把下列各式分解因式:(1)a4+64b4;(2)x4+x2y2+y4;(3)x2+(1+x)2+(x+x2)2;(4)(c-a)2-4(b-c)(a-b);(5
题型:解答题难度:一般来源:不详
把下列各式分解因式: (1)a4+64b4; (2)x4+x2y2+y4; (3)x2+(1+x)2+(x+x2)2; (4)(c-a)2-4(b-c)(a-b); (5)x3-9x+8; (6)x3+2x2-5x-6 |
答案
(1)a4+64b4 =a4+64b4+16a2b2-16a2b2 =(a2+8b2)2-(4ab)2 =(a2+8b2-4ab)(a2+8b2+4ab); (2)x4+x2y2+y4; =x4+2x2y2+y4-x2y2 =(x2+y2)2-(xy)2 =(x2+y2-xy)(x2+y2+xy); (3)x2+(1+x)2+(x+x2)2 =1+2(x+x2)+(x+x2)2 =(1+x+x2)2; (4)设b-c=x,a-b=y,则c-a=-(x+y), 则(c-a)2-4(b-c)(a-b) =[-(x+y)]2-4xy, =(x-y)2, 所以(c-a)2-4(b-c)(a-b) =(b-c-a+b)2 =(2b-a-c)2; (5)x3-9x+8; =x3-x-8x+8 =(x3-x)-(8x-8) =x(x2-1)-8(x-1) =x(x+1)(x-1)-8(x-1) =(x-1)(x2+x-8); (6)x3+2x2-5x-6 =x3+x2+x2+x-6x-6, =(x3+x2)+(x2+x)-(6x+6) =x2(x+1)+x(x+1)-6(x+1) =(x+1)(x2-x-6) =(x+1)(x+3)(x-2). |
举一反三
已知关于x、y的二次式x2+7xy+ay2-5x-45y-24可分解为两个一次因式的乘积,求a的值. |
把下列各式分解因式: (1)x4-7x2+1; (2)x4+x2+2ax+1-a2 (3)(1+y)2-2x2(1-y2)+x4(1-y)2 (4)x4+2x3+3x2+2x+1. |
把下列各式分解因式: (1)4x3-31x+15; (2)2a2b2+2a2c2+2b2c2-a4-b4-c4; (3)x5+x+1; (4)x3+5x2+3x-9; (5)2a4-a3-6a2-a+2. |
已知x2+2x+5是x4+ax2+b的一个因式,求a+b的值. |
把多项式x2y-xy+y分解因式所得的结果是 ______. |
最新试题
热门考点