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Polynomials 5: The Factor theorem

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Presentation on theme: "Polynomials 5: The Factor theorem"β€” Presentation transcript:

1 Polynomials 5: The Factor theorem
Silent Teacher Intelligent Practice Narration Your Turn Use the factor theorem to show that π‘₯βˆ’1 is a factor of 4 π‘₯ 3 βˆ’3 π‘₯ 2 βˆ’1. Use the factor theorem to show that π‘₯βˆ’4 is a factor of βˆ’3 π‘₯ π‘₯ 2 βˆ’6π‘₯+8. Use the factor theorem to show that π‘₯+1 is a factor of π‘₯ 3 +3 π‘₯ 2 βˆ’33π‘₯βˆ’35. Practice

2 Use the factor theorem to show that
Worked Example Your Turn Use the factor theorem to show that π‘₯βˆ’2 is a factor of π‘₯ 3 + π‘₯ 2 βˆ’4π‘₯βˆ’4. Use the factor theorem to show that π‘₯+3 is a factor of π‘₯ 3 + 2π‘₯ 2 βˆ’2π‘₯+3. Use the factor theorem to show that π‘₯βˆ’3 is NOT a factor of π‘₯ 3 + π‘₯ 2 βˆ’4π‘₯βˆ’4. Use the factor theorem to show that π‘₯+2 is NOT a factor of π‘₯ 3 + π‘₯ 2 +π‘₯βˆ’3.

3 Use the factor theorem to show that π‘₯βˆ’1 is a factor of 3 π‘₯ 2 βˆ’5π‘₯+2.
π‘₯βˆ’2 is NOT a factor of π‘₯ 3 βˆ’5 π‘₯ 2 βˆ’6π‘₯+18.

4 Use the factor theorem to show that π‘₯βˆ’1 is a factor of 3 π‘₯ 2 βˆ’5π‘₯+2.
If 𝑓 π‘₯ =3 π‘₯ 2 βˆ’5π‘₯+2, then f(1)=0. So (x-1) is a factor of 𝑓(π‘₯). Use the factor theorem to show that π‘₯βˆ’1 is a factor of 3 π‘₯ 3 +2 π‘₯ 2 βˆ’5. If 𝑓 π‘₯ =3 π‘₯ 3 +2 π‘₯ 2 βˆ’5, then f(1)=0. So (x-1) is a factor of 𝑓(π‘₯). If 𝑓 π‘₯ =4 π‘₯ 3 βˆ’3 π‘₯ 2 βˆ’1, then f(1)=0. So (x-1) is a factor of 𝑓(π‘₯). Use the factor theorem to show that π‘₯βˆ’1 is a factor of 4 π‘₯ 3 βˆ’3 π‘₯ 2 βˆ’1. Use the factor theorem to show that π‘₯βˆ’4 is a factor of βˆ’3 π‘₯ π‘₯ 2 βˆ’6π‘₯+8. If 𝑓 π‘₯ =βˆ’3 π‘₯ π‘₯ 2 βˆ’6π‘₯+8, then f(4)=0. So (x-4) is a factor of 𝑓(π‘₯). Use the factor theorem to show that π‘₯+1 is a factor of π‘₯ 3 +3 π‘₯ 2 βˆ’33π‘₯βˆ’35. If 𝑓 π‘₯ = π‘₯ 3 +3 π‘₯ 2 βˆ’33π‘₯βˆ’35, then f(-1)=0. So (x+1) is a factor of 𝑓(π‘₯). Use the factor theorem to show that π‘₯+3 is a factor of 5π‘₯ 4 βˆ’45 π‘₯ 2 βˆ’6π‘₯βˆ’18. If 𝑓 π‘₯ = 5π‘₯ 4 βˆ’45 π‘₯ 2 βˆ’6π‘₯βˆ’18, then f(-3)=0. So (x+3) is a factor of 𝑓(π‘₯). Use the factor theorem to show that π‘₯βˆ’2 is NOT a factor of π‘₯ 3 βˆ’5 π‘₯ 2 βˆ’6π‘₯+18. If 𝑓 π‘₯ = π‘₯ 3 βˆ’5 π‘₯ 2 βˆ’6π‘₯+18, then f(2)β‰ 0. So (x-2) is NOT a factor of 𝑓(π‘₯).


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