Question

已知x²-2xy-3y²+6x-14y+p能分解成两个一次因式之积，求常数P，并且因式分解

Answer 1

x²-2xy-3y²+6x-14y+p
= (x²-2xy-3y²) + 6x-14y + p
= (x+y)(x-3y) + 6x-14y + p
= (x+y)(x-3y) + (x+y) + 5x-15y + p
= (x+y)(x-3y+1) + 5x-15y + p
= (x+y)(x-3y+1) + (5x-15y+5) - 5 + p
= (x+y+5)(x-3y+1) - 5 + p
上式中，未知数x与y都已包含在两个因式之中。所以剩余的 -p+5 就应该等于0
-5+p = 0
p=5
原式=(x+y+5)(x-3y+1)

Answer 2

原式=(x-3y)(x+y)+6x-14y+p
设原式=(x-3y+a)(x+y+b)
=(x-3y)(x+y)+(a+b)x+(a-3b)y+ab
所以a+b=6
a-3b=-14
ab=p
所以
b=5,a=1
所以
p=ab=5

原式=(x-3y+1)(x+y+5)

Answer 3

x²-2xy-3y²+6x-14y+p
=(x-3y)(x+y)+6x-14y+p
p=A*B
A+B=6 (1)
-3A+B=-14 (2)
(1)-(2)得
4A=20
A=5,
则 p=5
=(x-3y)(x+y)+6x-14y+5
=(x-3y+1)(x+y+5)