Question

请帮忙因式分解这道题，谢谢。

f(x)=x四次方-11x三次方+36x二次方-16x-64

就是分解成几项乘积形式……
谢谢 T_T

Answer 1

x四次方-11x三次方+36x二次方-16x-64
=x^4-11x³+28x²+8x²-16x-64
=x²(x²-11x+28)+8(x²-2x-8)
=x²(x-4)(x-7)+8(x-4)(x+2)
=(x-4)(x³-7x²+8x+16)
=(x-4)(x³-7x²-8x+16x+16)
=(x-4)[(x(x²-7x-8)+16(x+1)]
=(x-4)[x(x-8)(x+1)+16(x+1)]
=(x-4)(x+1)(x²-8x+16)
=(x-4)³(x+1)

Answer 2

=(x-5y+3)(x-3y-1) 这题太难了!,但你弄清楚三项数相加减后的和差再平方就迎刃而解了. x^2-8xy+15y^2+2x-4y-3 =x^2-8xy+2x-8y+

Answer 3

f(x)=x⁴-11x³+36x²-16x-64
解：由于f(-1)=0，因此x+1必是其因式:
故f(x)=x⁴-11x³+36x²-16x-64=x³(x+1)-12x²(x+1)+48x(x+1)-64(x+1)
=(x+1)(x³-12x²+48x-64) (x=4是x³-12x²+48x-64=0的根，故必有x-4的因式)
=(x+1)[x²(x-4)-8x(x-4)+16(x-4)]
=(x+1)(x-4)(x²-8x+16)
=(x+1)(x-4)(x-4)²=(x+1)(x-4)³