因式分解⒈x^4+4x³+4x²-11﹙x²+2x﹚+24 ⒉﹙x+1﹚﹙x+2﹚﹙3x-1﹚﹙4x-1﹚+6x^4用换元法或添﹑拆项法⒊x^4+x³+﹙9/4﹚x²+x+1⒋﹙x+1﹚﹙x+﹚﹙x+3﹚﹙x+4﹚-2
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 03:08:33
![因式分解⒈x^4+4x³+4x²-11﹙x²+2x﹚+24 ⒉﹙x+1﹚﹙x+2﹚﹙3x-1﹚﹙4x-1﹚+6x^4用换元法或添﹑拆项法⒊x^4+x³+﹙9/4﹚x²+x+1⒋﹙x+1﹚﹙x+﹚﹙x+3﹚﹙x+4﹚-2](/uploads/image/z/469837-37-7.jpg?t=%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%E2%92%88x%5E4%EF%BC%8B4x%26%23179%3B%EF%BC%8B4x%26%23178%3B%EF%BC%8D11%EF%B9%99x%26%23178%3B%EF%BC%8B2x%EF%B9%9A%EF%BC%8B24+%E2%92%89%EF%B9%99x%EF%BC%8B1%EF%B9%9A%EF%B9%99x%EF%BC%8B2%EF%B9%9A%EF%B9%993x%EF%BC%8D1%EF%B9%9A%EF%B9%994x%EF%BC%8D1%EF%B9%9A%EF%BC%8B6x%5E4%E7%94%A8%E6%8D%A2%E5%85%83%E6%B3%95%E6%88%96%E6%B7%BB%EF%B9%91%E6%8B%86%E9%A1%B9%E6%B3%95%E2%92%8Ax%5E4%EF%BC%8Bx%26%23179%3B%EF%BC%8B%EF%B9%999%EF%BC%8F4%EF%B9%9Ax%26%23178%3B%EF%BC%8Bx%EF%BC%8B1%E2%92%8B%EF%B9%99x%EF%BC%8B1%EF%B9%9A%EF%B9%99x%EF%BC%8B%EF%B9%9A%EF%B9%99x%EF%BC%8B3%EF%B9%9A%EF%B9%99x%EF%BC%8B4%EF%B9%9A%EF%BC%8D2)
因式分解⒈x^4+4x³+4x²-11﹙x²+2x﹚+24 ⒉﹙x+1﹚﹙x+2﹚﹙3x-1﹚﹙4x-1﹚+6x^4用换元法或添﹑拆项法⒊x^4+x³+﹙9/4﹚x²+x+1⒋﹙x+1﹚﹙x+﹚﹙x+3﹚﹙x+4﹚-2
因式分解⒈x^4+4x³+4x²-11﹙x²+2x﹚+24 ⒉﹙x+1﹚﹙x+2﹚﹙3x-1﹚﹙4x-1﹚+6x^4
用换元法或添﹑拆项法
⒊x^4+x³+﹙9/4﹚x²+x+1
⒋﹙x+1﹚﹙x+﹚﹙x+3﹚﹙x+4﹚-24
因式分解⒈x^4+4x³+4x²-11﹙x²+2x﹚+24 ⒉﹙x+1﹚﹙x+2﹚﹙3x-1﹚﹙4x-1﹚+6x^4用换元法或添﹑拆项法⒊x^4+x³+﹙9/4﹚x²+x+1⒋﹙x+1﹚﹙x+﹚﹙x+3﹚﹙x+4﹚-2
如果是针对考试题目的话,试根法不失为一种不错的方法:
第一题:试根为x=1(一般就±1,±2,±0.5);所以(x-1)是其一个因式,然后第一个式子:x^4+4^3-7x^2-22x+24;
然后做多项式的除法,就跟代数的除法差不多(多项式按未知数的次数降序排列,如果没有x的几次项项,以0代替之).除以(x-1)得x^3+5x^2-2x-24;试根为2;所以又一因式为(x-2);剩下的就是一个一元二次多项式了;
第二题呢,没有想到什么好办法……如果有的话,请告诉我一下!
⒈x^4+4x³+4x²-11﹙x²+2x﹚+24
=x²(x+2)²-11x(x+2x)+24
=[x(x+2)-3][x(x+2)-8]
=(x²+2x-3)(x²+2x-8)
=(x-1)(x+3)(x-2)(x+4)
⒉﹙x+1﹚﹙x+2﹚﹙3x-1﹚﹙4x-1﹚+6...
全部展开
⒈x^4+4x³+4x²-11﹙x²+2x﹚+24
=x²(x+2)²-11x(x+2x)+24
=[x(x+2)-3][x(x+2)-8]
=(x²+2x-3)(x²+2x-8)
=(x-1)(x+3)(x-2)(x+4)
⒉﹙x+1﹚﹙x+2﹚﹙3x-1﹚﹙4x-1﹚+6x^4
=(4x²+3x-1)(3x²+5x-2)+6x^4
=12x^4+9x³-3x²+20x³+15x²-5x-8x²-6x+2+6x^4
=18x^4+29x³+4x²-11x+2
=
⒊x^4+x³+﹙9/4﹚x²+x+1
=
⒋﹙x+1﹚﹙x+2﹚﹙x+3﹚﹙x+4﹚-24
=(x²+5x+4)(x²+5x+6)-24
=(x²+5x)²+10(x²+5x)+24-24
=(x²+5x)(x²+5x+10)
收起