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Summary
for Chapter 6
Example
Topic
Reference
Common-Term Factoring
4x2 is the greatest common
monomial factor of
8x4 12x3 16x2.
8x4 12x3 16x2
2
Common Monomial Factor A single term that is a factor of every term of the polynomial. The greatest common factor
(GCF) is the common monomial factor that has the largest possible numerical coefficient and the largest possible exponents.
Factoring a Monomial from a Polynomial
6.1
p. 384
p. 385
2
4x (2x 3x 4)
1.
Determine the greatest common factor.
2.
Apply the distributive law in the form
ab ac a(b c)
The greatest common factor
The Difference of Squares
To factor: 16x2 25y2:
(4x)2 (5y)2
Think:
so
Factoring a Difference of Squares
ing form:
Use the follow-
6.2
p. 393
a2 b2 (a b)(a b)
2
16x 25y
2
(4x 5y)(4x 5y)
Factoring Trinomials
4x2 6x 10x 15
2x(2x 3) 5(2x 3)
(2x 3)(2x 5)
x2 3x 28
ac 28; b 3
mn 28; m n 3
m 7, n 4
2
x 7x 4x 28
x(x 7) 4(x 7)
(x 7)(x 4)
474
Factoring by Grouping When there are four terms of a
polynomial, factor the first pair and factor the last pair. If these
two pairs have a common binomial factor, factor that out. The
result will be the product of two binomials.
Factoring Trinomials To factor a trinomial, first use the
ac test to determine factorability. If the trinomial is factorable,
the ac test will yield two terms (which have as their sum the
middle term of the trinomial) that allow the factoring to be
completed by using the grouping method.
6.3
p. 400
p. 402
■
Summary for Chapter 6
Dividing Polynomials
9x4 6x3 15x2
3x
3x3 2x2 5x
x5
x 3
x2
2x
7
x2 3x
————
5x 17
5x 15
————
8
To Divide a Polynomial by a Monomial Divide
each term of the polynomial by the monomial. Then combine
the results.
6.4
p. 422
To Divide a Polynomial by a Polynomial Use the
long division method.
p. 424
Solving Quadratic Equations by Factoring
6.5
p. 432
8
The result is x 5 x3
To solve:
1.
Add or subtract the necessary terms on both sides of the
equation so that the equation is in standard form (set equal
to 0).
(x 10)(x 3) 0
2.
Factor the quadratic expression.
x 10 0 or x 3 0
x 10 and x 3 are
solutions.
3.
Set each factor equal to 0.
4.
Solve the resulting equations to find the solutions.
5.
Check each solution by substituting in the original equation.
2
x 7x 30
2
x 7x 30 0
475
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