因式分解

分解因式

(1). p2(a−1)+p(1−a)p^2(a-1)+p(1-a)p2(a1)+p(1a)

= p2(a−1)−(a−1)p^2(a-1)-(a-1)p2(a1)(a1)
= p(a−1)(p−1)p(a-1)(p-1)p(a1)(p1)

(2). (a2−a)2−14(a2−a)+24(a^2-a)^2-14(a^2-a)+24(a2a)214(a2a)+24

= (a2−a−2)(a2−a−12)(a^2-a-2)(a^2-a-12)(a2a2)(a2a12)
= (a−2)(a+1)(a−4)(a+3)(a-2)(a+1)(a-4)(a+3)(a2)(a+1)(a4)(a+3)

(3). ab+b2+a−b−2ab+b^2+a-b-2ab+b2+ab2

= b(a+b)+(a+b)−2(b+1)b(a+b)+(a+b)-2(b+1)b(a+b)+(a+b)2(b+1)
= (b+1)(a+b)−2(b+1)(b+1)(a+b)-2(b+1)(b+1)(a+b)2(b+1)
= (b+1)(a+b−2)(b+1)(a+b-2)(b+1)(a+b2)

(4). 2x4+13x3+20x2+11x+22x^4+13x^3+20x^2+11x+22x4+13x3+20x2+11x+2

= 2x4+13x3+6x2+14x2+11x+22x^4+13x^3+6x^2+14x^2+11x+22x4+13x3+6x2+14x2+11x+2
= x2(2x+1)(x+6)+(2x+1)(7x+2)x^2(2x+1)(x+6)+(2x+1)(7x+2)x2(2x+1)(x+6)+(2x+1)(7x+2)
= (2x+1)(x3+6x2+7x+2)(2x+1)(x^3+6x^2+7x+2)(2x+1)(x3+6x2+7x+2)
= (2x+1)[x(x2+6x+5)+2(x+1)](2x+1)[x(x^2+6x+5)+2(x+1)](2x+1)[x(x2+6x+5)+2(x+1)]
= (2x+1)(x+1)(x2+5x+2)(2x+1)(x+1)(x^2+5x+2)(2x+1)(x+1)(x2+5x+2)

(5). 6x2+xy−15y2+4x−25y−106x^2+xy-15y^2+4x-25y-106x2+xy15y2+4x25y10

= (2x−3y)(3x+5y)+(4x−25y)−10(2x-3y)(3x+5y)+(4x-25y)-10(2x3y)(3x+5y)+(4x25y)10
= (2x−3y−2)(3x+5y+5)(2x-3y-2)(3x+5y+5)(2x3y2)(3x+5y+5)

(6). 如果把多项式 x2−8x+mx^2-8x+mx28x+m 分解因式得 (x−10)(x+n)(x-10)(x+n)(x10)(x+n),求m、nm、nmn的值。

(x−10)(x+n)=x2+(10−n)x−10n(x-10)(x+n)=x^2+(10-n)x-10n(x10)(x+n)=x2+(10n)x10n
n−10=−8n-10=-8n10=8
−10n=m-10n=m10n=m
n=−2n=-2n=2
m=20m=20m=20

(7). 已知 3a+3b=−93a+3b=-93a+3b=9,求 2a2+4ab+2b2−62a^2+4ab+2b^2-62a2+4ab+2b26 的值。

a+b=−3a+b=-3a+b=3
2a2+4ab+2b2−62a^2+4ab+2b^2-62a2+4ab+2b26
= 2(a+b)2−62(a+b)^2-62(a+b)26
= 121212

(8). 设 a(a−1)−(a2−b)=2a(a-1)-(a^2-b)=2a(a1)(a2b)=2 ,求 a2+b22−ab\frac {a^2+b^2}2-ab2a2+b2ab 的值。

a(a−1)−(a2−b)=2a(a-1)-(a^2-b)=2a(a1)(a2b)=2
= a2−a−a2+ba^2-a-a^2+ba2aa2+b
= −a+b=2-a+b=2a+b=2
a2+b22−ab=a2+b2−2ab2=(a−b)22=2\frac {a^2+b^2}2-ab=\frac {a^2+b^2-2ab}2=\frac{(a-b)^2}{2}=22a2+b2ab=2a2+b22ab=2(ab)2=2

1.把下列各式分解因式:
(1) x2+2y−y2−2xx^2+2y-y^2-2xx2+2yy22x

= (x2−y2)−(2x−2y)(x^2-y^2)-(2x-2y)(x2y2)(2x2y)
= (x+y)(x−y)−2(x−y)(x+y)(x-y)-2(x-y)(x+y)(xy)2(xy)
= (x+y−2)(x−y)(x+y-2)(x-y)(x+y2)(xy)

(2) 2a2+bc2−a2b−2ac2a^2+bc^2-a^2b-2ac2a2+bc2a2b2ac

= (2a2−2ac)+(bc2−a2b)(2a^2-2ac)+(bc^2-a^2b)(2a22ac)+(bc2a2b)
= 2a(a−c)−b(a2−c2)2a(a-c)-b(a^2-c^2)2a(ac)b(a2c2)
= 2a(a−c)−b(a−c)(a+c)2a(a-c)-b(a-c)(a+c)2a(ac)b(ac)(a+c)
= (a−c)(2a−ab−bc)(a-c)(2a-ab-bc)(ac)(2aabbc)

(3) a2+b2+2ac+2bc+2aba^2+b^2+2ac+2bc+2aba2+b2+2ac+2bc+2ab

= (a2+2ab+b2)+(2ac+2bc)(a^2+2ab+b^2)+(2ac+2bc)(a2+2ab+b2)+(2ac+2bc)
= (a+b)2+2c(a+b)(a+b)^2+2c(a+b)(a+b)2+2c(a+b)
= (a+b)(a+b+2c)(a+b)(a+b+2c)(a+b)(a+b+2c)

(4) 4x2+y2−z2−4xy4x^2+y^2-z^2-4xy4x2+y2z24xy

= (2x−y)2−z2(2x-y)^2-z^2(2xy)2z2
= (2x−y+z)(2x−y−z)(2x-y+z)(2x-y-z)(2xy+z)(2xyz)

2.把下列关于 xxx 的二次多项式因式分解
(1) x2+x−1x^2+x-1x2+x1

= (x+1+52)(x−1+52)(x+\frac {1+\sqrt 5}2)(x-\frac {1+\sqrt 5}2)(x+21+5)(x21+5)

(2) x2+4xy−4y2x^2+4xy-4y^2x2+4xy4y2

= [x+2(1−2)y][x+2(1+2)y][x+2(1-\sqrt 2)y][x+2(1+\sqrt 2)y][x+2(12)y][x+2(1+2)y]

1.分解因式: x2−y2−2y−1x^2-y^2-2y-1x2y22y1

= x2−(y2+2y+1)x^2-(y^2+2y+1)x2(y2+2y+1)
= x2−(y+1)2x^2-(y+1)^2x2(y+1)2
= (x+y+1)(x−y−1)(x+y+1)(x-y-1)(x+y+1)(xy1)

2.把 2(a2−3mn)+a(4m−3n)2(a^2-3mn)+a(4m-3n)2(a23mn)+a(4m3n) 因式分解

= 2a2−6mn+4am−3an2a^2-6mn+4am-3an2a26mn+4am3an
= 2a(a+2m)−3n(a+2m)2a(a+2m)-3n(a+2m)2a(a+2m)3n(a+2m)
= (a+2m)(2a−3n)(a+2m)(2a-3n)(a+2m)(2a3n)

分组分解法

定义

先分组,再分组分解,然后提公因式

应用

能分组分解的式子有四项或六项或大于六项,一般的分组分解有两种形式:二二分法,三一分法。

比如:

ax+ay+bx+byax+ay+bx+byax+ay+bx+by
=a(x+y)+b(x+y)a(x+y)+b(x+y)a(x+y)+b(x+y)
=(a+b)(x+y)(a+b)(x+y)(a+b)(x+y)

1.用分组分解法把 4x2−2x−y2−y4x^2-2x-y^2-y4x22xy2y 因式分解。

= (4x2−y2)−(2x+y)(4x^2-y^2)-(2x+y)(4x2y2)(2x+y)
= (2x−y)(2x+y)−(2x+y)(2x-y)(2x+y)-(2x+y)(2xy)(2x+y)(2x+y)
= (2x+y)(2x−y−1)(2x+y)(2x-y-1)(2x+y)(2xy1)

(1)在实数范围内分解因式: x5−64x^5-64x564 .

= x(x2+8)(x+22)(x−22)x(x^2+8)(x+2\sqrt 2)(x-2\sqrt 2)x(x2+8)(x+22)(x22)

(2)已知 a,b,c,da,b,c,da,b,c,d为非负整数,且 ac+bd+ad+bc=1997ac+bd+ad+bc=1997ac+bd+ad+bc=1997 ,求 a+b+c+da+b+c+da+b+c+d 的值。

ac+bd+ad+bc=(a+b)(c+d)=1997ac+bd+ad+bc=(a+b)(c+d)=1997ac+bd+ad+bc=(a+b)(c+d)=1997
又因为1997为质数
不妨 a+b=1a+b=1a+b=1&&c+d=1997c+d=1997c+d=1997
a+b+c+d=1998a+b+c+d=1998a+b+c+d=1998

(3)已知 a,b,ca,b,ca,b,c 是一个三角形的三边,求证: a4+b4+c4−2a2b2−2b2c2−2c2a2&lt;0a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2&lt;0a4+b4+c42a2b22b2c22c2a2<0

= a4+b4+c4+2a2b2−2b2c2−2c2a2−4a2b2a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2-4a^2b^2a4+b4+c4+2a2b22b2c22c2a24a2b2
= (a2+b2−c2)2−4a2b2(a^2+b^2-c^2)^2-4a^2b^2(a2+b2c2)24a2b2
= (a+b+c)(a+b−c)(a−b+c)(a−b−c)(a+b+c)(a+b-c)(a-b+c)(a-b-c)(a+b+c)(a+bc)(ab+c)(abc)
由三角形两边之长大于第三边即判断正负

(4) 已知 x+y=3,x2+y2−xy=4x+y=3, x^2+y^2-xy=4x+y=3,x2+y2xy=4 ,求 x4+y4+x3y+xy3x^4+y^4+x^3y+xy^3x4+y4+x3y+xy3 的值。

x4+y4+x3y+xy3x^4+y^4+x^3y+xy^3x4+y4+x3y+xy3
= x3(x+y)+y3(x+y)x^3(x+y)+y^3(x+y)x3(x+y)+y3(x+y)
= (x3+y3)(x+y)(x^3+y^3)(x+y)(x3+y3)(x+y)
= (x+y)2(x2−xy+y2)(x+y)^2(x^2-xy+y^2)(x+y)2(x2xy+y2)
= 32⋅43^2\cdot4324
= 12

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