1. Slide 5- 1 Copyright © 2012 Pearson Education, Inc.

2. 5.7 Polynomials in Several Variables ■ Evaluating Polynomials ■ Like Terms and Degrees ■ Addition and Subtraction ■ Multiplication ■ Function Notation

3. Slide 5- 3 Copyright © 2012 Pearson Education, Inc. Evaluate the polynomial 5 + 4x + xy 2 + 9x 3 y 2 for x =  3 and y = 4. Solution We substitute  3 for x and 4 for y: 5 + 4x + xy 2 + 9x 3 y 2 = 5 + 4(  3) + (  3)(4 2 ) + 9(  3) 3 (4) 2 = 5  12  48  3888 =  3943 Example

4. Slide 5- 4 Copyright © 2012 Pearson Education, Inc. The surface area of a right circular cylinder is given by the polynomial 2  rh + 2  r 2 where h is the height and r is the radius of the base. A barn silo has a height of 50 feet and a radius of 9 feet. Approximate its surface area. Solution We evaluate the polynomial for h = 50 ft and r = 9 ft. If 3.14 is used to approximate , we have Example h r

5. Slide 5- 5 Copyright © 2012 Pearson Education, Inc. continued h = 50 ft and r = 9 ft 2  rh + 2  r 2  2(3.14)(9 ft)(50 ft) + 2(3.14)(9 ft) 2  2(3.14)(9 ft)(50 ft) + 2(3.14)(81 ft 2 )  2826 ft 2 + 508.68 ft 2  3334.68 ft 2 Note that the unit in the answer (square feet) is a unit of area. The surface area is about 3334.7 ft 2 (square feet).

6. Slide 5- 6 Copyright © 2012 Pearson Education, Inc. Recall that the degree of a monomial is the number of variable factors in the term. Example Identify the coefficient and the degree of each term and the degree of the polynomial 10x 3 y 2  15xy 3 z 4 + yz + 5y + 3x 2 + 9. TermCoefficientDegreeDegree of the Polynomial 10x 3 y 2 105 8  15xy 3 z 4  15 8 yz12 5y5y51 3x23x2 32 990

7. Slide 5- 7 Copyright © 2012 Pearson Education, Inc. Like Terms Like, or similar terms either have exactly the same variables with exactly the same exponents or are constants. For example, 9w 5 y 4 and 15w 5 y 4 are like terms and  12 and 14 are like terms, but  6x 2 y and 9xy 3 are not like terms.

8. Slide 5- 8 Copyright © 2012 Pearson Education, Inc. a) 10x 2 y + 4xy 3  6x 2 y  2xy 3 b) 8st  6st 2 + 4st 2 + 7s 3 + 10st  12s 3 + t  2 Solution a) 10x 2 y + 4xy 3  6x 2 y  2xy 3 = (10  6)x 2 y + (4  2)xy 3 = 4x 2 y + 2xy 3 b) 8st  6st 2 + 4st 2 + 7s 3 + 10st  12s 3 + t  2 =  5s 3  2st 2 + 18st + t  2 Example Combine like terms.

9. Slide 5- 9 Copyright © 2012 Pearson Education, Inc. Addition and Subtraction Example Add: (  6x 3 + 4y  6y 2 ) + (7x 3 + 5x 2 + 8y 2 ). Solution (  6x 3 + 4y  6y 2 ) + (7x 3 + 5x 2 + 8y 2 ) = (  6 + 7)x 3 + 5x 2 + 4y + (  6 + 8)y 2 = x 3 + 5x 2 + 4y + 2y 2

10. Slide 5- 10 Copyright © 2012 Pearson Education, Inc. Subtract: (5x 2 y + 2x 3 y 2 + 4x 2 y 3 + 7y)  (5x 2 y  7x 3 y 2 + x 2 y 2  6y). Solution (5x 2 y + 2x 3 y 2 + 4x 2 y 3 + 7y)  (5x 2 y  7x 3 y 2 + x 2 y 2  6y) = (5x 2 y + 2x 3 y 2 + 4x 2 y 3 + 7y)  5x 2 y + 7x 3 y 2  x 2 y 2 + 6y = 9x 3 y 2 + 4x 2 y 3  x 2 y 2 + 13y Example

11. Slide 5- 11 Copyright © 2012 Pearson Education, Inc. Multiplication Example Multiply: (4x 2 y  3xy + 4y)(xy + 3y). Solution 4x 2 y  3xy + 4y xy + 3y 12x 2 y 2  9xy 2 + 12y 2 4x 3 y 2  3x 2 y 2 + 4xy 2 4x 3 y 2 + 9x 2 y 2  5xy 2 + 12y 2

12. Slide 5- 12 Copyright © 2012 Pearson Education, Inc. Multiply. a) (x + 6y)(2x  3y) b) (5x + 7y) 2 c) (a 4  5a 2 b 2 ) 2 d) (7a 2 b + 3b)(7a 2 b  3b) e) (  3x 3 y 2 + 7t)(3x 3 y 2 + 7t) f) (3x + 1  4y)(3x + 1 + 4y) Solution a) (x + 6y)(2x  3y) = 2x 2  3xy + 12xy  18y 2 = 2x 2 + 9xy  18y 2 FOIL Example

13. Slide 5- 13 Copyright © 2012 Pearson Education, Inc. b) (5x + 7y) 2 = (5x) 2 + 2(5x)(7y) + (7y) 2 = 25x 2 + 70xy + 49y 2 c) (a 4  5a 2 b 2 ) 2 = (a 4 ) 2  2(a 4 )(5a 2 b 2 ) + (5a 2 b 2 ) 2 = a 8  10a 6 b 2 + 25a 4 b 4 d) (7a 2 b + 3b)(7a 2 b  3b) = (7a 2 b) 2  (3b) 2 = 49a 4 b 2  9b 2 Solution continued

14. Slide 5- 14 Copyright © 2012 Pearson Education, Inc. e) (  3x 3 y 2 + 7t)(3x 3 y 2 + 7t) = (7t  3x 3 y 2 )(7t + 3x 3 y 2 ) = (7t) 2  (3x 3 y 2 ) 2 = 49t 2  9x 6 y 4 f) (3x + 1  4y)(3x + 1 + 4y) = (3x + 1) 2  (4y) 2 = 9x 2 + 6x + 1  16y 2 Solution continued

15. Slide 5- 15 Copyright © 2012 Pearson Education, Inc. Function Notation Example Given f(x) = x 2 – 3x + 2, find and simplify f(a + 5). Solution To find f(a + 5), we replace x with a + 5 and simplify. f(a + 5) = (a + 5) 2 – 3 (a + 5) + 2 = a 2 + 10a + 25 – 3a – 15 + 2 = a 2 + 7a + 12