14.15 Notes 8.7 factoring special cases complete.notebook
December 09, 2014
Recall our conclusions from Section 8.3 when we foiled:
a.
(x + 3)(x + 3)
b.
(x ­ 4)(x + 4)
14.15 Notes 8.7 factoring special cases complete.notebook
December 09, 2014
Any polynomial of the form a2 + 2ab + b2 is called a perfect square trinomial. We can factor a perfect square trinomial into identical binomial factors:
a2 + 2ab + b2 = (a + b)(a + b)
a2 ­ 2ab + b2 = (a ­ b)(a ­ b)
So, x2 + 18x + 81 =
and 4n2 + 12n + 9 = 14.15 Notes 8.7 factoring special cases complete.notebook
December 09, 2014
Any polynomial of the form a2 ­ b2 is called a difference of squares. We can factor a difference of squares into binomial factors with opposite signs:
a2 ­ b2 = (a + b)(a ­ b)
So, x2 ­ 81 =
and 25a2 ­ 49 =
14.15 Notes 8.7 factoring special cases complete.notebook
December 09, 2014
Ex 1: Are the following polynomials perfect­square trinomials? If so, factor them. a.
n2 ­ 22n + 121
b.
x2 + 9x + 81
c.
9a2 + 24ab + 16b2 14.15 Notes 8.7 factoring special cases complete.notebook
December 09, 2014
Ex 2: Are the following polynomials differences of squares? If so, factor them.
a.
x2 ­ 100
b.
t2 + 81
c.
9u2 ­ 64v2 14.15 Notes 8.7 factoring special cases complete.notebook
Ex 3: Factor completely. Hint: Look for a GCF first!
a.
2x2 + 14x + 20
b.
y3 + 6y2 + 9y
c.
5a2 ­ 20b2 December 09, 2014
14.15 Notes 8.7 factoring special cases complete.notebook
Homework: 538/ 2­40 x2 and study for quiz.
December 09, 2014