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Factoring Polynomials of Higher Degree The Factor Theorem Part II.

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1 Factoring Polynomials of Higher Degree The Factor Theorem Part II

2 Factoring Polynomials of Higher Degree From last class, synthetic division is a quick process to divide polynomials by binomials of the form x – a and bx – a. 21 -4 -7 10 1 2 -2 -4 -11 -12 -22

3 Factoring Polynomials of Higher Degree Example 1: Use synthetic division to find the quotient and remainder when 6x 3 – 25x 2 – 29x + 25 is divided by 2x – 1. Solution: 6 -25 -29 25 6 3 -22 -11 -40 5 -20 bx – a is written in the form in front of the “L”.

4 Factoring Polynomials of Higher Degree The Factor Theorem (Part 2) If p(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 +…+ a x x 2 + a z x + a 0 is a polynomial function with integer coefficients, and if bx – a is a factor of p(x) where “b” and “a” are also integers, then “b” is a factor of the leading coefficient a n and “a” is a factor of the constant term a 0.  “b” and “a” should have no factors in common.

5 Factoring Polynomials of Higher Degree Example 2: Factor x 3 + 5x 2 + 2x – 8. Solution: If x 3 + 5x 2 + 2x – 8 has a factor of the form bx – a, then “a” is a factor of -8 and “b” is a factor of 1. A list of possible factors is x – 1, x – 2, x – 4, x – 8, x + 1, x + 2, x + 4, x + 8.  We will use synthetic division to try and find one factor. Start by testing x – 1.

6 Factoring Polynomials of Higher Degree 1 5 2 -8 1 1 6 6 8 0 8 1 We note that x – 1 must be a factor since the remainder is 0. Thus, we have: x 3 + 5x 2 + 2x – 8 = (x – 1)(x 2 + 6x + 8) = (x – 1)(x + 2)(x +4) Factor the trinomial

7 Factoring Polynomials of Higher Degree Note that only 3 of the possible 8 solutions for the polynomial were factors. It is unlikely that you will actually find a factor on your first try. It will take a few attempts to find the first one, and then factor the remaining quadratic by inspection.

8 Summary To factor a polynomial of higher degree: 1) Write the polynomial in decreasing degree inserting 0’s for any missing terms. 2) Make a list of all possible factors bx – a, where “a” is a factor of the constant term and “b” is a factor of the leading coefficent. “b” and “a” should have no factors in common. 3) Test the list of potential factors using synthetic division. 4) When you find a factor, factor the remaining quadratic. Factoring Polynomials of Higher Degree

9 Homework Do # 19, 25, 29, 31, 33, and 35 on page 130 from section 4.3 for Monday April 20 th Have a Safe and Fun Easter Break


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