Ontbinden in factoren (1)

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Ontbind in factoren door gebruik te maken van merkwaardige producten

  1. \(225b^{6}-420b^3q+196q^2\)
  2. \(64x^2-25q^{12}\)
  3. \(144p^{4}+168p^2+49\)
  4. \(64b^2+80b+25\)
  5. \(b^2-16\)
  6. \(121-256x^{8}\)
  7. \(16x^{8}-81\)
  8. \(225x^{10}-16\)
  9. \(64y^{10}+240y^5+225\)
  10. \(144q^{8}-49\)
  11. \(256a^{6}-225p^2\)
  12. \(169q^2-16a^{12}\)

Ontbind in factoren door gebruik te maken van merkwaardige producten

Verbetersleutel

  1. \(225b^{6}-420b^3q+196q^2=(15b^3-14q)^2\)
  2. \(64x^2-25q^{12}=(8x-5q^6)(8x+5q^6)\)
  3. \(144p^{4}+168p^2+49=(12p^2+7)^2\)
  4. \(64b^2+80b+25=(8b+5)^2\)
  5. \(b^2-16=(b+4)(b-4)\)
  6. \(121-256x^{8}=(11-16x^4)(11+16x^4)\)
  7. \(16x^{8}-81=(4x^4+9)(4x^4-9)\)
  8. \(225x^{10}-16=(15x^5+4)(15x^5-4)\)
  9. \(64y^{10}+240y^5+225=(8y^5+15)^2\)
  10. \(144q^{8}-49=(12q^4+7)(12q^4-7)\)
  11. \(256a^{6}-225p^2=(16a^3+15p)(16a^3-15p)\)
  12. \(169q^2-16a^{12}=(13q-4a^6)(13q+4a^6)\)
Oefeningengenerator wiskundeoefeningen.be 2026-04-25 15:43:54
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