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.

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 \( R(x) =0 \)

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 \( k \).

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)

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 (2): Bring down the first coefficient under the horizontal of the table.

Step (3): Multiply the coefficient under the horizontal by k = 3 and put the result under the next coefficient.
Step (4): Add the numbers that are in the same vertical.
Step (5): Multiply the coefficient under the horizontal by k = 3 and put the result under the next coefficient.
Step (6): Add the numbers that are in the same vertical line.
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: \( R(x) = 35 \)
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:

Quotient = \( - 2 x^4 + 4 x^3 - 2 x^2 + 3x - 6 \) , Remainder = \( 14 \)

Hence: \( \dfrac{-2x^5 + 6 x^3 - x^2 + 2}{x + 2 } = - 2 x^4 + 4 x^3 - 2 x^2 + 3x - 6 + \dfrac{14}{x + 2} \)

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:

The remainder is equal to \( 5 \) and is equal to \( P(-1) \) as stated by the remainder theorem above.

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 \( x + 2 \).

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.

.

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: \( (x - 1 - \sqrt 3) \)

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

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

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

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

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 \( P_4(x) \).

Synthetic Division Calculator

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