Worksheet 6



Worksheet 5.4 & 5.5

Algebra 2

|Factoring: Look for a greatest common factor and then look |Synthetic Division: Divide using coefficients. |

|at the number of terms of the problem for a clue on how to |Ex.: Simplify (x3 – 7x – 6) ÷ (x – 2) |

|start the factoring. See the notes on factoring (before Ch.|[pic] |

|5) for more examples. |The numbers at the bottom stand for the coefficients of the answer, which is [pic]. |

|Remainder Theorem: The remainder from synthetic division is|Factor Theorem: If a remainder is zero, the number outside the box in synthetic division |

|the value of the function at that point. |is a zero of the function; therefore, the related factor is a factor of the polynomial. |

|Ex: Use synthetic division to find the value of P(5) if |Ex: Factor f(x) = 2x3 + 11x2 + 18x + 9, given that f(-3) = 0. |

|P(x) = 3x4 – 5x2 + x – 9. |The value for x is -3 because of the f(x) notation. Put -3 outside the box, and the |

|The value for x is 5 because of the function notation. Put |coefficients of the polynomial inside the box. |

|5 outside the box and put the coefficients of the polynomial|[pic] |

|inside the box. Remember to use all descending powers, so |The remainder is 0, so x = -3 is a zero of the function. The remainder of the problem is |

|there’s a 0x3. |2x2 + 5x + 3, which factors as (2x + 3)(x + 1). x = -3 becomes the factor (x + 3) when |

|[pic] |you move the -3 over. |

|The remainder is 1746, so P(5)=1746. |The factors of 2x3 + 11x2 + 18x + 9 are (2x + 3),(x + 1), and (x + 3). |

Factor.

|1. 216x3 + 1 |2. 3x2 + 11x + 6 |3. x3 – 4x2 + 4x – 16 |

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|4. x4 + 3x2 + 2 |5. 2x7 – 32x3 |6. 4x4 – 5x2 – 9 |

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Solve by factoring. (Find real and/or imaginary answers.)

|7. x3 + 2x2 – x = 2 |8. x3 + 8 = 0 |9. 3x4 + 15x2 – 72 = 0 |

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Divide using synthetic division.

|10. ( x3 – 14x + 8) ÷ (x + 4) |11. (x4 – 6x3 – 40x + 33) ÷ (x – 7) |12. (3x2 – 10x) ÷ (x – 6) |

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Factor the polynomial.

|13. f(x) = x3 – 3x2 – 16x – 12, and f(6) = 0. |14. f(x) = x3 – 11x2 + 14x + 80, and f(8) = 0. |

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|15. f(x) = 2x3 + 7x2 – 33x – 18, and f(-6) = 0. |16. f(x) = x3 – x2 – 21x + 45, and f(-5) = 0. |

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Find the zeros of the function given that one of the zeros is ___.

|17. f(x) = 2x3 + 3x2 – 39x – 20, and 4 is a zero. |18. f(x) = x3 + 11x2 – 150x – 1512, & -14 is a zero. |

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|19. f(x) = x3 + x2 +2x + 24, and -3 is a zero. |20. f(x) = 4x3 + 9x2 – 52x + 15, and -5 is a zero. |

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