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Polynomial function 10
- 1. 11.9.20
- 2. EVALUATION – the process of replacing the variable with the numerical value. •DIRECT SUBSTITUTION •SYNTHETIC SUBSTITUTION
- 3. 1. Evaluate 𝑷 𝒙 = 𝟒𝒙 𝟐 + 𝟑𝒙 − 𝟐, 𝒙 = −𝟑 𝑃 −3 = 4(−3)2 + 3(−3) − 2 𝑃 −3 = 4 9 − 9 − 2 𝑃 −3 = 36 − 9 − 2 𝑷 −𝟑 = 𝟐𝟓
- 4. 2. Evaluate 𝑷 𝒙 = 𝟐𝒙 𝟒 + 𝟐𝒙 𝟑 + 𝟐𝒙 𝟐 + 𝟐𝒙 + 𝟐, 𝒙 = 𝟐 𝑃 2 = 2(2)4 + 2(2)3 + 2(2)2 + 2(2) + 2 𝑃 2 = 2 16 + 2 8 + 2 4 + 4 + 2 𝑃 2 = 32 + 16 + 8 + 4 + 2 𝑷 𝟐 = 𝟔𝟐
- 5. 1. Evaluate 𝑷 𝒙 = 𝟒𝒙 𝟐 + 𝟑𝒙 − 𝟐, 𝒙 = −𝟑 𝑷 −𝟑 = 𝟐𝟓 −𝟑 𝟒 𝟑 −𝟐 + 𝟒 −𝟏𝟐 −𝟗 𝟐𝟕 𝟐𝟓
- 6. 2. Evaluate 𝑷 𝒙 = 𝟐𝒙 𝟒 + 𝟐𝒙 𝟑 + 𝟐𝒙 𝟐 + 𝟐𝒙 + 𝟐, 𝒙 = 𝟐 𝑷 −𝟑 = 𝟔𝟐 𝟐 𝟐 𝟐 𝟐 + 𝟐 𝟒 𝟔 𝟏𝟐 𝟏𝟒 𝟐 𝟐 𝟑𝟎 𝟔𝟐 𝟐𝟖 𝟔𝟎
- 7. This represents the missing 𝒙 𝟑 𝐭𝐞𝐫𝐦. 𝑷 𝒙 = 𝟑𝒙 𝟒 − 𝟓𝒙 𝟐 − 𝟐𝒙 + 𝟐 𝟑 −𝟓 −𝟐𝟎 𝟐
- 8. Answer the following problems below. Use direct and synthetic substitution to find the value of the given polynomial function. 1. Evaluate 𝑷 𝒙 = 𝟑𝒙 𝟓 − 𝟐𝒙 𝟑 + 𝟒𝒙 − 𝟓, 𝒙 = 𝟐 2. Evaluate 𝑷 𝒙 = 𝒙 𝟒 − 𝟑𝒙 𝟑 + 𝟗𝒙 − 𝟔, 𝒙 = −𝟏
- 9. 11.11.20
- 10. DIVIDE 𝒙 𝟑 + 𝟔𝒙 𝟐 − 𝟕𝒙 − 𝟖, 𝒃𝒚 𝒙 − 𝟐
- 11. DIVIDE 𝟐𝒙 𝟒 + 𝟓𝒙 𝟑 − 𝟐𝒙 − 𝟖, 𝒃𝒚 𝒙 + 𝟐 𝟐𝒙 𝟑 + 𝒙 𝟐 − 𝟐𝒙 + 𝟐 + 𝟏𝟎 𝒙 + 𝟐 −𝟐 𝟐 𝟓 𝟎 + 𝟐 −𝟒 𝟏 −𝟐 −𝟐 −𝟐 −𝟖 𝟐 𝟏𝟎 𝟒 −𝟐
- 12. DIVIDE 𝒙 𝟑 − 𝟐𝒙 𝟐 + 𝟓𝒙 − 𝟒, 𝒃𝒚 𝒙 + 𝟏 𝒙 𝟐 − 𝟑𝒙 + 𝟖 + 𝟏𝟐 𝒙 + 𝟏 −𝟏 𝟏 −𝟐 𝟓 + 𝟏 −𝟏 −𝟑 𝟑 𝟖 −𝟒 𝟏𝟐 −𝟖
- 13. DIVIDE 𝟒𝒙 𝟑 + 𝟏𝟓𝒙 𝟐 + 𝟐𝒙 − 𝟐𝟏, 𝒃𝒚 𝒙 + 𝟑 𝟒𝒙 𝟐 + 𝟑𝒙 − 𝟕 −𝟑 𝟒 𝟏𝟓 𝟐 + 𝟒 −𝟏𝟐 𝟑 −𝟗 −𝟕 −𝟐𝟏 𝟎 𝟐𝟏