Algebra-Trigonometry-CynthiaYoung-3ed7.pdf - 11:16 AM Page 31 0.3 Polynomials Basic Operations EXAMPLE 5 31 Multiplying Two Polynomials Multiply and

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0.3 Polynomials: Basic Operations 31 First Inner Product of First Terms Product of Last Terms Outer Last Product of Outer Terms Product of Inner Terms Special Products The method outlined for multiplying polynomials works for all products of polynomials. For the special case when both polynomials are binomials, the FOIL method can also be used. W ORDS M ATH Apply the distributive property. (5 x 1)(2 x 3) 5 x (2 x 3) 1(2 x 3) Apply the distributive property. 5 x (2 x ) 5 x (3) 1(2 x ) 1(3) Multiply each set of monomials. 10 x 2 15 x 2 x 3 Combine like terms. 10 x 2 13 x 3 The FOIL method finds the products of the F irst terms, O uter terms, I nner terms, and (5 x 1)(2 x 3) 10 x 2 15 x 2 x 3 L ast terms. EXAMPLE 6 Multiplying Binomials Using the FOIL Method Multiply (3 x 1)(2 x 5) using the FOIL method. Solution: Multiply the first terms. (3 x )(2 x ) 6 x 2 Multiply the outer terms. (3 x )( 5) 15 x Multiply the inner terms. (1)(2 x ) 2 x Multiply the last terms. (1)( 5) 5 Add the first, outer, inner, and last terms, and identify the like terms. Combine like terms. 6 x 2 13 x 5 YOUR TURN Multiply (2 x 3)(5 x 2). (3 x + 1)(2 x - 5) = 6 x 2 - 15 x + 2 x - 5 Answer: 10 x 2 19 x 6 Study Tip When the binomials are of the form ( ax b )( cx d ), the outer and inner terms will be like terms and can be combined. EXAMPLE 5 Multiplying Two Polynomials Multiply and simplify . Solution: Multiply each term of the first trinomial by the entire second trinomial. Identify like terms. Combine like terms. 2 x 4 13 x 3 30 x 2 26 x 7 YOUR TURN Multiply and simplify ( - x 3 + 2 x - 4)(3 x 2 - x + 5). = 2 x 4 - 10 x 3 + 14 x 2 - 3 x 3 + 15 x 2 - 21 x + x 2 - 5 x + 7 = 2 x 2 ( x 2 - 5 x + 7) - 3 x ( x 2 - 5 x + 7) + 1( x 2 - 5 x + 7) ( x 2 - 5 x + 7) (2 x 2 - 3 x + 1) Answer: 3 x 5 x 4 x 3 14 x 2 14 x 20
EXAMPLE 7 Multiplying Binomials Resulting in Special Products Find the following: a. ( x 5)( x 5) b. ( x 5) 2 c. ( x 5) 2 Solution: a. ( x - 5)( x + 5) = x 2 + 5 x - 5 x - 5 2 = x 2 - 5 2 = x 2 - 25 32 CHAPTER 0 Prerequisites and Review Square of a binomial sum: ( a b ) 2 ( a b )( a b ) a 2 2 ab b 2 Square of a binomial difference: ( a b ) 2 ( a b )( a b ) a 2 2 ab b 2 P ERFECT SQUARES ( a + b )( a - b ) = a 2 - b 2 D IFFERENCE OF TWO SQUARES EXAMPLE 8 Finding the Square of a Binomial Sum Find ( x 3) 2 . INCORRECT ERROR: Don’t forget the middle term, which is twice the product of the two terms in the binomial. ( x + 3) 2 Z x 2 + 9 CORRECT ( x 3) 2 ( x 3)( x 3) x 2 3 x 3 x 9 x 2 6 x 9 Forgetting the middle term, which is twice the product of the two terms in the binomial. C O M M O N M I S TA K E Let a and b be any real number, variable, or algebraic expression in the following special products. Some products of binomials occur frequently in algebra and are given special names. Example 7 illustrates the difference of two squares and perfect squares . r First r Inner s Difference of two squares r Outer r Last r First r Inner r Outer r First r Inner r Outer r Last r Last b.

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• Summer '17
• juan alberto
• Algebra, Distributive Property, GCF