﻿ Synthetic Division Worksheet | Problems & Solutions

# Synthetic Division Worksheet

Synthetic Division Worksheet
• Page 1
1.
($\frac{dividend}{divisor}$) = quotient + ($\frac{remainder}{divisor}$).
($x$3 - 9$x$2 - 5$x$ + 4) ÷ ($x$ - 3)

 a. ($x$2 - 6$x$ - 23) - b. ($x$2 - 12$x$ + 23) - c. ($x$2 - 6$x$ - 23) + d. ($x$2 - 6$x$ - 23) -

#### Solution:

x3 - 9x2 - 5x + 4

x - 3 is the divisor.

Use the synthetic division.

3|     1     - 9       - 5       4
3     - 18    - 69
-----------------------------
1       - 6     - 23    - 65
-----------------------------

x3-9x2-5x+4x - 3 = (x2 - 6x - 23) - 65x - 3

2.
($\frac{dividend}{divisor}$) = quotient + ($\frac{remainder}{divisor}$).
(4$x$3 - 15$x$ + 17) ÷ ($x$ + 8)

 a. (4$x$2 - 32$x$ + 241) + b. (4$x$2 - 32$x$ + 241) - c. (4$x$2 + 32$x$ + 241) + d. none of these

#### Solution:

4x3 - 15x + 17

x + 8 is the divisor.

Use the synthetic division.

- 8|     4     0       - 15      17
- 32      256      - 1928
-------------------------------
4    - 32       241     - 1911
-------------------------------
[The x2 term is missing, so insert 0 for 0x2.]

4x3 - 15x + 17x + 8 = (4x2 - 32x + 241) - 1911x + 8

3.
$\frac{dividend}{divisor}$ = quotient + $\frac{remainder}{divisor}$.
($x$4 - 14$x$2 + 15$x$ - 16) ÷ ($x$ + 3)

 a. ($x$3 + 3$x$2 - 5$x$ + 0) - b. ($x$3 - 3$x$2 + 5$x$ - 30) - c. ($x$3 - 3$x$2 - 5$x$ + 30) - d. none of these

#### Solution:

x4 - 14x2 + 15x - 16

x + 3 is the divisor.

Use the synthetic division.

- 3|     1    0     - 14      15    - 16
- 3      9       15    - 90
-------------------------------
1   - 3      - 5    30    - 106
-------------------------------
[The x3 term is missing, so insert 0 for 0x3.]

x4 - 14x2 + 15x - 16x + 3 = (x3 - 3x2 - 5x + 30) - 106x + 3

4.
($\frac{dividend}{divisor}$) = quotient + ($\frac{remainder}{divisor}$).
($x$4 - 4$x$3 + 4$x$2 - 2$x$ + 3) ÷ ($x$ - 2)

 a. ($x$3 - 2$x$2 - 2) + b. ($x$3 - 2$x$2 - 2) - c. ($x$3 - 6$x$2 + 4$x$ - 2) + d. ($x$3 + 2$x$2 + 2) -

#### Solution:

x4 - 4x3 + 4x2 - 2x + 3

x - 2 is the divisor.

Use the synthetic division.

2|     1   - 4     4     - 2     3
2    - 4      0    - 4
--------------------------
1   - 2     0    - 2     - 1
--------------------------

x4 - 4x3 + 4x2 - 2x + 3x - 2 = (x3 - 2x2 - 2) - 1x - 2

5.
($\frac{dividend}{divisor}$) = quotient + ($\frac{remainder}{divisor}$).
($x$4 - $x$3 - 5$x$2 + 4$x$ - 6) ÷ ($x$ + 8)
 a. ($x$3 - 9$x$2 + 67$x$ + 532) + b. ($x$3 + 5$x$2 + 77$x$ + 540) + c. ($x$3 - 9$x$2 + 67$x$ - 532) + d. none of these

#### Solution:

x4 - x3 - 5x2 + 4x - 6

x + 8 is the divisor.

Use the synthetic division.

- 8|     1    -1     - 5        4       - 6
- 8     72   - 536    4256
----------------------------------
1    - 9     67   - 532    4250
---------------------------------

x4-x3-5x2+4x - 6x + 8 = (x3 - 9x2 + 67x - 532) + 4250x + 8

6.
($\frac{dividend}{divisor}$) = quotient + ($\frac{remainder}{divisor}$).
($y$4 - 32) ÷ ($y$ + 2)
 a. ($y$3 - 2$y$2 + 4$y$ - 8) - b. ($y$3 + 2$y$2 + 4$y$ + 8) - c. ($y$3 - 2$y$2 + 4$y$ - 8) + d. none of these

#### Solution:

y4 - 32

y + 2 is the divisor.

Use the synthetic division.

- 2|     1    0     0       0    - 32
- 2     4   - 8     16
----------------------------
1   - 2     4   - 8    - 16
----------------------------
[The x3, x2 and x terms are missing, so insert 0 for 0x3, 0x2 and 0x.]

y4 - 32y + 2 = (y3 - 2y2 + 4y - 8) - 16y + 2.

7.
Find the value of $c$ that will make the given divisor a factor of the dividend.

 a. 9 b. 15 c. 12

#### Solution:

x3 - 4x2 - x + c

x - 3 is the divisor.

A divisor of a polynomial is called a factor, when it divides the polynomial without leaving a remainder

Use the synthetic division

3|     1     - 4     -1     c
3     - 3     - 12
-----------------------
1     - 1    - 4      0
-----------------------

c - 12 = 0
[Equate remainder to zero.]

c = 12
[Add 12 to both sides of the equation.]

8.
Find the value of $d$ that will make the given divisor a factor of the dividend.

 a. 16 b. - 80 c. 160

#### Solution:

4x3 + 4x2 - 76x - d

x - 4 is the divisor.

A divisor of a polynomial is called a factor, when it divides the polynomial without leaving a remainder.

Use the synthetic division.

4|     4     4   - 76   - d
16   80     16
-----------------------
4    20     4   0
-----------------------

- d + 16 = 0
[Equate remainder to zero.]

- d = - 16
[Subtracting 16 from the two sides of the equation.]

d = 16
[Multiply throughout by - 1.]

9.
($\frac{dividend}{divisor}$) = quotient + ($\frac{remainder}{divisor}$).
($x$3 + 125) ÷ ($x$ - 5)

 a. ($x$2 + 5$x$ + 25) - b. ($x$2 + 5$x$ + 25) + c. ($x$2 - 5$x$ + 25) - d. ($x$2 - 5$x$ + 25)

#### Solution:

x3 + 125

x - 5 is the divisor.

Use the synthetic division.

5|     1     0      0      125
5      25      125
------------------------
1     5      25    250
-----------------------
[The x2, x terms are missing, so insert 0 for 0x2, 0x.]

x3 + 125x - 5 = (x2 + 5x + 25) + 250x - 5

10.
$\frac{dividend}{divisor}$ = quotient + $\frac{remainder}{divisor}$.
(9$x$3 + 9$x$2 + 11$x$ - 4) ÷ (3$x$ + 1)
 a. 9$x$2 +6$x$ + 9 + b. 3$x$2 +2$x$ + 3 + c. 9$x$2 +6$x$ + 9 - d. 3$x$2 +2$x$ + 3 -

#### Solution:

9x3 + 9x2 + 11x - 43x + 1 = 1 / 3(9x3 + 9x2 +11x - 4 x +13)

9x3 + 9x2 + 11x - 4

x + 1 / 3 is the divisor.

Use the synthetic division.

- 1/3|     9     9     11    -4
-3      -2     -3
-------------------------
9     6      9     -7
-------------------------

Then 9x3 + 9x2 + 11x - 43x + 1 = 1 / 3(9x2 + 6x + 9 -  7 x +13)

= 3x2 + 2x + 3 - 7 3x + 1

So, 9x3 + 9x2 + 11x - 43x + 1 = 3x2 + 2x + 3 - 7 3x + 1