初二因式分解题SOS!
分解因式:x^2-2xy-3y^2+2x+10y-8
参考答案:答案是(X-3Y+4)*(X+Y-2)
x^2-2xy-3y^2+2x+10y-8
=x^2+2(1-y)x+(1-y)^2-(1-y)^2-3y^2+10y-8
=[x+(1-y)]^2-4y^2+12y-9
=[x+(1-y)]^2-(2y-3)^2
=[x+(1-y)-(2y-3)]*[x+(1-y)+(2y-3)]
=(X-3Y+4)*(X+Y-2)
分解因式:x^2-2xy-3y^2+2x+10y-8
参考答案:答案是(X-3Y+4)*(X+Y-2)
x^2-2xy-3y^2+2x+10y-8
=x^2+2(1-y)x+(1-y)^2-(1-y)^2-3y^2+10y-8
=[x+(1-y)]^2-4y^2+12y-9
=[x+(1-y)]^2-(2y-3)^2
=[x+(1-y)-(2y-3)]*[x+(1-y)+(2y-3)]
=(X-3Y+4)*(X+Y-2)