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