Question

因式分解2题

Answer 1

(1)原式=20y³+20xy²+xy-x-y
= 20y²（x+y)+y(x+y)-(x+y)
=(x+y)(20y²+y-1)
=（x+y)(4y+1)(5y-1)
十字相乘（最后一步)(2)用待定系数法，先分解x^2-2xy-3y^2=(x-3y)(x+y)
那么原式=（x-3y+A)(x+y+B)=x^2-2xy-3y^2+Ax+Ay+Bx-3By+AB
A+B=3
A-3B=-5
A=1
B=2
原式=（x-3y+1)(x+y+2)赞同2| 评论

Answer 2

20xy²+y²-x+20y3+xy-y
=20y²(x+y+y(x+y)-(x+y)
=(20y²+y-1)*(x+y)
=(4y-1)(5y+1)(x+y)

x²-2xy-3y²+3x-5y+2
=(x-3y)(x+y)+3x-5y+2

x²-2xy-3y²+3x-5y+2
=x(x+2y+3)-(3y²+5y-2)
=x(x+2y+3)-(y+2)(3y-1)