三道因式分解的题目,请写出详细过程

1个回答

  • (1)a^5+a+1

    =a^5-a²+a²+a+1

    =a²(a³-1)+(a²+a+1)

    =a²(a-1)(a²+a+1)+(a²+a+1)

    =(a²+a+1)(a³-a²+1)

    (2)用十字相乘法法,把y作为常数,x 做降幂排列.

    原式=2x2+(y-4)x+(-y2+5y-6)

    =2x2+(y-4)x+[-(y2-5y+6)]

    =2x2+(y-4)x+[-(y-2)(y-3)]

    作十字分解,如下:

    1 y-3

    2 -y+2

    则:

    原式=[1x+(y-3)][2x+(-y+2)]

    =(x+y-3)(2x-y+2)

    验算,结果=2x2-xy+2x+2xy-y2+2y-6x+3y-6

    =2x2+xy-y2+5y-6=题目的式子 无误

    (3)设x^4-x^3+4x^2+3x+5

    =(x^2+ax+1)(x^2+bx+5)

    =x^4+(a+b)x^3+(ab+b)x^2+(5a+b)x+5

    根据对应项系数相等,得

    a+b=-1 ①

    ab+b=4 ②

    5a+b=3 ③

    由①③得a=1,b=-2

    代入②中,成立

    ∴x^4-x^3+4x^2+3x+5=(x^2+x+1)(x^2-2x+5)