Question

化简问题求救！！！

请问(5-β)(6-β)(4-β)-4(4-β)-4(6-β)=(5-β)(2-β)(8-β),如何从等式左边化到等式右边？像3次方程如何分解成积的形式？？

Answer 1

(5-β)(6-β)(4-β)-4(4-β)-4(6-β)

=(5-β)(6-β)(4-β)-4(5-β-1)-4(5-β+1)
=(5-β)(6-β)(4-β)-4(5-β)+4-4(5-β)-4
=(5-β)(6-β)(4-β)-8(5-β)
=(5-β)[(6-β)(4-β)-8]
=(5-β)(β^2-10β+16)
=(5-β)(2-β)(8-β)

Answer 2

(5-β)(6-β)(4-β)-4(4-β)-4(6-β)
=(5-β)(6-β)(4-β)-40+8β
=(5-β)(6-β)(4-β)-8(5-β)
=(5-β)[(6-β)(4-β)-8]
=(5-β)(β^2-10-β-16)
=(5-β)(2-β)(8-β),

Answer 3

我想本题的出题目的是让你去熟悉平方差公式和等价替换因式分解的：
思路：
令 5-β = x
上式左边 = x(x+1)(x-1)-4(x-1)-4(x+1)
= x(x²-1)-4((x-1)+(x+1))
= x³-x-8x
= x(x-3)(x+3)
再将 5-β = x带回上式，得到等式右边：(5-β)(2-β)(8-β)

追问

很好，那 x³-x-8x这一步如何化成 x(x-3)(x+3)呢？？

追答

x³-x-8x = x³-9x
=x(x²-9)(提取公因子)
=x(x-3)(x+3)(平方差公式)