Question

在实数范围内因式分解

（X²-2x）²-7（x²-2x）+12 详细步骤求解 谢谢
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Answer 1

将x²-2x看成一个整体将12拆分成3*4，用十字交叉法得出，原式=(x-3)(x+1)(X²-2x-4)
可以将2x³+3x²-1=0看出x=-1为一解，所以有一式为x+1，用2x³+3x²-1除以x+1得2x²+x-1再分为(x+1)(2x-1)所以原式=(x+1)²(2x-1)

Answer 2

解：
（1）4x^4-13x²+9
=(x²-1)(4x²-9)
=(x-1)(x+1)(2x-3)(2x+3)

(2) 4(x-y+1)+y(y-2x)
=4x-4y+4+y²-2xy+x²-x²
=(y-x)²-4(y-x)+4-x²
=(y-x-2)²-x²
=(y-x-2+x)(y-x-2-x)
=(y-2)(y-2-2x)

(3)3x²+5xy-2y²+x+9y-4
=(x+2y)(3x-y)+x+9y-4
=(x+2y)(3x-y)+4(x+2y)-(3x-y)-4
=(x+2y)(3x-y+4)-(3x-y+4)
=(x+2y-1)(3x-y+4)

(4)x^5+x+1
=x^5-x²+x²+x+1
=x²(x³-1)+(x²+x+1)
=x²(x-1)(x²+x+1)+(x²+x+1)
=(x²+x+1)(x³-x²+1)

(5) (x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x²+8x+7)(x²+8x+15)+15
=(x²+8x)²+22(x²+8x)+105+15
=(x²+8x)²+22(x²+8x)+120
=(x²+8x+10)(x²+8x+12)
=(x+4+√6)(x+4-√6)(x+2)(x+6)
7月i9

Answer 3

（X²-2x）²-7（x²-2x）+12
=[（X²-2x）-3]*[（X²-2x）-4]
=(x-3)(x+1)(X²-2x-4)

Answer 4

2x^3 + 3x^2 - 1
= (x+1)(2x^2 +x -1 )
= (x+1)(x+1)(2x-1)
=(x+1)^2 (2x-1)

(x²-2x）²-7（x²-2x）+12
= (x^2 - 2x - 3)(x^2 - 2x - 4)
= (x+1)(x-3)(x - 2 - 5^(1/2))(x - 2 + 5^(1/2))