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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
6-3 Dividing Polynomials
Use long division and synthetic division to divide polynomials.
Objective
Holt Algebra 2
6-3 Dividing Polynomials
synthetic division
Vocabulary
Holt Algebra 2
6-3 Dividing Polynomials
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
6-3 Dividing Polynomials
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
6-3 Dividing Polynomials
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
subtract. Bring down the next
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
6-3 Dividing Polynomials
Step 4 Write the final answer.
Example 1 Continued
–y2 + 2y3 + 25
y – 3 = 2y2 + 5y + 15 + 70
y – 3
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 1a
Divide using long division.
(15x2 + 8x – 12) ÷ (3x + 1)
15x2 + 8x – 12
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.
3x + 1 15x2 + 8x – 12
Holt Algebra 2
6-3 Dividing Polynomials
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
subtract. Bring down the next
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
6-3 Dividing Polynomials
Check It Out! Example 1a Continued
Step 4 Write the final answer.
15x2 + 8x – 12
3x + 1 = 5x + 1 – 13
3x + 1
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 1b
Divide using long division.
(x2 + 5x – 28) ÷ (x – 3)
x2 + 5x – 28
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.
x – 3 x2 + 5x – 28
Holt Algebra 2
6-3 Dividing Polynomials
Notice that x times x is x2.
Write x above x2.
Step 3 Divide.
x
–(x2 – 3x) Multiply x – 3 by x. Then
subtract. Bring down the next
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
6-3 Dividing Polynomials
Check It Out! Example 1b Continued
Step 4 Write the final answer.
x2 + 5x – 28
x – 3 = x + 8 –
4
x – 3
Holt Algebra 2
6-3 Dividing Polynomials
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
6-3 Dividing Polynomials
Holt Algebra 2
6-3 Dividing Polynomials
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
6-3 Dividing Polynomials
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
6-3 Dividing Polynomials
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
6-3 Dividing Polynomials
Divide using synthetic division.
(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
6-3 Dividing Polynomials
Example 2B Continued
Draw a box around the
remainder, 45.
3 –1 0 5 –1 –2
Step 3 Bring down the first coefficient. Then multiply and add for each column.
–6
3 45
Step 4 Write the quotient.
3x3 – 7x2 + 14x – 23 + 45
x + 2 Write the remainder over
the divisor.
46 –28 14
–23 14 –7
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 2a
Divide using synthetic division.
(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
Step 2 Write the coefficients and a in the synthetic division format.
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 2a Continued
Draw a box around the
remainder, 63.
6 –5 –6 –3
Step 3 Bring down the first coefficient. Then multiply and add for each column.
–18
6 63
Step 4 Write the quotient.
6x – 23 + 63
x + 3 Write the remainder over
the divisor.
–23
69
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 2b
Divide using synthetic division.
(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
Step 2 Write the coefficients and a in the synthetic division format.
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 2b Continued
There is no remainder. 1 –3 –18 6
Step 3 Bring down the first coefficient. Then multiply and add for each column.
6
1 0
Step 4 Write the quotient.
x + 3
18
3
Holt Algebra 2
6-3 Dividing Polynomials
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
6-3 Dividing Polynomials
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
6-3 Dividing Polynomials
Example 3B: Using Synthetic Substitution
Use synthetic substitution to evaluate the polynomial for the given value.
P(x) = 6x4 – 25x3 – 3x + 5 for x = – .
6 –25 0 –3 5
–2
6 7
1 3
– 1 3
Write the coefficients of
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
6-3 Dividing Polynomials
Check It Out! Example 3a
Use synthetic substitution to evaluate the polynomial for the given value.
P(x) = x3 + 3x2 + 4 for x = –3.
Write the coefficients of
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
6-3 Dividing Polynomials
Check It Out! Example 3b
Use synthetic substitution to evaluate the polynomial for the given value.
P(x) = 5x2 + 9x + 3 for x = .
5 9 3
1
5 5
1 5
1 5
Write the coefficients of
the dividend. Use a = . 1 5
P( ) = 5 1 5
2
10
Holt Algebra 2
6-3 Dividing Polynomials
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
6-3 Dividing Polynomials
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
6-3 Dividing Polynomials
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
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