Question

分解因式3x^4+4x^2+28x+5

Answer 1

3x^4+4x^2+28x+5=[3x^4-6x^3+15x^2]+[6x^3-12x^2+30x]+[x^2-2x+5]=3x^2(x^2-2x+5)+6x(x^2-2x+5)+(x^2-2x+5)==(x^2-2x+5)(3x^2+6x+1).就是添加了6x^3中间项,把二次项分了几部分

Answer 2

你好！详细如下：首先x^4+6x^2+25=x^4+10x^2+25-4x^2=(x^2+5)^2-4x^2=(x^2+2x+5)(x^2-2x+5)其次3x^4+4x^2+28x+5=（9x^4+12x^2+84x+15)/3=[(9x^4+48x^2+64)-(36x^2-84x+49)]/3=[(3x^2+8)^2-(6x-7)^2]/3=(3x^2+8+6x-7)(3x^2+8-6x+7)/3=(3x^2+6x+1)(3x^2-6x+15)/3=(3x^2+6x+1)(x^2-2x+5)（仅供参考）希望可以帮到你！

Answer 3

3x^4+4x^2+28x+5=3x^2(x^2-2x+5)+6x(x^2-2x+5)+(x^2-2x+5)

=(x^2-2x+5)(3x^2+6x+1).