因式分解x的五次方-3x的三次方-x的三次方+11x的平方-12x+4
答案:1 悬赏:50 手机版
解决时间 2021-01-25 02:37
- 提问者网友：火车头
- 2021-01-24 17:09
因式分解x的五次方-3x的三次方-x的三次方+11x的平方-12x+4
最佳答案
- 五星知识达人网友：独行浪子会拥风
- 2021-01-24 17:34
x^5-3x^4-x^3+11x^2-12x+4
=(x^5-x^4)-(2x^4-2x^3)-(3x^3-3x^2)+(8x^2-8x)-(4x-4)
=x^4(x-1)-2x^3(x-1)-3x^2(x-1)+8x(x-1)-4(x-1)
=(x-1)(x^4-2x^3-3x^2+8x-4)
=(x-1)[(x^4-x^3)-(x^3-x^2)-(4x^2-4x)+(4x-4)]
=(x-1)[x^3(x-1)-x^2(x-1)-4x(x-1)+4(x-1)]
=(x-1)^2(x^3-x^2-4x+4)
=(x-1)^2[x^2(x-1)-4(x-1)]
=(x-1)^3(x^2-4)
=(x-1)^3(x+2)(x-2)
=(x^5-x^4)-(2x^4-2x^3)-(3x^3-3x^2)+(8x^2-8x)-(4x-4)
=x^4(x-1)-2x^3(x-1)-3x^2(x-1)+8x(x-1)-4(x-1)
=(x-1)(x^4-2x^3-3x^2+8x-4)
=(x-1)[(x^4-x^3)-(x^3-x^2)-(4x^2-4x)+(4x-4)]
=(x-1)[x^3(x-1)-x^2(x-1)-4x(x-1)+4(x-1)]
=(x-1)^2(x^3-x^2-4x+4)
=(x-1)^2[x^2(x-1)-4(x-1)]
=(x-1)^3(x^2-4)
=(x-1)^3(x+2)(x-2)
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