■ Hello School 高校数学 数T(ハロT) 因数分解(1) 解答■
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(1) 3x2+7x+2
   =(3x+1)(x+2)
 
(2) 4x2−11xy+6y2
   =(4x−3y)(x−2y) 
(3) 27x3+8y3
   =(3x)3+(2y)3
   =(3x+2y)(9x2−6xy+4y2
 
(4) 216x3−1
   =(6x)3−(1)3
   =(6x−1)(36x2+6x+1)
(5) −64x3+144x2y−108xy2+27y3
   =−(64x3−144x2y+108xy2−27y3
   =−{(4x)3−3・(4x)2・3
              +3・(4x)・32−(3y)3

   =−(4x−3)3
 
(6) x3y+256−16xy−16x2
   =x(x2−16)y−16(x2−16)
   =(x2−16)(xy−16)
   =(x+4)(x−4)(xy−16)
(7) 2(2x−1)2−11(2x−1)+15
   =2A2−11A+15
   =(2A−5)(A−3)
   =(4x−2−5)(2x−1−3)
   =(4x−7)(2x−4)
 
(8) x2−y2+6y−9
   =x2−(y2−6y+9)
   =2−(y−3)2
   ={(x+(y−3)}{x−(y−3)}
   =(x+y−3)(x−y+3)
 
(9) x4−29x2+100
   =(x2−4)(x2−25)
   =(x+2)(x−2)(x+5)(x−5)
(10) x4−16y4
   =(x22−(4y22
   =(x2+4y2)(x2−4y2
   =(x2+4y2)(x+2y)(x−2y)
 
(11) x4−x2−12
   =(x2−4)(x2+3)
   =(x+2)(x−2)(x2+3)
 
(12) 2x6+3x3−2
   =2(x32+3x3−2
   =(2x3−1)(x3+2)
(13) x6−1
   =(x32−1
   =(x3+1)(x3−1)
   =(x+1)(x2−x+1)(x−1)(x2+x+1)
   =(x+1)(x−1)(x2+x+1)(x2−x+1)
 
(14) (x2−7x)+4(x2−7x)−96
   =(x2−7x+12)(x2−7x−8)
   =(x−3)(x−4)(x−8)(x+1)
 
(15) 3x2y−xy2−2xy+3x−y−2
   =xy(3x−y−2)+3x−y−2
   =(xy+1)(3x−y−2)
 
(16) 6x2+5xy+y2+x+y−2
   =6x2+(5y+1)x+y2+y−2
   =6x2+(5y+1)x+(y−1)(y+2)
   =(2x+y−1)(3x+y+2)
(17) 7x2+4xy−3y2+15x−5y+2
   =7x2+(4y+15)x−(3y2+5y−2)
   =7x2+(4y+15)x−(3y−1)(y+2)
   =(x+y+2)(7x−3y+1)
 
(18) 8x2−yz+xz−8xy
  =8x2+xz−8xy−yz
  =(8x+z)x−(8x+z)y
  =(8x+z)(x−y)
 
(19) 2x3+5x2y+4x2z−12xy2+10xyz−12y2
   =2x3+5x2y−12xy2+4x2z+10xyz−12y2
   =(2x3+5x2y−12xy2)x+2z(2x3+5x2y−12xy2
   =(2x−3y)(x+4y)x+2z(2x−3y)(x+4y)
   =(2x−3y)(x+4y)(x+2z)
 
(20) (2x+y)(y+z)(2x+z)+2xyz
   ={(y+z)(2x+z)}(2x+y)+2xyz
   ={z2+(2x+y)z+2xy}(2x+y)+2xyz
   =(2x+y)z2+(4x2+4xy+y2)z+4x2y+2xy2+2xyz
   =(2x+y)z2+(4x2+6xy+y2)z+2xy(2x+y)
   =(2x+y+z){(2x+y)z+2xy}
   =(2x+y+z)(2xz+yz+2xy)
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