Applications of Synthetic Division of Polynomials

Q: How do you use synthetic division to factor a polynomial?

First, find one root k using the factor theorem (P(k) = 0). Then divide the polynomial by (x - k) using synthetic division. The quotient is a lower-degree polynomial. Repeat the process with the quotient until it can be factored completely.

Q: What is the factor theorem?

The factor theorem states that (x - k) is a factor of a polynomial P(x) if and only if P(k) = 0. In other words, k is a root of the polynomial.

Synthetic division is a fast and efficient method used to factor polynomials and simplify rational functions. In this lesson, you will see worked examples demonstrating these applications, followed by practice questions with detailed, step-by-step solutions.

You may also use this synthetic division calculator to check your results.

When can you use synthetic division? Synthetic division only works when the divisor is a linear expression of the form \( x - k \) (with a leading coefficient of 1). For divisors like \( 2x - 3 \) or quadratic expressions, use polynomial long division instead.

Worked Examples

Example 1: Factoring a Polynomial Completely

a) Show that \(x - 1\) and \(x + 3\) are factors of the polynomial: \[ P(x) = x^4 + 2x^3 - 7x^2 - 8x + 12 \] b) Use synthetic division to factor \(P(x)\) completely.

Solution to Example 1

Part a) Using the Factor Theorem:

For \(x - 1 = x - k\), we have \(k = 1\).

\[ P(1) = 1^4 + 2(1)^3 - 7(1)^2 - 8(1) + 12 = 1 + 2 - 7 - 8 + 12 = 0 \]

Since \( P(1) = 0 \), \(x - 1\) is a factor of \(P(x)\).

For \(x + 3 = x - k\), we have \(k = -3\).

\[ P(-3) = (-3)^4 + 2(-3)^3 - 7(-3)^2 - 8(-3) + 12 = 81 - 54 - 63 + 24 + 12 = 0 \]

Since \( P(-3) = 0 \), \(x + 3\) is also a factor of \(P(x)\).

Part b) Factoring completely using synthetic division:

Divide \(P(x)\) by \(x - 1\) using synthetic division.

Synthetic division of x^4 + 2x^3 - 7x^2 - 8x + 12 by x - 1

This division yields a quotient of:

\[ \frac{P(x)}{x - 1} = x^3 + 3x^2 - 4x - 12 \]

We can temporarily write \(P(x)\) as:

\[ P(x) = (x - 1)(x^3 + 3x^2 - 4x - 12) \]

Now divide the cubic quotient \( x^3 + 3x^2 - 4x - 12 \) by \(x + 3\).

Synthetic division of x^3 + 3x^2 - 4x - 12 by x + 3

This division yields a quotient of:

\[ \frac{x^3 + 3x^2 - 4x - 12}{x + 3} = x^2 - 4 \]

We now write \( P(x) \) in factored form as follows:

\[ P(x) = (x - 1)(x + 3)(x^2 - 4) \]

Using the difference of squares to factor \( x^2 - 4 \), we write \( P(x) \) in complete factored form:

\[ P(x) = (x - 1)(x + 3)(x - 2)(x + 2) \]

Example 2: Simplifying a Rational Function

A rational function of the form \(\frac{P(x)}{D(x)}\) can be simplified when the numerator and denominator share common factors.

a) Show that \(x - 2\) is a factor of both: \[ P(x) = x^3 + 2x^2 - x - 14 \quad \text{and} \quad D(x) = x^2 - 4 \] b) Simplify the rational function: \[ \frac{P(x)}{D(x)} = \frac{x^3 + 2x^2 - x - 14}{x^2 - 4} \]

Solution to Example 2

Part a) Test \( x = 2 \) using the factor theorem:

\[ P(2) = (2)^3 + 2(2)^2 - 2 - 14 = 8 + 8 - 2 - 14 = 0 \] \[ D(2) = (2)^2 - 4 = 4 - 4 = 0 \]

Because both evaluate to \( 0 \), \(x - 2\) is a common factor.

Part b) Use synthetic division to divide \( P(x) \) by \( x - 2 \):

Synthetic division of x^3 + 2x^2 - x - 14 by x - 2

From the division, we obtain the factored forms of the polynomials:

\[ P(x) = (x - 2)(x^2 + 4x + 7) \] \[ D(x) = x^2 - 4 = (x - 2)(x + 2) \]

Substitute these back into the rational function:

\[ \frac{P(x)}{D(x)} = \frac{(x - 2)(x^2 + 4x + 7)}{(x - 2)(x + 2)} \]

Cancel out the common \( (x - 2) \) factor:

\[ \frac{P(x)}{D(x)} = \frac{x^2 + 4x + 7}{x + 2} \]

Since \(x + 2\) is not a factor of the numerator (testing \( x = -2 \) gives \( (-2)^2 + 4(-2) + 7 = 3 \neq 0 \)), the rational function cannot be simplified further.

Domain restriction: The original denominator is zero when \( x = 2 \) or \( x = -2 \). The simplified form is valid for all \( x \neq -2 \), and the original function also excludes \( x = 2 \).

Practice Questions

Try each question on your own before opening the solution.

Part A: Identifying Factors

Which of the following are factors of the polynomial \( P(x) = x^4 - 4x^3 - 31x^2 + 70x \)?

\(x + 2\)
\(x - 7\)
\(x - 3\)
\(x + 5\)

Solution to Part A

For a linear term of the form \( x - k \) to be a factor of \( P(x) \), we need \( P(k) = 0 \).

a) Check if \( x + 2 \) is a factor.
Here, \( k = -2 \).
\[ P(-2) = (-2)^4 - 4(-2)^3 - 31(-2)^2 + 70(-2) = 16 + 32 - 124 - 140 = -216 \] Hence, \( x + 2 \) is not a factor of \( P(x) \).

b) Check if \( x - 7 \) is a factor.
Here, \( k = 7 \).
\[ P(7) = (7)^4 - 4(7)^3 - 31(7)^2 + 70(7) = 2401 - 1372 - 1519 + 490 = 0 \] Hence, \( x - 7 \) is a factor of \( P(x) \).

c) Check if \( x - 3 \) is a factor.
Here, \( k = 3 \).
\[ P(3) = (3)^4 - 4(3)^3 - 31(3)^2 + 70(3) = 81 - 108 - 279 + 210 = -96 \] Hence, \( x - 3 \) is not a factor of \( P(x) \).

d) Check if \( x + 5 \) is a factor.
Here, \( k = -5 \).
\[ P(-5) = (-5)^4 - 4(-5)^3 - 31(-5)^2 + 70(-5) = 625 + 500 - 775 - 350 = 0 \] Hence, \( x + 5 \) is a factor of \( P(x) \).

Part B: Finding Unknown Coefficients

Find \(r\) so that \(P_1(x) = x^3 + r x^2 - x - 4\) has a factor of \(x - 1\).
Find \(r\) and \(s\) so that \(P_2(x) = x^5 - r x^4 - 3x^3 + 3x + s\) has factors of \(x + 1\) and \(x - 2\).

Solution to Part B

a) For \( P_1(x) = x^3 + r x^2 - x - 4 \) to have \( x - 1 \) as a factor, according to the factor theorem we must have \( P_1(1) = 0 \).

\[ (1)^3 + r(1)^2 - (1) - 4 = 0 \] \[ 1 + r - 1 - 4 = 0 \] \[ r - 4 = 0 \implies r = 4 \]

b) For \( P_2(x) = x^5 - r x^4 - 3x^3 + 3x + s \) to have \( x + 1 \) and \( x - 2 \) as factors, we must have \( P_2(-1) = 0 \) and \( P_2(2) = 0 \). This sets up a system of two equations:

\[ (-1)^5 - r(-1)^4 - 3(-1)^3 + 3(-1) + s = 0 \] \[ (2)^5 - r(2)^4 - 3(2)^3 + 3(2) + s = 0 \]

Simplifying both equations gives:

\[ -1 - r + 3 - 3 + s = 0 \implies -r + s - 1 = 0 \] \[ 32 - 16r - 24 + 6 + s = 0 \implies -16r + s + 14 = 0 \]

Solve the above system of linear equations. From the first equation, \( s = r + 1 \). Substitute this into the second equation:

\[ -16r + (r + 1) + 14 = 0 \] \[ -15r + 15 = 0 \implies r = 1 \]

If \( r = 1 \), then \( s = 1 + 1 = 2 \). Therefore, \( r = 1 \) and \( s = 2 \).

Part C: Factoring Given a Known Factor

Factor \(P_1(x) = x^3 + 3x^2 - 10x - 24\), given the factor \(x + 4\).
Factor \(P_2(x) = x^4 - x^3 - 22x^2 + 16x + 96\), given the factors \(x + 2\) and \(x - 3\).

Solution to Part C

a) Use synthetic division to divide \( P_1(x) \) by \( (x + 4) \).

Synthetic division of x^3 + 3x^2 - 10x - 24 by x + 4

The result of the division gives the factored form:

\[ P_1(x) = (x^2 - x - 6)(x + 4) \]

Factor the quadratic term \( x^2 - x - 6 \) completely to finish factoring \( P_1(x) \):

\[ P_1(x) = (x - 3)(x + 2)(x + 4) \]

b) Use synthetic division to divide \( P_2(x) \) by \( (x + 2) \).

Synthetic division of x^4 - x^3 - 22x^2 + 16x + 96 by x + 2

This allows us to rewrite \( P_2(x) \) as:

\[ P_2(x) = (x^3 - 3x^2 - 16x + 48)(x + 2) \quad \text{--- (I)} \]

Now, use synthetic division to divide the cubic quotient \( x^3 - 3x^2 - 16x + 48 \) by the next given factor \( (x - 3) \).

Synthetic division of x^3 - 3x^2 - 16x + 48 by x - 3

This division factors the cubic polynomial into:

\[ x^3 - 3x^2 - 16x + 48 = (x^2 - 16)(x - 3) \]

Factor the difference of squares \( x^2 - 16 \) completely:

\[ (x - 4)(x + 4)(x - 3) \]

Substitute this back into equation (I) to find \( P_2(x) \) completely factored:

\[ P_2(x) = (x - 4)(x + 4)(x - 3)(x + 2) \]

Part D: Simplifying Rational Functions

Simplify \(F_1(x) = \frac{x^3 + 5x^2 - 7x - 35}{x^2 + 4x - 5}\), given a common factor of \(x + 5\).
Simplify \(F_2(x) = \frac{x^4 - 3x^3 - 16x^2 - 6x - 36}{x^2 - 3x - 18}\), given common factors of \(x - 6\) and \(x + 3\).

Solution to Part D

a) Use synthetic division to divide the numerator of \( F_1(x) \) by \( (x + 5) \).

Synthetic division of x^3 + 5x^2 - 7x - 35 by x + 5

Factor the numerator using the division result, and factor the quadratic denominator:

\[ F_1(x) = \frac{(x^2 - 7)(x + 5)}{(x - 1)(x + 5)} \]

Divide the numerator and denominator by \( (x + 5) \) to simplify:

\[ F_1(x) = \frac{x^2 - 7}{x - 1} \]

It is easy to check that \( x - 1 \) is not a factor of \( x^2 - 7 \), meaning we cannot simplify \( F_1(x) \) any further.

b) Use synthetic division to divide the numerator of \( F_2(x) \) by \( (x - 6) \).

Synthetic division of x^4 - 3x^3 - 16x^2 - 6x - 36 by x - 6

Rewrite the rational function using the factored numerator:

\[ F_2(x) = \frac{(x^3 + 3x^2 + 2x + 6)(x - 6)}{x^2 - 3x - 18} \]

Next, use synthetic division to divide the new cubic expression \( x^3 + 3x^2 + 2x + 6 \) by \( (x + 3) \).

Synthetic division of x^3 + 3x^2 + 2x + 6 by x + 3

Rewrite the factored numerator:

\[ F_2(x) = \frac{(x^2 + 2)(x + 3)(x - 6)}{x^2 - 3x - 18} \]

Factor the quadratic expression in the denominator \( x^2 - 3x - 18 \):

\[ F_2(x) = \frac{(x^2 + 2)(x + 3)(x - 6)}{(x + 3)(x - 6)} \]

Divide the numerator and the denominator by \( (x + 3)(x - 6) \) to completely simplify:

\[ F_2(x) = x^2 + 2 \]

Common Mistakes to Avoid

Forgetting to include zero coefficients for missing terms. For example, when dividing \( x^3 + 2x + 1 \) by \( x - 1 \), write the coefficients as \( 1,\; 0,\; 2,\; 1 \), not \( 1,\; 2,\; 1 \).
Using synthetic division with a non-linear divisor. Synthetic division only works for divisors of the form \( x - k \). For divisors like \( 2x - 3 \) or \( x^2 + 1 \), use polynomial long division.
Sign errors when the divisor is \( x + k \). Remember that \( x + 3 = x - (-3) \), so you use \( k = -3 \) in the synthetic division setup.
Forgetting to state domain restrictions. When simplifying a rational function, note the values of \( x \) that make the original denominator zero, even if they cancel in the simplification.
Stopping too early when factoring. After one synthetic division, the quotient may still be factorable. Keep factoring until you reach linear factors or irreducible quadratics.

Frequently Asked Questions

When can you use synthetic division?

Synthetic division can only be used when the divisor is a linear polynomial of the form \( x - k \), where the leading coefficient is 1. For divisors like \( 2x - 3 \) or quadratic expressions, use polynomial long division instead.

How do you use synthetic division to factor a polynomial?

First, find one root \( k \) using the factor theorem (\( P(k) = 0 \)). Then divide the polynomial by \( (x - k) \) using synthetic division. The quotient is a lower-degree polynomial. Repeat the process with the quotient until it can be factored completely.

What is the factor theorem?

The factor theorem states that \( (x - k) \) is a factor of a polynomial \( P(x) \) if and only if \( P(k) = 0 \). In other words, \( k \) is a root of the polynomial.

How do you simplify a rational function using synthetic division?

Factor the numerator and denominator completely. If they share a common factor, cancel it out. Synthetic division helps factor the numerator when you already know one of its roots. Remember to state any domain restrictions (values that make the original denominator zero).

What should you do if a term is missing when using synthetic division?

You must include a zero coefficient for any missing term. For example, when dividing \( x^3 + 2x + 1 \) by \( x - 1 \), write the coefficients as \( 1,\; 0,\; 2,\; 1 \) — the 0 holds the place for the missing \( x^2 \) term.