Chapter 5 Polynomials
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1 Department of Mathematics Grossmont College October 7, 2012
2 Multiplying Polynomials
3 Multiplying Binomials using the Distributive Property We can multiply two binomials using the Distributive Property, a (b + c) = a b + a c. Suppose we want to multiply (2x 3) times (x + 4). Then we treat (2x 3) like it is playing the role of a in the distributive property: (2x 3) (x + 4) = (2x 3) x +(2x 3) 4
4 Multiplying Binomials using the Distributive Property We can multiply two binomials using the Distributive Property, a (b + c) = a b + a c. Suppose we want to multiply (2x 3) times (x + 4). Then we treat (2x 3) like it is playing the role of a in the distributive property: (2x 3) (x + 4) = (2x 3) x +(2x 3) 4
5 Multiplying Binomials using the Distributive Property We can multiply two binomials using the Distributive Property, a (b + c) = a b + a c. Suppose we want to multiply (2x 3) times (x + 4). Then we treat (2x 3) like it is playing the role of a in the distributive property: (2x 3) (x + 4) = (2x 3) x +(2x 3) 4
6 Multiplying Binomials using the Distributive Property We can multiply two binomials using the Distributive Property, a (b + c) = a b + a c. Suppose we want to multiply (2x 3) times (x + 4). Then we treat (2x 3) like it is playing the role of a in the distributive property: (2x 3) (x + 4) = (2x 3) x +(2x 3) 4
7 Multiplying Binomials using the Distributive Property We can multiply two binomials using the Distributive Property, a (b + c) = a b + a c. Suppose we want to multiply (2x 3) times (x + 4). Then we treat (2x 3) like it is playing the role of a in the distributive property: (2x 3) (x + 4) = (2x 3) x +(2x 3) 4
8 Multiplying Binomials using the Distributive Property We can multiply two binomials using the Distributive Property, a (b + c) = a b + a c. Suppose we want to multiply (2x 3) times (x + 4). Then we treat (2x 3) like it is playing the role of a in the distributive property: (2x 3) (x + 4) = (2x 3) x +(2x 3) 4
9 Multiplying Binomials using the Distributive Property We can multiply two binomials using the Distributive Property, a (b + c) = a b + a c. Suppose we want to multiply (2x 3) times (x + 4). Then we treat (2x 3) like it is playing the role of a in the distributive property: (2x 3) (x + 4) = (2x 3) x +(2x 3) 4
10 Multiplying Binomials using the Distributive Property We can multiply two binomials using the Distributive Property, a (b + c) = a b + a c. Suppose we want to multiply (2x 3) times (x + 4). Then we treat (2x 3) like it is playing the role of a in the distributive property: (2x 3) (x + 4) = (2x 3) x +(2x 3) 4 = x (2x 3)+4 (2x 3) since a b = b a = x (2x)+x ( 3)+4 (2x)+4 ( 3) Distributive Prop. = (2x) x +( 3) x + 4 (2x)+( 3) 4 Commutative. Prop. = 2x x +( 3) x + 4 2x +( 3) 4 Associative Prop. = 2x 2 +( 3x)+8x +( 12) Closure Prop. = 2x 2 + 5x 12 Combine Like Terms
11 Multiplying Binomials using the FOIL Technique The product of two binomials results in four terms before the like terms are combined. The acronym Foil stands for FIRST, OUTSIDE, INSIDE, LAST, and should remind you how to compute the product of two binomials. Consider the following product: (2x 3) (x + 4) = 2x 2 }{{} + 8x }{{} +( 3x) }{{} +( 12) }{{} F O I L
12 Multiplying Binomials using the FOIL Technique 2x 2 comes from multiplying the first terms of each binomial. (2x 3) (x + 4) = }{{} 2x 2 }{{} 8x }{{} +( 12) }{{} F O I L I R S T
13 Multiplying Binomials using the FOIL Technique 8x comes from multiplying the outside terms of each binomial. (2x 3) (x + 4) = }{{} 2x 2 }{{} 8x }{{} +( 12) }{{} F O I L U T S I D E
14 Multiplying Binomials using the FOIL Technique 3x comes from multiplying the inside terms of each binomial. (2x 3) (x + 4) = }{{} 2x 2 }{{} 8x }{{} +( 12) }{{} F O I L N S I D E
15 Multiplying Binomials using the FOIL Technique 12 comes from multiplying the last terms of each binomial. (2x 3) (x + 4) = }{{} 2x 2 }{{} 8x }{{} +( 12) }{{} F O I L A S T
16 Trinomial times Binomial Example: Multiply (x 2 3x + 4) (2x 3) Solution (x 2 3x + 4) (2x 3) = = (2x 3) (x 2 3x + 4) comm prop = 2x (x 2 3x + 4)+( 3) (x 2 3x + 4) distr. prop = 2x 3 6x 2 + 8x 3x 2 + 9x 12 distr. prop = 2x 3 +( 6x 2 3x 2 )+(8x + 9x) 12 comm., assoc. + = 2x 3 9x x 12 addn closure prop
17 p5 Factoring Polynomials
18 #1 RULE: FACTOR OUT THE GCF Factoring reverses multiplication. Consider the polynomial expression 6x 2 3x, whose two terms have a greatest common factor, 3x. 6x 2 3x = (3x) (2x) (3x) (1) = 3x (2x 1) since a b a c = a (b c) We can rewrite 6x 2 3x as a difference of two products. Afterwards, we can rewrite an equivalent expression using the distributive property. We call this process factoring out the gcf. Definition (The #1 Rule of Factoring) The first step to factoring any algebraic expression is to factor out the gcf (if there is one).
19 Definition The greatest common factor (GCF) for a polynomial is the largest monomial that divides (is a factor of) each term of the polynomial. Example: The greatest common factor for 25x x 4 30x 3 is 5x 3 since it is the largest monomial that is a factor of each term. 25x x 4 30x 3 = 5x 3 (5x 2 )+5x 3 (4x) 5x 3 (6) = 5x 3 (5x 2 + 4x 6)
20 How to find the GCF of a polynomial 1 Find the GCF of the coefficients of each variable factor.
21 How to find the GCF of a polynomial 1 Find the GCF of the coefficients of each variable factor. 2 For each variable factor common to all the terms, determine the smallest exponent that the variable factor is raised to.
22 How to find the GCF of a polynomial 1 Find the GCF of the coefficients of each variable factor. 2 For each variable factor common to all the terms, determine the smallest exponent that the variable factor is raised to. 3 Compute the product of the common factors found in Steps 1 and 2. This expression is the GCF of the polynomial.
23 How to find the GCF of a polynomial 1 Find the GCF of the coefficients of each variable factor. 2 For each variable factor common to all the terms, determine the smallest exponent that the variable factor is raised to. 3 Compute the product of the common factors found in Steps 1 and 2. This expression is the GCF of the polynomial. Factor the greatest common factor from each of the following. 8x 3 8x 2 48x 15a 7 25a a 3 12x 4 y 5 9x 3 y 4 15x 5 y 3 4(a + b) 4 + 6(a + b) (a + b) 2 x(x + 7)+2(x + 7)
24 Factoring Trinomials with a Leading Coefficient of 1 Earlier in the chapter, we multiplied binomials. (x + 2) (x + 8) = x x + 16 (x + 6)(x + 3) = x 2 + 9x + 18 In each case, the product of the two binomials is a trinomial.
25 Factoring Trinomials with a Leading Coefficient of 1 Earlier in the chapter, we multiplied binomials. (x + 2) (x + 8) = x x + 16 (x + 6)(x + 3) = x 2 + 9x + 18 In each case, the product of the two binomials is a trinomial. The first term in the resulting trinomial is obtained by multiplying the first term in each binomial.
26 Factoring Trinomials with a Leading Coefficient of 1 Earlier in the chapter, we multiplied binomials. (x + 2) (x + 8) = x x + 16 (x + 6)(x + 3) = x 2 + 9x + 18 In each case, the product of the two binomials is a trinomial. The first term in the resulting trinomial is obtained by multiplying the first term in each binomial. The middle term arises from adding the product of the two inside terms with the product of the two outside terms.
27 Factoring Trinomials with a Leading Coefficient of 1 Earlier in the chapter, we multiplied binomials. (x + 2) (x + 8) = x x + 16 (x + 6)(x + 3) = x 2 + 9x + 18 In each case, the product of the two binomials is a trinomial. The first term in the resulting trinomial is obtained by multiplying the first term in each binomial. The middle term arises from adding the product of the two inside terms with the product of the two outside terms. The last term is the product of the two outside terms. The last term is the product of the last terms in each binomial.
28 Factoring Trinomials with a Leading Coefficient of 1 In general, (x + a) (x + b) = x 2 + ax + bx + a b
29 Factoring Trinomials with a Leading Coefficient of 1 In general, (x + a) (x + b) = x 2 + ax + bx + a b = x 2 +(a + b)x + a b
30 Factoring Trinomials with a Leading Coefficient of 1 In general, (x + a) (x + b) = x 2 + ax + bx + a b = x 2 +(a + b)x + a b We can view this generalization as a factoring problem x 2 +(a + b)x + a b = (x + a) (x + b)
31 Factoring Trinomials with a Leading Coefficient of 1 In general, (x + a) (x + b) = x 2 + ax + bx + a b = x 2 +(a + b)x + a b We can view this generalization as a factoring problem x 2 +(a + b)x + a b = (x + a) (x + b) To factor a trinomial with a leading coefficient of 1, we simply find the two numbers a and b whose sum is the coefficient of the middle term, and whose product is the constant term.
32 Factoring Trinomials with a Leading Coefficient of 1 Factor. x 2 + 5x + 4 x 2 + 7x + 6 x 2 + 9x + 14 x x + 24 x x + 34 x x + 27 x x + 64 x x + 65 x 2 x + 5 x 2 + 5xy + 4y 2 x 2 + 5xy + 6y 2 x xy + 27y 2 m mn+60n 2 x 2 + 2x 15 x 2 7x 18 x 2 + x 20 x x 24
33 Factor By Grouping Method Polynomials with four terms can sometimes be factored by grouping. Example Factor x 4 2x 3 8x + 16 Solution: x 4 2x 3 8x + 16 = (x 4 2x 3 )+( 8x + 16) (assoc. prop +) [ ] [ ] = x (x 3 )+( 2) (x 3 ) + ( 8) x +( 8) ( 2) = x 3 (x 2) 8(x 2) (distr. prop.) = x 3 (x 2) 8(x 2) (identify common factor) = (x 2)(x 3 8) (distr. prop) **Technically we are not Professor done. Tim Busken This is not Chapter the5 Polynomials prime factorization of the
34 Factor By Grouping Method Factor each using the grouping technique. 5x + 5y + x 2 + xy ab 3 + b 3 + 6a y 4 + 3x 3 + x 3 y 4 x 3 + 5x 2 + 3x + 15
35 ac-grouping method for quadratics Problem: Factor ax 2 + bx + c Problem: Factor 10x 2 11x 6 (1) Multiply a times c. (1) a = 10, b = 11, c = 6, so clearly a c = 60 (2) List all possible pairs of numbers whose product is ac ւց ւց ( 10) 1 ( 60) 3 ( 20) ( 12) ( 4) (3) Box the pair whose sum is bր (3) b = 11, and 15+4 = 11
36 ac-grouping method for quadratics Problem: Factor ax 2 + bx + c Problem: Factor 10x 2 11x 6 (4) Replace b with the sum (4) 10x 2 11x 6 of the circled pair. Distribute = 10x 2 +( 15x + 4x) 6 x into this quantity = 10x 2 15x + 4x 6 (5) Now factor by grouping: (5) (10x 2 15x)+(4x 6) Use parenthesis to group the first two terms, and another () = 5x(2x 3)+2(2x 3) to group the second two terms. = 5x(2x 3)+2(2x 3) = (2x 3) (5x + 2)
Factoring completely is factoring a product down to a product of prime factors. 24 (2)(12) (2)(2)(6) (2)(2)(2)(3)
Factoring Contents Introduction... 2 Factoring Polynomials... 4 Greatest Common Factor... 4 Factoring by Grouping... 5 Factoring a Trinomial with a Table... 5 Factoring a Trinomial with a Leading Coefficient
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