Ontbinden in factoren (1)

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Ontbind in factoren door gebruik te maken van merkwaardige producten

  1. \(4b^2+12b+9\)
  2. \(25a^{10}+140a^5q+196q^2\)
  3. \(a^2-9\)
  4. \(100p^{8}+220p^4+121\)
  5. \(100a^2-260a+169\)
  6. \(144x^{12}-169y^2\)
  7. \(81b^2-144b+64\)
  8. \(1-196b^{8}\)
  9. \(25q^2-90q+81\)
  10. \(9-4p^{8}\)
  11. \(81x^2-25b^{16}\)
  12. \(25a^{6}-120a^3s+144s^2\)

Ontbind in factoren door gebruik te maken van merkwaardige producten

Verbetersleutel

  1. \(4b^2+12b+9=(2b+3)^2\)
  2. \(25a^{10}+140a^5q+196q^2=(5a^5+14q)^2\)
  3. \(a^2-9=(a-3)(a+3)\)
  4. \(100p^{8}+220p^4+121=(10p^4+11)^2\)
  5. \(100a^2-260a+169=(10a-13)^2\)
  6. \(144x^{12}-169y^2=(12x^6+13y)(12x^6-13y)\)
  7. \(81b^2-144b+64=(9b-8)^2\)
  8. \(1-196b^{8}=(1-14b^4)(1+14b^4)\)
  9. \(25q^2-90q+81=(5q-9)^2\)
  10. \(9-4p^{8}=(3-2p^4)(3+2p^4)\)
  11. \(81x^2-25b^{16}=(9x-5b^8)(9x+5b^8)\)
  12. \(25a^{6}-120a^3s+144s^2=(5a^3-12s)^2\)
Oefeningengenerator wiskundeoefeningen.be 2026-07-04 14:34:33
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