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Algebra 1 Section 10.5.

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1 Algebra 1 Section 10.5

2 Factoring Completely Some factoring involves numbers that are not integers. We consider a polynomial completely factored when it cannot be factored further over the set of integers.

3 Factoring Completely In order for a polynomial to be completely factored, each polynomial factor must itself be completely factored.

4 Example 1 Factor 5ax5 – 5axy4. Factor out the common factor.
5ax(x4 – y4) Factor the difference of squares. 5ax(x2 + y2)(x2 – y2)

5 Example 1 5ax(x2 + y2)(x2 – y2) Factor the resulting difference of squares. 5ax(x2 + y2)(x + y)(x – y) Always be sure to check your answer by multiplying.

6 Example 2 Factor x4 – 9x2 + 8. Note that each binomial will start with x2. (x2 – 1)(x2 – 8) Factor the first binomial, since it is a difference of squares. (x + 1)(x – 1)(x2 – 8)

7 Example 3 Factor 2xy – 7x + 10y – 35.
Group the first two and the last two terms since there are no common monomials and no special patterns. (2xy – 7x) + (10y – 35)

8 Example 3 (2xy – 7x) + (10y – 35) Factor a common monomial from each binomial. x(2y – 7) + 5(2y – 7) (x + 5)(2y – 7)

9 Example 4 Factor am – an – pm + pn.
Group the terms with a common factor. (am – an) + (-pm + pn) a(m – n) – p(m – n) (a – p)(m – n)

10 Example 5 Factor 2y3 + 5y2 – 8y – 20. Group the first two and the last two terms since there are no common monomials and no special patterns. (2y3 + 5y2) + (-8y – 20)

11 Is this the final answer?
Example 5 (2y3 + 5y2) + (-8y – 20) Factor a common monomial factor from each binomial. y2(2y + 5) – 4(2y + 5) (y2 – 4)(2y + 5) Is this the final answer?

12 Example 5 (y2 – 4)(2y + 5) We need to factor the difference of squares. (y – 2)(y + 2)(2y + 5)

13 Homework: pp


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