1. Dividing Polynomials 6-3 Warm Up Lesson Presentation Lesson Quiz
Holt Algebra 2

2. 2. Use Pascal’s Triangle to expand the expression. (y – 5)4
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 – 20y y2 – 500y + 625
3. Use Pascal’s Triangle to expand the expression.
(3a – b)3
27a3 – 27a2b + 9ab2 – b3

3. Objective
Use long division and synthetic division to divide polynomials.

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

5. 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

6. Example 1 Continued
Step 3 Divide.
2y2
+ 5y
+ 15
Notice that y times 2y2 is 2y3. Write 2y2 above 2y3.
y – 3 2y3 – y2 + 0y + 25
–(2y3 – 6y2)
Multiply y – 3 by 2y2. Then subtract. Bring down the next term. Divide 5y2 by y.
5y2 + 0y
–(5y2 – 15y)
Multiply y – 3 by 5y. Then subtract. Bring down the next term. Divide 15y by y.
15y + 25
Multiply y – 3 by 15. Then subtract.
–(15y – 45) 70
Find the remainder.

7. Example 1 Continued
Step 4 Write the final answer.
–y2 + 2y3 + 25
y – 3
= 2y2 + 5y 70
y – 3

8. 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 x2 + 8x – 12

9. Check It Out! Example 1a Continued
Step 3 Divide. 5x
+ 1
Notice that 3x times 5x is 15x2. Write 5x above 15x2.
3x x2 + 8x – 12
–(15x2 + 5x)
Multiply 3x + 1 by 5x. Then subtract. Bring down the next term. Divide 3x by 3x.
3x – 12
–(3x + 1)
Multiply 3x + 1 by 1. Then subtract.
–13
Find the remainder.

10. Check It Out! Example 1a Continued
Step 4 Write the final answer.
15x2 + 8x – 12
3x + 1
= 5x + 1 – 13
3x + 1

11. 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).

13. 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.
3 – –1 –2
Use 0 for the coefficient of x2.

14. Example 2B Continued
Step 3 Bring down the first coefficient. Then multiply and add for each column. –2
3 – –1
Draw a box around the remainder, 45. –6 14
–28 46 3 –7 14
–23 45
Step 4 Write the quotient.
3x3 – 7x2 + 14x – 23 + 45
x + 2
Write the remainder over the divisor.

15. 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
Write the coefficients of 6x2 – 5x – 6.

16. Check It Out! Example 2a Continued
Step 3 Bring down the first coefficient. Then multiply and add for each column. –3
6 –5 –6
Draw a box around the remainder, 63.
–18 69 6
–23 63
Step 4 Write the quotient.
6x – 23 + 63
x + 3
Write the remainder over the divisor.

17. You can use synthetic division to evaluate 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.

18. 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
–1 7
Write the coefficients of the dividend. Use a = 2. 4 18 34 2 9 17 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 

19. 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
Write the coefficients of the dividend. Use 0 for the coefficient of x2 Use a = –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 

20. 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. (x3 – 3x + 5) ÷ (x + 2)
x2 – 2x + 1 + 3
x + 2
3. Use synthetic substitution to evaluate P(x) = x3 + 3x2 – 6 for x = 5 and x = –1.
194; –4
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