Synthetic Division of Polynomials with Examples

Synthetic division is a simplified method of dividing polynomials. This page also covers the Remainder Theorem and Factor Theorem, provides examples, and includes practice questions with solutions. An online synthetic division calculator is also available to check your results.

Division Algorithm for Polynomials

For polynomials \(P(x)\) and \(D(x)\) (with \(D(x) \neq 0\)), division can be expressed as:

\[ \frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)} \] or equivalently: \[ P(x) = Q(x) D(x) + R(x) \] where \(Q(x)\) is the quotient, \(R(x)\) is the remainder, and the degree of \(R(x)\) is less than the degree of \(D(x)\). If \(R(x) = 0\), then \(D(x)\) is a factor of \(P(x)\).

Steps of Synthetic Division

Synthetic division is applied when the divisor is of the form \(x - k\), in which case the remainder \(R(x)\) is a constant.

To divide a polynomial \(ax^2 + bx + c\) by \(x - k\):

Step 1: Create a synthetic division table with coefficients \(a, b, c\) in decreasing powers, and place \(k\) on the left.

Fig.1 – Starting Table for Synthetic Division
Step 2: Perform substeps:
1. Bring down the first coefficient \(a\) under the horizontal line.
2. Multiply this number by \(k\) and place above the horizontal line.
3. Add to the next coefficient and place the result below the line.
4. Repeat multiplication by \(k\) and addition until complete.
Fig.2 – Steps in Synthetic Division
Step 3: The numbers under the horizontal line (except the last) are the coefficients of the quotient, and the last number is the remainder.

Example 1

Divide \(2x^2 + 6x - 1\) by \(x-3\) using synthetic division.

Set \(k = 3\) (from \(x - 3 = x - k\)) and create the table:
Bring down the first coefficient:
Multiply by \(k\) and add for remaining coefficients:
Continue until all coefficients are processed:
Final quotient and remainder: \[ Q(x) = 2x + 12, \quad R(x) = 35 \] \[ \frac{2x^2 + 6x - 1}{x-3} = 2x + 12 + \frac{35}{x-3} \]

Remainder Theorem

Theorem: If a polynomial \(P(x)\) is divided by \(x - k\), the remainder is \(R = P(k)\).

Example 2

Find the remainder of \(\frac{4x^4 - x^3 + x^2 - 1}{x+1}\):

Using the theorem: \(k = -1\) \[ R = P(-1) = 4(-1)^4 - (-1)^3 + (-1)^2 - 1 = 5 \]
Using synthetic division:

Remainder = 5 (matches theorem)

Factor Theorem

Theorem: \(x - k\) is a factor of \(P(x)\) if and only if \(P(k) = 0\).

Example 3

Verify \(x+2\) is a factor of \(P(x) = x^3 + x + 10\):

Evaluate \(P(-2) = (-2)^3 + (-2) + 10 = 0\)
Use synthetic division:

\[ \frac{x^3 + x + 10}{x+2} = x^2 - 2x + 5, \quad x^3 + x + 10 = (x+2)(x^2-2x+5) \]

Practice Questions

Solutions included below

Divide and find quotient and remainder:
1. \( \frac{-4x^4 + 2x^2 - x}{x+5} \)
2. \( \frac{x^5 - 5x^2 - x + 4}{x-3} \)
Use the remainder theorem to find \(P(k)\):
1. \(P(x) = -x^3 + 2x^2 + 3x - 8, k = 2\)
2. \(P(x) = 0.2x^2 + 0.4x + 0.5, k = 0.3\)
3. \(P(x) = x^3 - 2x^2 + 9, k = 1+\sqrt{2}\)
Use the factor theorem:
1. \(P_1(x) = x^3 + 4x^2 - 11x - 30, \text{factor } x+5\)
2. \(P_2(x) = x^4 - x^3 -22x^2 + 16x + 96, \text{factors } x+2, x-3\)
3. \(P_3(x) = x^3 -0.4x^2 -0.09x + 0.036, \text{factor } x-0.4\)
4. \(P_4(x) = x^3-(5+2\sqrt3)x^2 + (11+7\sqrt3)x - 10 - 6\sqrt3, \text{factor } x-1-\sqrt3\)

Solutions to the Above Questions on Synthetic Division

Part A: Division of Polynomials

a) Divide \(\dfrac{-4x^4 + 2 x^2 - x}{x+5}\)

Rewrite numerator with all coefficients: \[ -4x^4 + 2 x^2 - x = -4x^4 + 0 x^3 + 2 x^2 - x + 0 \] Determine \( k \): \( x - k = x + 5 \implies k = -5 \) Set up the synthetic division table and divide:

Synthetic division solution for -4x^4 + 2x^2 - x divided by x+5

Quotient: \( Q(x) = -4x^3 + 20x^2 - 98 x + 489\)
Remainder: \( R = -2445 \)

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

Rewrite numerator with all coefficients: \[ x^5 - 5 x^2 - x + 4 = x^5 + 0 x^4 + 0 x^3 - 5 x^2 - x + 4 \] Determine \( k \): \( x - k = x - 3 \implies k = 3 \) Set up the synthetic division table and divide:

Synthetic division solution for x^5 - 5x^2 - x + 4 divided by x-3

Quotient: \( Q(x) = x^4 + 3 x^3 + 9 x^2 + 22 x + 65\)
Remainder: \( R = 199 \)

Part B: Using the Remainder Theorem

a) Evaluate \( P(2) \) for \( P(x) = -x^3 + 2 x^2 + 3x - 8 \)

Synthetic division: \[ \dfrac{P(x)}{x-2} = \dfrac{-x^3 + 2 x^2 + 3x - 8}{x-2} \] Synthetic division solution for P(x)/x-2

Remainder according to the remainder theorem: \( P(2) = -2 \)

b) Evaluate \( P(0.3) \) for \( P(x) = 0.2 x^2 + 0.4 x + 0.5 \)

Synthetic division: \[ \dfrac{P(x)}{x-0.3} = \dfrac{0.2 x^2 + 0.4 x + 0.5}{x-0.3} \] Synthetic division solution for P(x)/x-0.3

Remainder: \( P(0.3) = 0.638 \)

c) Evaluate \( P(1+\sqrt{2}) \) for \( P(x) = x^3 - 2 x^2 + 9 \)

Synthetic division: \[ \dfrac{P(x)}{x-(1+\sqrt{2})} = \dfrac{x^3 - 2 x^2 + 9}{x-(1+\sqrt{2})} \] Synthetic division solution for P(x)/x-(1+sqrt2)

Remainder: \( P(1+\sqrt 2) = 10+\sqrt 2 \)

Part C: Using the Factor Theorem

a) Verify \( x + 5 \) is a factor of \( P_1(x) = x^3+4x^2-11 x-30 \)

Find \( k \): \( x + 5 = x - k \implies k = -5 \) \[ P_1(-5) = (-5)^3 + 4(-5)^2 - 11(-5) - 30 = 0 \] Hence, \( x + 5 \) is a factor of \( P_1(x) \)

b) Verify \( x + 2 \) and \( x - 3 \) are factors of \( P_2(x) = x^4-x^3-22x^2+16x+96 \)

Check \( x + 2 \) (\( k = -2 \)): \[ P_2(-2) = 16 + 8 - 88 - 32 + 96 = 0 \] Check \( x - 3 \) (\( k = 3 \)): \[ P_2(3) = 81 - 27 - 198 + 48 + 96 = 0 \] Hence, both \( x + 2 \) and \( x - 3 \) are factors.

c) Verify \( x - 0.4 \) is a factor of \( P_3(x) = x^3-0.4x^2-0.09 x + 0.036 \)

\[ P_3(0.4) = (0.4)^3 - 0.4(0.4)^2 - 0.09(0.4) + 0.036 = 0 \] Hence, \( x - 0.4 \) is a factor.

d) Verify \( x - 1 - \sqrt{3} \) is a factor of \( P_4(x) = x^3-(5+2\sqrt{3})x^2+(11+7\sqrt{3})x - 10 - 6\sqrt{3} \)

\[ P_4(1 + \sqrt 3) = (1 + \sqrt 3)^3-(5+2\sqrt{3})(1 + \sqrt 3)^2+(11+7\sqrt{3})(1 + \sqrt 3) - 10 - 6\sqrt{3} = 0 \] Hence, \( x - 1 - \sqrt 3 \) is a factor.