Kopertina e librit Matematika 10 - 11: Pjesa I

Zgjidhja e ushtrimit 10

Zgjidhja e ushtrimit 10 të mësimit 6.4A në librin Matematika 10 - 11: Pjesa I nga shtëpia botuese Botime Pegi me autorë Steve Fearnley, June Haighton, Steve Lomax, Peter Mullarkey, James Nicholson dhe Matt Nixon.

Pyetja

Faktorizoni plotësisht shprehjet kuadratike të mëposhtme.

  1. 2x2+7x+32x^2+7x+3
  2. 2y2+5y+32y^2+5y+3
  3. 3b2+9b+63b^2+9b+6
  4. 3b2+11b+63b^2+11b+6
  5. 2x2+x−62x^2+x-6
  6. 2x2+4x−62x^2+4x-6
  7. 19x+4x2+1219x+4x^2+12
  8. 4x2−6x−44x^2-6x-4
  9. 15x2−31x+1415x^2-31x+14
  10. 18x2+37x−2018x^2+37x-20

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Zgjidhja

  1. 2x2+x+6x+3=x(2x+1)+3(2x+1)=(x+3)(2x+1)2x^2 + x + 6x + 3 = x(2x+1)+3(2x+1)=(x+3)(2x+1)
  2. 2y2+2y+3y+3=2y(y+1)+3(y+1)=(2y+3)(y+1)2y^2 + 2y + 3y + 3 = 2y(y+1)+3(y+1) = (2y+3)(y+1)
  3. 3b2+3b+6b+6=3b(b+1)+6(b+1)=(3b+6)(b+1)3b^2+3b+6b+6 = 3b(b+1)+6(b+1) = (3b+6)(b+1) == 3(b+2)(b+1)3(b+2)(b+1)
  4. 3b2+9b+2b+6=3b(b+3)+2(b+3)=(3b+2)(b+3)3b^2+9b+2b+6=3b(b+3)+2(b+3)=(3b+2)(b+3)
  5. 2x2+4x−3x−6=2x(x+2)−3(x+2)=(2x−3)(x+2)2x^2+4x-3x-6=2x(x+2)-3(x+2)=(2x-3)(x+2)
  6. 2x2+6x−2x−6=2x(x+3)−2(x+3)=(2x−2)(x+3)=2(x−1)(x+3)2x^2+6x-2x-6 = 2x(x+3)-2(x+3) = (2x-2)(x+3)=2(x-1)(x+3)
  7. 4x2+19x+12=4x2+16x+3x+12=4x(x+4)+3(x+4)=(4x+3)(x+4)4x^2+19x+12 = 4x^2+16x+3x+12 = 4x(x+4)+3(x+4) = (4x+3)(x+4)
  8. 4x2+2x−8x−4=2x(2x+1)−4(2x+1)=(2x−4)(2x+1)=2(x−2)(2x+1)4x^2+2x-8x-4 = 2x(2x+1)-4(2x+1)=(2x-4)(2x+1)=2(x-2)(2x+1)
  9. 15x2−10x−21x+14=5x(3x−2)−7(3x−2)=(5x−7)(3x−2)15x^2 -10x-21x+14 = 5x(3x-2)-7(3x-2)=(5x-7)(3x-2)
  10. 18x2+45x−8x−20=9x(2x+5)−4(2x+5)=(9x−4)(2x+5)18x^2+45x-8x-20=9x(2x+5)-4(2x+5)=(9x-4)(2x+5)