Slide 1 :
CHAPTER R: Basic Concepts of Algebra : Copyright © 2008 Pearson Education, Inc. CHAPTER R: Basic Concepts of Algebra R.1 The Real-Number System
R.2 Integer Exponents, Scientific Notation, and Order of
Operations
R.3 Addition, Subtraction, and Multiplication of Polynomials
R.4 Factoring
R.5 Rational Expressions
R.6 Radical Notation and Rational Exponents
R.7 The Basics of Equation Solving
R.4 Factoring : Copyright © 2008 Pearson Education, Inc. R.4 Factoring Factor polynomials by removing a common factor.
Factor polynomials by grouping.
Factor trinomials of the type x2 + bx + c.
Factor trinomials of the type ax2 + bx + c, a ? 1, using the FOIL method and the grouping method.
Factor special products of polynomials.
Terms with Common Factors : Copyright © 2008 Pearson Education, Inc. Terms with Common Factors When factoring, we should always look first to factor out a factor that is common to all the terms.
Example: 18 + 12x ? 6x2
= 6 • 3 + 6 • 2x ? 6 • x2
= 6(3 + 2x ? x2)
Factoring by Grouping : Copyright © 2008 Pearson Education, Inc. Factoring by Grouping In some polynomials, pairs of terms have a common binomial factor that can be removed in the process called factoring by grouping.
Example: x3 + 5x2 ? 10x ? 50
= (x3 + 5x2) + (?10x ? 50)
= x2(x + 5) ? 10(x + 5)
= (x2 ? 10)(x + 5)
Trinomials of the Type x2 + bx + c : Copyright © 2008 Pearson Education, Inc. Trinomials of the Type x2 + bx + c Factor: x2 + 9x + 14.
Solution:
1. Look for a common factor.
2. Find the factors of 14, whose sum is 9.
Pairs of Factors Sum
1, 14 15
2, 7 9 The numbers we need.
3. The factorization is (x + 2)(x + 7).
Another Example : Copyright © 2008 Pearson Education, Inc. Another Example Factor: 2y2 ? 20y + 48.
1. First, we look for a common factor.
2(y2 ? 10y + 24)
2. Look for two numbers whose product is 24 and whose sum is ?10.
Pairs Sum Pairs Sum
?1, ?24 ?25 ?2, ?12 ?14
?3, ?8 ?11 ?4, ?6 ?10
3. Complete the factorization: 2(y ? 4)(y ? 6).
Trinomials of the Typeax2 + bx + c, a ? 1 : Copyright © 2008 Pearson Education, Inc. Trinomials of the Typeax2 + bx + c, a ? 1 FOIL method
1. Factor out the largest common factor.
2. Find two First terms whose product is ax2.
3. Find two Last terms whose product is c.
4. Repeat steps (2) and (3) until a combination is found
for which the sum of the Outside and Inside products
is bx.
Example : Copyright © 2008 Pearson Education, Inc. Example Factor: 8x2 + 10x + 3.
(8x + )(x + )
(8x + 1)(x + 3) middle terms are wrong 24x + x = 25x
(4x + )(2x + )
(4x + 1)(2x + 3) middle terms are wrong 12x + 2x = 14x
(4x + 3)(2x + 1) Correct! 4x + 6x = 10x
Grouping Method ax2 + bx + c : Copyright © 2008 Pearson Education, Inc. Grouping Method ax2 + bx + c 1. Factor out the largest common factor.
2. Multiply the leading coefficient a and the constant c.
3. Try to factor the product ac so that the sum of the factors is b. That is, find integers p and q such that pq = ac and p + q = b.
4. Split the middle term. That is, write it as a sum using the factors found in step (3).
5. Factor by grouping.
Example : Copyright © 2008 Pearson Education, Inc. Example Factor: 12a3 ? 4a2 ? 16a.
1. Factor out the largest common factor, 4a.
4a(3a2 ? a ? 4)
2. Multiply a and c: (3)(?4) = ?12.
3. Try to factor ?12 so that the sum of the factors is the coefficient of the middle term, ?1.
(3)(?4) = ?12 and 3 + (?4) = ?1
4. Split the middle term using the numbers found in (3).
3a2 + 3a ? 4a ? 4
5. Factor by grouping. 3a2 + 3a ? 4a ? 4 = (3a2 + 3a) + (?4a ? 4)
= 3a(a + 1) ? 4(a + 1)
= (3a ? 4)(a + 1)
Be sure to include the common factor to get the complete factorization.
4a(3a ? 4)(a + 1)
Special Factorizations : Copyright © 2008 Pearson Education, Inc. Special Factorizations Difference of Squares
A2 ? B2 = (A + B)(A ? B)
Example: x2 ? 25 = (x + 5)(x ? 5)
Squares of Binomials
A2 + 2AB + B2 = (A + B)2
A2 ? 2AB + B2 = (A ? B)2
Example: x2 + 12x + 36 = (x + 6)2
More Factorizations : Copyright © 2008 Pearson Education, Inc. More Factorizations Sum or Difference of Cubes
A3 + B3 = (A + B)(A2 ? AB + B2)
A3 ? B3 = (A ? B)(A2 + AB + B2)
Example: 8y3 + 125 = (2y)3 + (5)3
= (2y + 5)(4y2 ? 10y + 25)