1. Slide 7- 1 Copyright © 2012 Pearson Education, Inc.

2. 5.2 Multiplication of Polynomials ■ Multiplying Monomials ■ Multiplying Monomials and Binomials ■ Multiply Any Two Polynomials ■ The Product of Two Binomials: FOIL ■ Squares of Binomials ■ Products of Sums and Differences ■ Function Notation

3. Slide 5- 3 Copyright © 2012 Pearson Education, Inc. Multiply: a) (6x)(7x) b) (5a)(  a) c) (  8x 6 )(3x 4 ) Solution a) (6x)(7x) = (6  7) (x  x) = 42x 2 b) (5a)(  a) = (5a)(  1a) = (5)(  1)(a  a) =  5a 2 c) (  8x 6 )(3x 4 ) = (  8  3) (x 6  x 4 ) =  24x 6 + 4 =  24x 10 Example

4. Slide 5- 4 Copyright © 2012 Pearson Education, Inc. Multiply: a) x and x + 7 b) 6x(x 2  4x + 5) Solution a) x(x + 7) = x  x + x  7 = x 2 + 7x b) 6x(x 2  4x + 5) = (6x)(x 2 )  (6x)(4x) + (6x)(5) = 6x 3  24x 2 + 30x Example

5. Slide 5- 5 Copyright © 2012 Pearson Education, Inc. Multiply each of the following. a) x + 3 and x + 5b) 3x  2 and x  1 Solution a) ( x + 3)(x + 5) = (x + 3)x + (x + 3)5 = x(x + 3) + 5(x + 3) = x  x + x  3 + 5  x + 5  3 = x 2 + 3x + 5x + 15 = x 2 + 8x + 15 Example

6. Slide 5- 6 Copyright © 2012 Pearson Education, Inc. Solution b) (3x  2)(x  1) = (3x  2)x  (3x  2)1 = x(3x  2)  1(3x  2) = x  3x  x  2  1  3x  1(  2) = 3x 2  2x  3x + 2 = 3x 2  5x + 2 continued

7. Slide 5- 7 Copyright © 2012 Pearson Education, Inc. The Product of Two Polynomials The product of two polynomials P and Q, is found by multiplying each term of P by every term of Q and combining like terms.

8. Slide 5- 8 Copyright © 2012 Pearson Education, Inc. Multiply: (5x 3 + x 2 + 4x)(x 2 + 3x). Solution 5x 3 + x 2 + 4x x 2 + 3x 15x 4 + 3x 3 + 12x 2 5x 5 + x 4 + 4x 3 5x 5 + 16x 4 + 7x 3 + 12x 2 Example

9. Slide 5- 9 Copyright © 2012 Pearson Education, Inc. The FOIL Method To multiply two binomials, A + B and C + D, multiply the First terms AC, the Outer terms AD, the Inner terms BC, and then the Last terms BD. Then combine like terms, if possible. (A + B)(C + D) = AC + AD + BC + BD Multiply First terms: AC. Multiply Outer terms: AD. Multiply Inner terms: BC Multiply Last terms: BD ↓ FOIL (A + B)(C + D) O I F L

10. Slide 5- 10 Copyright © 2012 Pearson Education, Inc. Multiply: (x + 4)(x 2 + 3). Solution F O I L (x + 4)(x 2 + 3) = x 3 + 3x + 4x 2 + 12 = x 3 + 4x 2 + 3x + 12 Example O I F L The terms are rearranged in descending order for the final answer.

11. Slide 5- 11 Copyright © 2012 Pearson Education, Inc. Multiply. a) (x + 8)(x + 5)b) (y + 4) (y  3) c) (5t 3 + 4t)(2t 2  1)d) (4  3x)(8  5x 3 ) Solution a) (x + 8)(x + 5)= x 2 + 5x + 8x + 40 = x 2 + 13x + 40 b) (y + 4) (y  3)= y 2  3y + 4y  12 = y 2 + y  12 Example

12. Slide 5- 12 Copyright © 2012 Pearson Education, Inc. Solution c) (5t 3 + 4t)(2t 2  1) = 10t 5  5t 3 + 8t 3  4t = 10t 5 + 3t 3  4t d) (4  3x)(8  5x 3 ) = 32  20x 3  24x + 15x 4 = 32  24x  20x 3 + 15x 4 Example continued In general, if the original binomials are written in ascending order, the answer is also written that way.

13. Slide 5- 13 Copyright © 2012 Pearson Education, Inc. Squaring a Binomial (A + B) 2 = A 2 + 2AB + B 2 ; (A – B) 2 = A 2 – 2AB + B 2 The square of a binomial is the square of the first term, plus twice the product of the two terms, plus the square of the last term. Trinomials that can be written in the form A 2 + 2AB + B 2 or A 2 – 2AB + B 2 are called perfect-square trinomials.

14. Slide 5- 14 Copyright © 2012 Pearson Education, Inc. Multiply. a) (x + 8) 2 b) (y  7) 2 c) (4x  3x 5 ) 2 Solution (A + B) 2 = A 2 +2  A  B + B 2 a) (x + 8) 2 = x 2 + 2  x  8 + 8 2 = x 2 + 16x + 64 Example

15. Slide 5- 15 Copyright © 2012 Pearson Education, Inc. Example continued Solution (A – B) 2 = A 2  2AB + B 2 b) (y  7) 2 = y 2  2  y  7 + 7 2 = y 2  14y + 49 c) (4x  3x 5 ) 2 = (4x) 2  2  4x  3x 5 + (3x 5 ) 2 = 16x 2  24x 6 + 9x 10

16. Slide 5- 16 Copyright © 2012 Pearson Education, Inc. The Product of a Sum and Difference The product of the sum and difference of the same two terms is the square of the first term minus the square of the second term. (A + B)(A – B) = A 2 – B 2. This is called a difference of squares.

17. Slide 5- 17 Copyright © 2012 Pearson Education, Inc. Multiply. a) (x + 8)(x  8) b) (6 + 5w) (6  5w) c) (4t 3  3)(4t 3 + 3) Solution (A + B)(A  B) = A 2  B 2 a) (x + 8)(x  8)= x 2  8 2 = x 2  64 Example

18. Slide 5- 18 Copyright © 2012 Pearson Education, Inc. Example continued Solution b) (6 + 5w) (6  5w) = 6 2  (5w) 2 = 36  25w 2 c) (4t 3  3)(4t 3 + 3) = (4t 3 ) 2  3 2 = 16t 6  9

19. Slide 5- 19 Copyright © 2012 Pearson Education, Inc. Function Notation Example Given f(x) = x 2 – 6x + 7, find and simplify each of the following. a. f(a) + 4b) f(a + 3) Solution a. To find f(a) + 4, we replace x with a to find f(a). Then we add 4 to the result. f(a) + 4 = a 2 – 6a + 7 + 4 = a 2 – 6a + 11

20. Slide 5- 20 Copyright © 2012 Pearson Education, Inc. continued f(x) = x 2 – 6x + 7 b) f(a + 3) f(a + 3) = (a + 3) 2 – 6(a + 3) + 7 = a 2 + 6a + 9 – 6a – 18 + 7 = a 2 – 2.