6-3-3 6 Dividing Polynomials - Plain Local Schools · 2016. 2. 18. · Holt Algebra 2 6-3 Dividing Polynomials Example 4: Geometry Application Write an expression that represents

Holt Algebra 2

6-3 Dividing Polynomials 6-3 Dividing Polynomials

Holt Algebra 2

Warm Up

Lesson Presentation

Lesson Quiz

Holt Algebra 2

6-3 Dividing Polynomials

Warm Up Divide using long division.

1. 161 ÷ 7

2. 12.18 ÷ 2.1

3.

4.

2x + 5y

23

5.8

7a – b

Divide.

6x – 15y 3

7a2 – ab a

Holt Algebra 2


Use long division and synthetic division to divide polynomials.

Objective

Holt Algebra 2


synthetic division

Vocabulary

Holt Algebra 2


Polynomial long division is a method for dividing a polynomial by another polynomials of a lower degree. It is very similar to dividing numbers.

Holt Algebra 2


Divide using long division.

Example 1: Using Long Division to Divide a

Polynomial

(–y2 + 2y3 + 25) ÷ (y – 3)

2y3 – y2 + 0y + 25

Step 1 Write the dividend in standard form, including terms with a coefficient of 0.

Step 2 Write division in the same way you would when dividing numbers.

y – 3 2y3 – y2 + 0y + 25

Holt Algebra 2


Notice that y times 2y2 is 2y3.

Write 2y2 above 2y3.

Step 3 Divide.

2y2

–(2y3 – 6y2) Multiply y – 3 by 2y2. Then

subtract. Bring down the next

term. Divide 5y2 by y. 5y2 + 0y

+ 5y

–(5y2 – 15y) Multiply y – 3 by 5y. Then


term. Divide 15y by y. 15y + 25

–(15y – 45)

70 Find the remainder.

+ 15

Multiply y – 3 by 15. Then

subtract.

Example 1 Continued

y – 3 2y3 – y2 + 0y + 25

Holt Algebra 2


Step 4 Write the final answer.

Example 1 Continued

–y2 + 2y3 + 25

y – 3 = 2y2 + 5y + 15 + 70

y – 3

Holt Algebra 2


Check It Out! Example 1a


(15x2 + 8x – 12) ÷ (3x + 1)

15x2 + 8x – 12



3x + 1 15x2 + 8x – 12

Holt Algebra 2


Check It Out! Example 1a Continued

Notice that 3x times 5x is 15x2.

Write 5x above 15x2.

Step 3 Divide.

5x

–(15x2 + 5x) Multiply 3x + 1 by 5x. Then


term. Divide 3x by 3x. 3x – 12

+ 1

–(3x + 1)

–13 Find the remainder.

Multiply 3x + 1 by 1. Then

subtract.

3x + 1 15x2 + 8x – 12

Holt Algebra 2




15x2 + 8x – 12

3x + 1 = 5x + 1 – 13

3x + 1

Holt Algebra 2


Check It Out! Example 1b


(x2 + 5x – 28) ÷ (x – 3)

x2 + 5x – 28



x – 3 x2 + 5x – 28

Holt Algebra 2


Notice that x times x is x2.

Write x above x2.

Step 3 Divide.

x

–(x2 – 3x) Multiply x – 3 by x. Then


term. Divide 8x by x. 8x – 28

+ 8

–(8x – 24)

–4 Find the remainder.

Multiply x – 3 by 8. Then

subtract.

Check It Out! Example 1b Continued

x – 3 x2 + 5x – 28

Holt Algebra 2




x2 + 5x – 28

x – 3 = x + 8 –

4

x – 3

Holt Algebra 2


Synthetic division is a shorthand method of dividing a polynomial by a linear binomial by using only the coefficients. For synthetic division to work, the polynomial must be written in standard form, using 0 and a coefficient for any missing terms, and the divisor must be in the form (x – a).

Holt Algebra 2


Holt Algebra 2


Divide using synthetic division.

Example 2A: Using Synthetic Division to Divide by a

Linear Binomial

(3x2 + 9x – 2) ÷ (x – )

Step 1 Find a. Then write the coefficients and a in the synthetic division format.

Write the coefficients of 3x2 + 9x – 2.

1 3

For (x – ), a = . 1 3

1 3

1 3

a =

1 3

3 9 –2

Holt Algebra 2


Example 2A Continued

Step 2 Bring down the first coefficient. Then multiply and add for each column.

Draw a box around the remainder, 1 . 1 3

1 3

3 9 –2

1

3

Step 3 Write the quotient.

3x + 10 + 1

1 3 1 3

x –

10 1 3

1

1 3

3

Holt Algebra 2


Example 2A Continued

3x + 10 + 1

1 3 1 3

x – Check Multiply (x – )

1 3

= 3x2 + 9x – 2

(x – ) 1 3

(x – ) 1 3

(x – ) 1 3

3x + 10 +

1 1 3 1 3

x –

Holt Algebra 2



(3x4 – x3 + 5x – 1) ÷ (x + 2)

Step 1 Find a.

Use 0 for the coefficient

of x2.

For (x + 2), a = –2. a = –2

Example 2B: Using Synthetic Division to Divide by a

Linear Binomial

3 – 1 0 5 –1 –2

Step 2 Write the coefficients and a in the synthetic division format.

Holt Algebra 2


Example 2B Continued

Draw a box around the

remainder, 45.

3 –1 0 5 –1 –2


–6

3 45


3x3 – 7x2 + 14x – 23 + 45

x + 2 Write the remainder over

the divisor.

46 –28 14

–23 14 –7

Holt Algebra 2




(6x2 – 5x – 6) ÷ (x + 3)

Step 1 Find a.

Write the coefficients of 6x2 – 5x – 6.

For (x + 3), a = –3. a = –3

–3 6 –5 –6


Holt Algebra 2



Draw a box around the

remainder, 63.

6 –5 –6 –3


–18

6 63


6x – 23 + 63

x + 3 Write the remainder over

the divisor.

–23

69

Holt Algebra 2




(x2 – 3x – 18) ÷ (x – 6)

Step 1 Find a.

Write the coefficients of x2 – 3x – 18.

For (x – 6), a = 6. a = 6

6 1 –3 –18


Holt Algebra 2



There is no remainder. 1 –3 –18 6


6

1 0


x + 3

18

3

Holt Algebra 2


You can use synthetic division to evaluate polynomials. This process is called synthetic substitution. The process of synthetic substitution is exactly the same as the process of synthetic division, but the final answer is interpreted differently, as described by the Remainder Theorem.

Holt Algebra 2


Example 3A: Using Synthetic Substitution

Use synthetic substitution to evaluate the polynomial for the given value.

P(x) = 2x3 + 5x2 – x + 7 for x = 2.

Write the coefficients of

the dividend. Use a = 2.

2 5 –1 7 2

4

2 41

P(2) = 41

Check Substitute 2 for x in P(x) = 2x3 + 5x2 – x + 7.

P(2) = 2(2)3 + 5(2)2 – (2) + 7

P(2) = 41

34 18

17 9

Holt Algebra 2


Example 3B: Using Synthetic Substitution


P(x) = 6x4 – 25x3 – 3x + 5 for x = – .

6 –25 0 –3 5

–2

6 7

1 3

– 1 3


the dividend. Use 0 for

the coefficient of x2. Use

a = . 1 3

P( ) = 7 1 3

2 –3 9

–6 9 –27

Holt Algebra 2




P(x) = x3 + 3x2 + 4 for x = –3.


the dividend. Use 0 for

the coefficient of x2 Use a

= –3.

1 3 0 4 –3

–3

1 4

P(–3) = 4

Check Substitute –3 for x in P(x) = x3 + 3x2 + 4.

P(–3) = (–3)3 + 3(–3)2 + 4

P(–3) = 4

0 0

0 0

Holt Algebra 2




P(x) = 5x2 + 9x + 3 for x = .

5 9 3

1

5 5

1 5

1 5


the dividend. Use a = . 1 5

P( ) = 5 1 5

2

10

Holt Algebra 2


Example 4: Geometry Application

Write an expression that represents the area of the top face of a rectangular prism when the height is x + 2 and the volume of the prism is x3 – x2 – 6x.

Substitute.

Use synthetic division.

The volume V is related to the area A and the height h by the equation V = A h. Rearranging for A gives A = . V

h x3 – x2 – 6x

x + 2 A(x) =

1 –1 –6 0 –2

–2

1 0

The area of the face of the rectangular prism can be represented by A(x)= x2 – 3x.

0 6

0 –3

Holt Algebra 2


Check It Out! Example 4

Write an expression for the length of a rectangle with width y – 9 and area y2 – 14y + 45.

Substitute.

Use synthetic division.

The area A is related to the width w and the length l by the equation A = l w.

y2 – 14y + 45

y – 9 l(x) =

1 –14 45 9

9

1 0 The length of the rectangle can be represented by l(x)= y – 5.

–45

–5

Holt Algebra 2


4. Find an expression for the height of a parallelogram whose area is represented by 2x3 – x2 – 20x + 3 and whose base is represented by (x + 3).

Lesson Quiz

2. Divide by using synthetic division. (x3 – 3x + 5) ÷ (x + 2)

1. Divide by using long division. (8x3 + 6x2 + 7) ÷ (x + 2)

2x2 – 7x + 1

194; –4 3. Use synthetic substitution to evaluate P(x) = x3 + 3x2 – 6 for x = 5 and x = –1.

8x2 – 10x + 20 – 33

x + 2

x2 – 2x + 1 + 3

x + 2

Documents

6-3-3 6 Dividing Polynomials - Plain Local Schools · 2016. 2. 18. · Holt Algebra 2 6-3 Dividing Polynomials Example 4: Geometry Application Write an expression that represents