Ontbinden in factoren (1)

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Ontbind in factoren door gebruik te maken van merkwaardige producten

  1. \(256a^{10}-1\)
  2. \(p^2-28p+196\)
  3. \(169-64y^{6}\)
  4. \(196b^{6}+84b^3q+9q^2\)
  5. \(36a^{16}-25\)
  6. \(-36b^2+1\)
  7. \(196a^{8}+28a^4+1\)
  8. \(x^2-81\)
  9. \(121x^{4}+22x^2+1\)
  10. \(16s^{10}+72s^5+81\)
  11. \(b^2-20b+100\)
  12. \(169a^{6}-78a^3p+9p^2\)

Ontbind in factoren door gebruik te maken van merkwaardige producten

Verbetersleutel

  1. \(256a^{10}-1=(16a^5+1)(16a^5-1)\)
  2. \(p^2-28p+196=(p-14)^2\)
  3. \(169-64y^{6}=(13-8y^3)(13+8y^3)\)
  4. \(196b^{6}+84b^3q+9q^2=(14b^3+3q)^2\)
  5. \(36a^{16}-25=(6a^8+5)(6a^8-5)\)
  6. \(-36b^2+1=(1-6b)(1+6b)\)
  7. \(196a^{8}+28a^4+1=(14a^4+1)^2\)
  8. \(x^2-81=(x+9)(x-9)\)
  9. \(121x^{4}+22x^2+1=(11x^2+1)^2\)
  10. \(16s^{10}+72s^5+81=(4s^5+9)^2\)
  11. \(b^2-20b+100=(b-10)^2\)
  12. \(169a^{6}-78a^3p+9p^2=(13a^3-3p)^2\)
Oefeningengenerator wiskundeoefeningen.be 2026-08-23 12:16:51
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