Question

在实数范围内分解因式：

χ²－3y²＝____
χ²＋6χ＋7＝____
4χ²－4χ－1＝_____

Answer 1

x²-3y²
=x²-(√3y)²
=(x+√3y)(x-√3y)

x²+6x+7
=(x²+6x+9)-2
=(x+3)²-√2²
=[(x+3)+√2] [(x+3)-√2]
=(x+3+√2)(x+3-√2)

4x²-4x-1
=(4x²-4x+1)-2
=(2x-1)²-√2²
=[(2x-1)+√2] [(2x-1)-√2]
=(2x-1+√2)(2x-1-√2)

Answer 2

χ²－3y²
=(x-y√3)(x+y√3)

χ²＋6χ＋7
=x²+6x+9-9+7
=(x+3)²-2
=(x+3-√2)(x+3+√2)

4χ²－4χ－1
=4x²-4x+1-1-1
=(2x-1)²-2
=(2x-1-√2)(2x-1+√2)