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分解下列因式:
1. $(2x^2+5x)^2 - 2x^2 - 5x - 6$
解答: $$(2x^2+5x)^2 - 2x^2 - 5x - 6 = (2x^2+5x)^2 - (2x^2 + 5x) - 6$$ $$= (2x^2 + 5x - 3)(2x^2 + 5x + 2)$$ $$= (2x - 1)(x + 3)(2x + 1)(x + 2).$$
2. $x^4 + 4x^3 + 4x^2 - 11(x^2 + 2x) + 24$
解答: $$x^4 + 4x^3 + 4x^2 - 11(x^2 + 2x) + 24 = (x^2 + 2x)^2 - 11(x^2 + 2x) + 24$$ $$= (x^2 + 2x - 3)(x^2 + 2x - 8)$$ $$= (x+3)(x - 1)(x - 2)(x + 4).$$
3. $(x+1)(2x + 1)(3x-1)(4x-1) + 6x^4$
解答: $$(x+1)(2x + 1)(3x-1)(4x-1) + 6x^4 = [(x + 1)(3x - 1)][(2x + 1)(4x - 1)] + 6x^4$$ $$= (3x^2 + 2x - 1)(8x^2 + 2x - 1) + 6x^4$$ $$= 24x^4 + 11x^2(2x - 1) + (2x-1)^2 + 6x^4$$ $$= 30x^4 + 11x^2(2x - 1) + (2x - 1)^2$$ $$= (5x^2 + 2x - 1)(6x^2 + 2x - 1).$$
4. $a(b+c-a)^2 + b(c+a-b)^2 + c(a+b-c)^2 + (b+c-a)(c+a-b)(a+b-c)$
解答:
设 $b + c - a = x$, $c + a - b = y$, $a + b - c = z$,
易知 $a + b + c = x + y + z$, $2a = y + z$, $2b = z + x$, $2c = x + y$.
由此原式可变形 $${1\over2}(y + z)x^2 + {1\over2}(z + x)y^2 + {1\over2}(x + y)z^2 + xyz$$ $$= {1\over2}\left[(y+z)x^2 + (z+x)y^2 + (x + y)z^2 + 2xyz\right]$$ $$= {1\over2}\left(x^2y + x^2z + y^2z + xy^2 + xz^2 + yz^2 + 2xyz\right)$$ $$= {1\over2}\left[xy(x + y + z) + yz(x + y + z) + xz(x + z)\right]$$ $$= {1\over2}\left[y(x + y + z)(x + z) + xz(x + z)\right]$$ $$= {1\over2}(x + z)\left(xy + y^2 + yz + xz\right)$$ $$= {1\over2}(x + z)(x + y)(y + z) = 4abc.$$
5. $2x^4 - x^3 - 6x^2 - x + 2$
解答: $$2x^4 - x^3 - 6x^2 - x + 2 = 2(x^4 + 1) - (x^3 + x) - 6x^2$$ $$= 2(x^2 + 1)^2 - 4x^2 - x(x^2 + 1) - 6x^2$$ $$= 2(x^2 + 1)^2 - x(x^2 + 1) - 10x^2$$ $$= \left[2(x^2+1) - 5x\right]\left[(x^2 + 1) + 2x\right]$$ $$= (2x^2 - 5x + 2)(x^2 + 2x + 1) = (2x - 1)(x - 2)(x + 1)^2.$$ 注: 本题还可采取拆项 $x^3(2x - 1) - 3x(2x - 1)- 2(2x - 1)$ 或由系数对称直接提取 $x^2$ 后进行换元求解(如以下第6题之解法).
6. $x^4 + x^3 + \displaystyle{9\over4}x^2 + x + 1$
解答: $$x^4 + x^3 + \displaystyle{9\over4}x^2 + x + 1 = x^2\left(x^2 + x + {9\over4} + {1\over x} + {1\over x^2}\right)$$ $$= x^2\left[(x + {1\over x})^2 - 2 + (x + {1\over x}) + {9\over4}\right]$$ $$= x^2\left[(x + {1\over x})^2 + (x + {1\over x}) + {1\over 4}\right]$$ $$= x^2\left(x + {1\over x} + {1\over2}\right)^2$$ $$= (x^2 + {1\over2}x + 1)^2 = {1\over4}(2x^2 + x + 2)^2.$$
7. $(x+y+z)^3 + (3x - 2y - 3z)^3 - (4x - y - 2z)^3$
解答:
注意到三项立方(代数)和, 且 $(x + y + z) + (3x - 2y - 3z) + (-4x + y + 2z) = 0$, 因此 $$(x+y+z)^3 + (3x - 2y - 3z)^3 - (4x - y - 2z)^3 = 3(x + y + z)(3x - 2y - 3z)(-4x + y + 2z)$$ $$= -3(x + y + z)(3x - 2y - 3z)(4x - y - 2z).$$
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原文地址：http://www.cnblogs.com/zhaoyin/p/5926721.html
