6-3
6-3 Dividing Polynomials
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
22
6-3
Dividing Polynomials
Lesson Quiz
Find each product.
1. (2a3 – a + 3)(a2 + 3a – 5)
2a5 + 6a4 – 11a3 + 14a – 15
2. Use Pascal’s Triangle to expand the expression.
(y – 5)4
y4 – 20y3 + 150y2 – 500y + 625
3. Use Pascal’s Triangle to expand the expression.
(3a – b)3
27a3 – 27a2b + 9ab2 – b3
Holt Algebra 2
6-3
Dividing Polynomials
Objective
Use long division and synthetic division
to divide polynomials.
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
Example 1: Using Long Division to Divide a
Polynomial
Divide using long division.
(–y2 + 2y3 + 25) ÷ (y – 3)
Step 1 Write the dividend in standard form, including
terms with a coefficient of 0.
2y3 – y2 + 0y + 25
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
Example 1 Continued
Step 3 Divide.
2y2+ 5y + 15
y – 3 2y3 – y2 + 0y + 25
–(2y3 – 6y2)
5y2 + 0y
–(5y2 – 15y)
15y + 25
–(15y – 45)
70
Holt Algebra 2
Notice that y times 2y2 is 2y3.
Write 2y2 above 2y3.
Multiply y – 3 by 2y2. Then
subtract. Bring down the next
term. Divide 5y2 by y.
Multiply y – 3 by 5y. Then
subtract. Bring down the next
term. Divide 15y by y.
Multiply y – 3 by 15. Then
subtract.
Find the remainder.
6-3
Dividing Polynomials
Example 1 Continued
Step 4 Write the final answer.
–y2 + 2y3 + 25
y–3
Holt Algebra 2
=
2y2
70
+ 5y + 15 + y – 3
6-3
Dividing Polynomials
Check It Out! Example 1a
Divide using long division.
(15x2 + 8x – 12) ÷ (3x + 1)
Step 1 Write the dividend in standard form, including
terms with a coefficient of 0.
15x2 + 8x – 12
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
Step 3 Divide.
5x + 1
3x + 1 15x2 + 8x – 12
–(15x2 + 5x)
3x – 12
–(3x + 1)
–13
Holt Algebra 2
Notice that 3x times 5x is 15x2.
Write 5x above 15x2.
Multiply 3x + 1 by 5x. Then
subtract. Bring down the next
term. Divide 3x by 3x.
Multiply 3x + 1 by 1. Then
subtract.
Find the remainder.
6-3
Dividing Polynomials
Check It Out! Example 1a Continued
Step 4 Write the final answer.
15x2 + 8x – 12
13
= 5x + 1 – 3x + 1
3x + 1
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
Example 2B: Using Synthetic Division to Divide by a
Linear Binomial
Divide using synthetic division.
(3x4 – x3 + 5x – 1) ÷ (x + 2)
Step 1 Find a.
a = –2
For (x + 2), a = –2.
Step 2 Write the coefficients and a in the synthetic
division format.
–2 3 – 1
Holt Algebra 2
0
5 –1
Use 0 for the coefficient
of x2.
6-3
Dividing Polynomials
Example 2B Continued
Step 3 Bring down the first coefficient. Then
multiply and add for each column.
–2 3 –1
0
5 –1
–6 14 –28 46
3 –7 14 –23 45
Draw a box around the
remainder, 45.
Step 4 Write the quotient.
3x3
–
7x2
Holt Algebra 2
45
+ 14x – 23 +
x+2
Write the remainder over
the divisor.
6-3
Dividing Polynomials
Check It Out! Example 2a
Divide using synthetic division.
(6x2 – 5x – 6) ÷ (x + 3)
Step 1 Find a.
a = –3
For (x + 3), a = –3.
Step 2 Write the coefficients and a in the synthetic
division format.
–3 6 –5 –6
Holt Algebra 2
Write the coefficients of 6x2 – 5x – 6.
6-3
Dividing Polynomials
Check It Out! Example 2a Continued
Step 3 Bring down the first coefficient. Then
multiply and add for each column.
–3
6
–5
–6
6
–18 69
–23 63
Draw a box around the
remainder, 63.
Step 4 Write the quotient.
63
6x – 23 +
x+3
Holt Algebra 2
Write the remainder over
the divisor.
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.
2
2
5 –1
2
4
9
7
18 34
17 41
Write the coefficients of
the dividend. Use a = 2.
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 
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.
–3
1
3 0 4
–3 0 0
1 0 0 4
P(–3) = 4
Write the coefficients of
the dividend. Use 0 for
the coefficient of x2 Use a
= –3.
Check Substitute –3 for x in P(x) = x3 + 3x2 + 4.
P(–3) = (–3)3 + 3(–3)2 + 4
P(–3) = 4 
Holt Algebra 2
6-3
Dividing Polynomials
Lesson Quiz
1. Divide by using long division.
(8x3 + 6x2 + 7) ÷ (x + 2) 8x2 – 10x + 20 –
33
x+2
2. Divide by using synthetic division.
3
2
(x3 – 3x + 5) ÷ (x + 2)
x – 2x + 1 + x + 2
3. Use synthetic substitution to evaluate
194; –4
3
2
P(x) = x + 3x – 6 for x = 5 and x = –1.
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).
2x2 – 7x + 1
Holt Algebra 2