-4)+(y-2)(y+5)=(2y^2-8y+3y-12)+(y^2+5y-2y-10)
=2y^2-8y+3y-12+y^2+5y-2y-10
=3y^2-2y-22
By comparing like terms, we have
Ay^2=3y^2
A=3
By=-2y
B=-2
C=-22
希望你明啦
參考: 自己
2007-01-03 3:53 am
(2y+3)(y-4)+(y-2)(y+5)=Ay^2+By+C
2y^2-5y-12+y^2+3y-10=Ay^2+By+C
3y^2-2y-22=Ay^2+By+C
A=3,B=-2,C=-22
2y^2-5y-12+y^2+3y-10=Ay^2+By+C
3y^2-2y-22=Ay^2+By+C
A=3,B=-2,C=-22
2007-01-03 2:46 am
(2y+3)(y-4)+(y-2)(y+5)=Ay^2+By+C
2y^2 - 8y + 3y - 12 + y^2 - 2y + 5y - 10 = Ay^2+By+C
3y^2 - 2y - 22 = Ay^2+By+C
so, A = 3
B = -2
C = -22
2y^2 - 8y + 3y - 12 + y^2 - 2y + 5y - 10 = Ay^2+By+C
3y^2 - 2y - 22 = Ay^2+By+C
so, A = 3
B = -2
C = -22
2007-01-03 2:46 am
(2y+3)(y-4)+(y-2)(y+5)=Ay^2+By+C
2y^2-5y-12+y^2+3y-10=Ay^2+By+C
3y^2-2y-22=Ay^2+By+C
A=3,B=-2,C=-22
2y^2-5y-12+y^2+3y-10=Ay^2+By+C
3y^2-2y-22=Ay^2+By+C
A=3,B=-2,C=-22
收錄日期: 2021-04-12 21:14:38
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