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§ 5.4 Special Products. Martin-Gay, Beginning and Intermediate Algebra, 4ed 22 The FOIL Method When multiplying two binomials, the distributive property.

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Presentation on theme: "§ 5.4 Special Products. Martin-Gay, Beginning and Intermediate Algebra, 4ed 22 The FOIL Method When multiplying two binomials, the distributive property."— Presentation transcript:

1 § 5.4 Special Products

2 Martin-Gay, Beginning and Intermediate Algebra, 4ed 22 The FOIL Method When multiplying two binomials, the distributive property can be easily remembered as the FOIL method. F – product of First terms O – product of Outside terms I – product of Inside terms L – product of Last terms

3 Martin-Gay, Beginning and Intermediate Algebra, 4ed 33 = y 2 – 8y – 48 Multiply (y – 12)(y + 4). (y – 12)(y + 4) Product of First terms is y 2 Product of Outside terms is 4y Product of Inside terms is – 12y Product of Last terms is – 48 (y – 12)(y + 4) = y 2 + 4y – 12y – 48 F O I L Using the FOIL Method Example:

4 Martin-Gay, Beginning and Intermediate Algebra, 4ed 44 Multiply (2x – 4)(7x + 5). (2x – 4)(7x + 5) = = 14x 2 + 10x – 28x – 20 F 2x(7x) F + 2x(5) O – 4(7x) I – 4(5) L O I L = 14x 2 – 18x – 20 We multiplied these same two binomials together in the previous section, using a different technique, but arrived at the same product. Using the FOIL Method Example:

5 Martin-Gay, Beginning and Intermediate Algebra, 4ed 55 Squaring a Binomial A binomial squared is equal to the square of the first term plus or minus twice the product of both terms plus the square of the second term. (a + b) 2 = a 2 + 2ab + b 2 (a – b) 2 = a 2 – 2ab + b 2 Example: Multiply (12a  3) 2. (12a  3) 2 = 144a 2  72a + 9 = (12a) 2  2(12a)(3) + (3) 2

6 Martin-Gay, Beginning and Intermediate Algebra, 4ed 66 Sum and Difference of Two Terms Multiplying the Sum and Difference of Two Terms The product of the sum and difference of two terms is the square of the first term minus the square of the second term. (a + b)(a – b) = a 2 – b 2 Example: Multiply (5a + 3)(5a  3). (5a + 3)(5a  3) = 25a 2  9 = (5a) 2  3 2

7 Martin-Gay, Beginning and Intermediate Algebra, 4ed 77 Although you will arrive at the same results for the special products by using the techniques of this section or last section, memorizing these products can save you some time in multiplying polynomials. Special Products


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