■ Hello School 高校数学 数T(ハロT) 因数分解(2) 解答■
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(1) 6x2−11x−10
   =(2x−5)(3x+2)
 
(2) (xy+4)2+(4x−y)2
   =x22+8xy+16+16x2−8xy+y2
   =x22+16+16x2+y2
   =x2(y2+16)+y2+16
   =(x2+1)(y2+16)
 
(3) (x2+y2)(p2+q2)−(px+qy)2
   =x22+x22+p22+q22−(p22+2pqxy+q22
   =x22−2pqxy+p22
   =(xq−py)2
 
(4) 2(2x−12+3(2x+3)−11
   =22+3(+4)−11
   =2A2+3A+12−11
   =2A+3A+1
   =(A+1)(2A+1)
   ={(2x−1)+1}{2(2x−1)+1}
   =2x(4x−1)
 
(5) x4−81y4
  =(x22−(9y22
  =(x2+9y2)(x2−9y2
  =(x2+9y2)(x+3y)(x−3y)
 
(6) x4−68x22+256y4
   =(x2−64y2)(x2−4y2
   =(x+8y)(x−8y)(x+2y)(x−2y)
 
(7) 8x6+215x33−27y6
   =(8x3−y3)(x3+27y3
   =(2x−y)(4x2+2xy+y2)(x−3y)(x2−3xy+9y2
   =(2x−y)(x−3y)(4x2+2xy+y2)(x2−3xy+9y2
 
(8) (x2−x)2−14x2+14x+24
   =A2−14A+24
   =(A−12)(A−2)
   =(x2−x−12)(x2−x−2)
   =(x+3)(x−4)(x−2)(x+1)
 
(9) xy+x+y+1
   =x(y+1)+(y+1)
   =(x+1)(y+1)
 
(10) xy−yz+pz−px
   =(x−z)y−(x−z)p
   =(x−z)(y−p)
 
(11) x2−4xy+4y2−x−2y
   =(x+2y)2−(x+2y)
   =(x+2y)(x+2y−1)
 
(12) x3+5x2−9x−45
   =x2(x+5)−9(x+5)
   =(x2−9)(x+5)
   =(x+3)(x−3)(x+5)
 
(13)  3−x2
64
(x−
3
 
(14) x4−2x3−7x2+8x+12
   =(x4−7x2+12)−2x(x2−4)
   =(x2−3)(x2−4)−2x(x2−4)
   =(x2−2x−3)(x2−4)
   =(x−3)(x+1)(x+2)(x−2)
 
 
(15) x2−(2y−3)−(3y2+y−2)
   =x2−(2y−3)−(3y−2)(y+1)
   =x2+{(3y−2)−(y+1)}x+(3y−2){−(y+1)}
   ={x+(3y−2)}{x−(y+1)}
   =(x+3y−2)(x−y−1)
 
(16) 3x2+7xy+x−3y+2y2−2
   =3x2+(7y+1)x+2y2−3y−2
   =3x2+(7y+1)x+(y−2)(2y+1)
   =(3x+y−2)(x+2y+1)
 
(17) 2x2−3y2+5xy+7x+7y+6
   =2x2+(5y+7)x−3y2+7y+6
   =2x2+(5y+7)x−(3y2−7y−6)
   =2x2+(5y+7)x−(3y+2)(y−3)
   ={2x−(y−3)}{x+(3y+2)}
   =(2x−y+3)(x+3y+2)
 
(18) x2y−5xyz−y−xy2+x−5z
   =−5z(xy+1)+xy(x−y)+(x−y)
   =−5z(xy+1)+((xy+1)(x−y)
   =(x−y−5z)(xy+1)
 
(19) x22−x2y−2x2−9y2+9y+18
   =x2(y2−y−2)−9(y2−y−2)
   =(x2−9)(y2−y−2)
   =(x+3)(x−3)(y+1)(y−2)
 
(20) x(y+z)2+y(z+x)2+z(x+y)2−4xyz
   =(y+z)2x+(x2+2xz+z2)y+(x2+2xy+y2)z−4xyz
   =(y+z)x2+{(y+z)2+2yz+2yz−4yz}x+y2z+yz2
   =(y+z)x2+(y+z)2x+y2z+yz2
   =(y+z)x2+(y+z)2x+yz(y+z)
   =(y+z){x2+(y+z)+yz}
   =(y+z)(x+y)(x+z)
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