Synthetic Division of Polynomials with Examples

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The synthetic division of polynomials is presented along with the remainder and factor theorems. Examples and questions and their solutions are also included.
An online synthetic division calculator is included and may be used to check results.

Division Algorithm for Polynomials

The division of polynonomial \( P(x) \) by the polynomial \( D(x) \) may be written as follows
\( \dfrac{P(x)}{D(x)} = Q(x) + \dfrac{R(x)}{D(x)} \)
or
\( P(x) = Q(x) D(x) + R(x) \)
where the quotient \( Q(x) \) and the remainder \( R(x) \) are unique and the degree of \( R(x) \) is less than the degree of \( D(x) \).
If \( Q(x) \) is a factor of \( P(x) \), then \( R(x) =0 \)

Steps of Synthetic Division of Polynomials

The synthetic division may be applied when the divisor \( D(x) \) is a polynomial of the form \( x - k \). In this case the remainder \( R(x) \) of the division is a constant.
To divide the polynomial \( ax^2 + bx + c \) by the polynomial \( x - k \), we follow the steps:
Step 1
Set a table of synthetic division as shown below. The table includes the coefficients \( a, b \) and \( c \), in the order of decreasing power of the polynomial and the constant \( k \).

Starting table for synthetic divsion — Fig.1 - Starting Table for Synthetic Division

Step 2
comprises several substeps, shown in the diagram below, depending on the number of coefficients included in the polynomial.
Bring down coefficient \( a \) (substep 1).
Multiply \( a \) below the horizontal line by \( k \) and put the result \( ka \) above the horizontal line. (substep 2)
Add \( b \) and \( ka \) and put the result \( b + k a \) below the horizontal line. (substep 3)
Multiply \( b + k a \) below the horizontal line by \( k \) and put the result \( k(b + k a) \) above the horizontal line. (substep 4)
Add \( c \) and \( k(b + k a) \) and put the result \( c + k(b + k a) \) below the horizontal line. (substep 5)

Steps for synthetic division — Fig.2 - Steps in Synthetic Division

Step 3
When all substeps above are achieved, all coefficients on the left side and under the horizontal line, except the last one on the right, are the coefficient of the quotient. The constant on the right, under the horizontal line is the remainder.
The synthetic division uses coefficients and is may therefore easily be programmed in many of the computer languages available.

Example 1
Use synthetic division to divide \( 2x^2 + 6 x - 1 \) by \( x - 3 \) and show all steps.
Solution
Find the constant \( k \) by setting the divisor \( x - 3 = x - k \) and solve for k, which gives \( k = 3 \)
Dividing the \( 2x^2 + 6 x - 1 \) by a polynomial \( x - 3 \) is done following the steps:
Step (1): Set up the synthetic division by arranging the coefficients of the polynomial \( 2x^2 + 6 x - 1 \) and the constant \( k = 3 \) in an "L" shaped table with the coefficients inside the table and \( k \) outside the table on the left side as shown below

Step (1) for synthetic division
Step (2): Bring down the first coefficient under the horizontal of the table.

Step (2) for synthetic division Step (3): Multiply the coefficient under the horizontal by k = 3 and put the result under the next coefficient. Step (3) for synthetic division Step (4): Add the numbers that are in the same vertical. Step (4) for synthetic division Step (5): Multiply the coefficient under the horizontal by k = 3 and put the result under the next coefficient. Step (5) for synthetic division Step (6): Add the numbers that are in the same vertical line. Step (6) for synthetic division Step (7): Write the quotient and the remainder of the division. The coefficients of the quotient are the first numbers below the horizontal line on the left and the remainder on the right.
Quotient: \( Q(x) = 2 x + 12 \) , Remainder: \( R(x) = 35 \) Step (7) for synthetic division Hence
\( \dfrac{2x^2 + 6 x - 1}{x-3} = 2 x + 12 + \dfrac{35}{x - 3} \)

Example 2
Use synthetic division to divide \( -2x^5 + 6 x^3 - x^2 + 2 \) by \( x + 2 \).
Solution
Find the constant \( k \) by setting the divisor \( x + 2 = x - k \) and solve for k, which gives \( k = - 2 \)
Rewrite the polynomial with all powers of \( x \) starting from the highest power as follows:
\( -2x^5 + 6 x^3 - x^2 + 2 = -2x^5 + 0 x^4 + 6 x^3 - x^2 + 0 x + 2 \)
Dividing the \( -2x^5 + 6 x^3 - x^2 + 2 \) by a polynomial \( x + 2 \) is done following the steps:

Solution of synthetic division example 2
Quotient = \( - 2 x^4 + 4 x^3 - 2 x^2 + 3x - 6 \) , Remainder = \( 14 \)
Hence: \( \dfrac{-2x^5 + 6 x^3 - x^2 + 2}{x + 2 } = - 2 x^4 + 4 x^3 - 2 x^2 + 3x - 6 + \dfrac{14}{x + 2} \)

Remainder Theorem

Theorem: If polynomial \( P(x) \) is divided by \( x - k\), then the remainder of the division is equal to \( P(k) \).
Example 3
a) Use the remainder theorem to find the remainder of the polynomial division \( \dfrac{4x^4 - x^3 + x^2 - 1}{x + 1} \).

b) Use the synthetic division to find the remainder of the polynomial division \( \dfrac{4x^4 - x^3 + x^2 - 1}{x + 1} \) and compare with part a)

Solution
a)
Let: \( \dfrac{P(x)}{x - k} = \dfrac{4x^4 - x^3 + x^2 - 1}{x + 1} \)
Hence \( P(x) = 4x^4 - x^3 + x^2 - 1 \) and \( x - k = x + 1 \) which gives \( k = - 1\)
The remainder: \( R = P(-1) = 4(-1)^4 - (-1)^3 + (-1)^2 - 1 = 4 + 1 + 1 - 1 = 5 \)
b)
Use the synthetic division to obtain: Solution of synthetic division example 3
The remainder is equal to \( 5 \) and is equal to \( P(-1) \) as stated by the remainder theorem above.

Factor Theorem

Theorem: \( x - k\) is a factor of the polynomial \( P(x) \) if and only if \( P(k) = 0 \).
Example 4
a) Evaluate \( P(-2) \) given that \( P(x) = x^3 + x + 10 \).
b) Use the synthetic division to show that \( x + 2 \) is a factor of \( P(x) \)

Solution
a)
\( P(-2) = (-2)^3 + (-2) + 10 = - 8 - 2 + 10 = 0 \)
b)
Hence, according to the theorem above, \( x + 2 \) is a factor of \( P(x) \)
Use the synthetic to divide \( P(x) = x^3 + x + 10 \) by \( x + 2 \).
Solution of synthetic division example 4
Note the remainder is equal \( P(-2) = 0 \). From the results of the division, we write
\( \dfrac{x^3 + x + 10}{x+2} = x^2-2x+5 \)
or
\( x^3 + x + 10 = (x+2)(x^2-2x+5) \)
which shows that \( x + 2 \) is a factor of \( x^3 + x + 10 \) as predicted by the result in part a) and the factor theorem above.

.

Questions

(with solutions)

Part A
Divide the polynomials and find the quotient and the remainder
a) \( \dfrac{-4x^4 + 2 x^2 - x}{x+5} \)

b) \( \dfrac{x^5 - 5 x^2 - x + 4}{x-3} \)

Part B
Use remainder theorem and synthetic division to find \( P(k) \)
a) \( P(x) = -x^3 + 2 x^2 + 3x - 8 \quad \) , \( \quad k = 2 \)
b) \( P(x) = 0.2 x^2 + 0. 4 x + 0.5 \quad \) , \( \quad k = 0.3 \)
c) \( P(x) = x^3 - 2 x^2 + 9 \quad \) , \( \quad k = 1 + \sqrt 2 \)

Part C
Use the factor theorem to show that each of the given polynomial has the given factor.
a) \( P_1(x) = x^3+4x^2-11 x-30 \quad \) , Factor: \( (x + 5) \)
b) \( P_2(x) = x^4-x^3-22x^2+16x+96 \quad \) , Factors: \( (x +2) \) and \( (x - 3) \)
c) \( P_3(x) = x^3-0.4x^2-0.09 x + 0.036 \quad \) , Factor: \( (x - 0.4) \)
d) \( P_4(x) = x^3-(5+2\sqrt{3})x^2+(11+7\sqrt{3})x - 10 - 6\sqrt{3} \) , Factor: \( (x - 1 - \sqrt 3) \)

Solutions to the Above Questions

Part A
a) \( \quad \dfrac{-4x^4 + 2 x^2 - x}{x+5} \)
Rewrite numerator with all coefficients and its constant:
\( -4x^4 + 2 x^2 - x = -4x^4 + 0 x^3 + 2 x^2 - x + 0 \)
Determine \( k \) : \( x - k = x + 5 \) which gives \( k = - 5 \)
Set up the table and divide
Solution of synthetic division in question a) Part A
Quotient: \( Q(x) = -4x^3 + 20x^2 - 98 x + 489\)
Remainder: \( R = -2445 \)

b) \( \quad \dfrac{x^5 - 5 x^2 - x + 4}{x-3} \)
Rewrite numerator with all coefficients and its constant:
\( x^5 - 5 x^2 - x + 4 = x^5 + 0 x^4 + 0 x^3 - 5 x^2 - x + 4 \)
Determine \( k \) : \( x - k = x - 3 \) which gives \( k = 3\)
Set up the table and divide
Solution of synthetic division in question b) Part A
Quotient: \( Q(x) = x^4 + 3 x^3 + 9 x^2 + 22 x + 65\)
Remainder: \( R = 199 \)

Part B
a)
Synthetic division of \( \dfrac{P(x)}{x-2} = \dfrac{-x^3 + 2 x^2 + 3x - 8}{x-2} \) is as shown below
Solution of synthetic division in question a) Part B
According to the remainder theorem: \( P(2) = \text{remainder of division} = - 2 \)

b)
Synthetic division of \( \dfrac{P(x)}{x-0.3} = \dfrac{0.2 x^2 + 0. 4 x + 0.5}{x-0.3} \) is as shown below

According to the remainder theorem: \( P(0.3) = \text{remainder of division} = 0.638 \)

c)
Synthetic division of \( \dfrac{P(x)}{x-(1+\sqrt 2)} = \dfrac{ x^3 - 2 x^2 + 9 }{x-(1+\sqrt 2)} \) is as shown below
Solution of synthetic division in question c) Part B
According to the remainder theorem: \( P(1+\sqrt 2) = \text{remainder of division} = 10+\sqrt 2 \)

Part C
a)
Find \( k \) by setting the equation \( x + 5 = x - k \) , which gives \( k = - 5 \)
Given: \( P_1(x) = x^3+4x^2-11 x-30 \)
Hence \( P_1(-5) = (-5)^3+4(-5)^2-11(-5) -30 = -125 +100 +55-30 = 0 \)
According to the factor theorem \( x + 5 \) is a factor of \( P_1(x) \).

b)
\( x + 2 = x - k\) gives \( k = -2 \)
Given: \( P_2(x) = x^4-x^3-22x^2+16x+96 \)
Hence: \( P_2(-2) = (-2)^4-(-2)^3-22(-2)^2+16(-2)+96 = 16 + 8 - 88 - 32 + 96 = 0\)
According to the factor theorem \( x + 2 \) is a factor of \( P_2(x) \).

\( x - 3 = x - k \) gives \( k = 3 \)
\( P_2(3) = (3)^4-(3)^3-22(3)^2+16(3)+96 = 81 - 27 - 198 + 48 + 96 = 0 \)
According to the factor theorem \( x - 2 \) is also a factor of \( P_2(x) \).

c)
\( x - 0.4 = x - k \) gives \( k = 0.4\)
Given: \( P_3(x) = x^3-0.4x^2-0.09 x + 0.036 \)
\( P_3(0.4) = (0.4)^3-0.4(0.4)^2-0.09 (0.4) + 0.036 = 0\)
According to the factor theorem \( x - 0.4\) is also a factor of \( P_3(x) \).

d)
\( x - 1 - \sqrt 3 = x - k \) gives \( k = 1 + \sqrt 3\)
Given: \( P_4(x) = x^3-(5+2\sqrt{3})x^2+(11+7\sqrt{3})x - 10 - 6\sqrt{3} \)
\( P_4(1 + \sqrt 3) = (1 + \sqrt 3)^3-(5+2\sqrt{3})(1 + \sqrt 3)^2+(11+7\sqrt{3})(1 + \sqrt 3) - 10 - 6\sqrt{3} = 0 \)
According to the factor theorem \( x - 1 - \sqrt 3 \) is also a factor of \( P_4(x) \).