9.1

Warm UpWarm Up

Lesson QuizLesson Quiz

Lesson PresentationLesson Presentation

Add and Subtract Polynomials

9.1 Warm-Up

Simplify the expression.

1. 5x + 4(2x + 7)

2. 9x – 6(x + 2) + 3

ANSWER 13x + 28

ANSWER 3x – 9

3. Imported square tiles used for a kitchen floor measure 18 centimeters on one side. What is the area of a floorcomposed of 50 tiles? Use A = s2 for the area of a tile.

ANSWER 16,200 cm2

9.1 Example 1

Write 15x – x3 + 3 so that the exponents decrease from left to right. Identify the degree and leading coefficient of the polynomial.

SOLUTION

Consider the degree of each of the polynomial’s terms.

The polynomial can be written as – x3 +15 + 3. The greatest degree is 3, so the degree of the polynomial is 3, and the leading coefficient is –1.

15x – x3 + 3

9.1 Guided Practice

Write 5y – 2y2 + 9 so that the exponents decrease from left to right. Identify the degree and leading coefficient of the polynomial.

1.

– 2y2 +5y + 9 Degree: 2, Leading Coefficient: –2

ANSWER

9.1 Example 2

Tell whether is a polynomial. If it is a polynomial, find its degree and classify it by the number of its terms. Otherwise, tell why it is not a polynomial.

5th degree binomialYes

No; variable exponent

No; variable exponent

2nd degree trinomialYes

0 degree monomialYes

7bc3 + 4b4c

n– 2 – 3

6n4 – 8n

2x2 + x – 5

9

Classify by degree and number of terms

Is it a polynomial?Expression

a.

b.c.d.e.

9.1 Guided Practice

Tell whether y3 – 4y + 3 is a polynomial. If it is a polynomial, find its degree and classify it by the number of its terms. Otherwise, tell why it is not a polynomial.

2.

ANSWER

polynomial Degree: 3, trinomial

9.1 Example 2

Find the sum.

a. (2x3 – 5x2 + x) + (2x2 + x3 – 1)

b. (3x2 + x – 6) + (x2 + 4x + 10)

SOLUTION

a. Vertical format: Align like terms in vertical columns. (2x3 – 5x2 + x)

+ x3 + 2x2 – 1

3x3 – 3x2 + x – 1

9.1 Example 2

b. Horizontal format: Use the associative and commutative properties to group like terms. Then simplify.

(3x2 + x – 6) + (x2 + 4x + 10) =

= 4x2 + 5x + 4

(3x2 + x2) + (x + 4x) + (– 6 + 10)

9.1 Guided Practice

(5x3 + 4x – 2x) + (4x2 +3x3 – 6)Find the sum .3.

= 8x3 + 4x2 + 2x – 6ANSWER

9.1 Example 4

Find the difference.

a. (4n2 + 5) – (–2n2 + 2n – 4)

b. (4x2 – 3x + 5) – (3x2 – x – 8)

SOLUTION

a. (4n2 + 5) 4n2 + 5

–(–2n2 + 2n – 4) 2n2 – 2n + 4

6n2 – 2n + 9

9.1 Example 4

b. (4x2 – 3x + 5) – (3x2 – x – 8) =

= (4x2 – 3x2) + (–3x + x) + (5 + 8)

= x2 – 2x + 13

4x2 – 3x + 5 – 3x2 + x + 8

9.1 Guided Practice

(4x2 – 7x) – (5x2 + 4x – 9)Find the difference .4.

–x2 – 11x + 9ANSWER

9.1 Example 5

BASEBALL ATTENDANCE

Major League Baseball teams are divided into two leagues. During the period 1995–2001, the attendance N and A (in thousands) at National and American League baseball games, respectively, can be modeled by

N = –488t2 + 5430t + 24,700 and

where t is the number of years since 1995. About how many people attended Major League Baseball games in 2001?

A = –318t2 + 3040t + 25,600

9.1 Example 5

SOLUTION

STEP 1

Add the models for the attendance in each league to find a model for M, the total attendance (in thousands).

M = (–488t2 + 5430t + 24,700) + (–318t2 + 3040t + 25,600)

= (–488t2 – 318t2) + (5430t + 3040t) + (24,700 + 25,600)

= –806t2 + 8470t + 50,300

9.1 Example 5

STEP 2

Substitute 6 for t in the model, because 2001 is 6 years after 1995.

M = –806(6)2 + 8470(6) + 50,300 72,100

ANSWER

About 72,100,000 people attended Major League Baseball games in 2001.

9.1 Guided Practice

BASEBALL ATTENDNCE Look back at Example 5. Find the difference in attendance at National and American League baseball games in 2001.

5.

ANSWER about 7,320,000 people

9.1 Lesson Quiz

If the expression is a polynomial, find its degree and classify it by the number of terms. Otherwise, tell why it is not a polynomial.

1. m3 + n4m2 + m–2 No; one exponent is not a whole

number.ANSWER

2. – 3b3c4 – 4b2c + c8

ANSWER 8th degree trinomial

9.1 Lesson Quiz

Find the sum or difference.

3. (3m2 – 2m + 9) + (m2 + 2m – 4)

4m2 + 5ANSWER

4. (– 4a2 + 3a – 1) – (a2 + 2a – 6)

ANSWER –5a2 + a + 5

5. The number of dog adoptions D and cat adoptions C can be modeled by D = 1.35 t2 – 9.8t + 131 and C= 0.1t2 – 3t + 79 where t represents the years since 1998. About how many dogs and cats were adopted in 2004?

about 185 dogs and catsANSWER