Applications of Synthetic Division of Polynomials

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

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

Factor Polynomials Using Synthetic Division

Example 1

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

Solution

a) Using the factor theorem:

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

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

Hence, \(x - 1\) is a factor of \(P(x)\).

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

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

Hence, \(x + 3\) is also a factor of \(P(x)\).

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

Synthetic division of P(x) by x-1
\[ \frac{P(x)}{x - 1} = x^3 + 3x^2 - 4x - 12. \]
\[ P(x) = (x - 1)(x^3 + 3x^2 - 4x - 12). \]

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

Synthetic division of cubic by x+3
\[ \frac{x^3 + 3x^2 - 4x - 12}{x + 3} = x^2 - 4. \]
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 \( x^2 - 4 \) and write \( P(x) \) in complete factored form:

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

Simplify Rational Functions Using Synthetic Division

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

Example 2

  1. Show that \(x - 2\) is a factor of
    \[ P(x) = x^3 + 2x^2 - x - 14 \quad \text{and} \quad D(x) = x^2 - 4. \]
  2. Simplify
    \[ \frac{P(x)}{D(x)} = \frac{x^3 + 2x^2 - x - 14}{x^2 - 4}. \]

Solution

a)

\[ P(2) = 8 + 8 - 2 - 14 = 0, \quad D(2) = 4 - 4 = 0. \]

Thus, \(x - 2\) is a common factor.

b)

Use synthetic division to divide \( P(x) \) by \( D(x) \): Synthetic division example
\[ P(x) = (x - 2)(x^2 + 4x + 7), \quad x^2 - 4 = (x - 2)(x + 2). \]
\[ \frac{P(x)}{D(x)} = \frac{(x - 2)(x^2 + 4x + 7)}{(x - 2)(x + 2)}. \]
\[ \frac{P(x)}{D(x)} = \frac{x^2 + 4x + 7}{x + 2}. \]

Since \(x + 2\) is not a factor of the numerator, the rational function cannot be simplified further.

More Questions

Part A

Which of the following are factors of

\[ P(x) = x^4 - 4x^3 - 31x^2 + 70x? \]
  1. \(x + 2\)
  2. \(x - 7\)
  3. \(x - 3\)
  4. \(x + 5\)
Part B
  1. Find \(r\) so that \(P_1(x) = x^3 + r x^2 - x - 4\) has factor \(x - 1\).
  2. Find \(r\) and \(s\) so that \(P_2(x) = x^5 - r x^4 - 3x^3 + 3x + s\) has factors \(x + 1\) and \(x - 2\).
Part C
  1. Factor \(P_1(x) = x^3 + 3x^2 - 10x - 24\), given factor \(x + 4\).
  2. Factor \(P_2(x) = x^4 - x^3 - 22x^2 + 16x + 96\), given factors \(x + 2\) and \(x - 3\).
Part D
  1. Simplify \(F_1(x) = \frac{x^3 + 5x^2 - 7x - 35}{x^2 + 4x - 5}\), given common factor \(x + 5\).
  2. Simplify \(F_2(x) = \frac{x^4 - 3x^3 - 16x^2 - 6x - 36}{x^2 - 3x - 18}\), given common factors \(x - 6\) and \(x + 3\).

Solutions to the Above Questions

Part A
Given \( P(x) = x^4-4x^3-31x^2+70x \),
For a linear term of the form \( x - k \) to be a factor of \( P(x) \), we need to have \( P(k) = 0 \)
a)
check if \( x + 2 \) is a factor of \( P(x) \).
\( x + 2 = x - k \) gives \( k = - 2 \) \[ P(-2) = (-2)^4-4(-2)^3-31(-2)^2+70(-2) = -126 \] Hence \( x + 2 \) IS NOT a factor of \( P(x) \)

b)
check if \( x - 7 \) is a factor of \( P(x) \).
\( x - 7 = x - k \) gives \( k = 7 \) \[ P(7) = (7)^4-4(7)^3-31(7)^2+70(7) = 0\] Hence \( x - 7 \) IS a factor of \( P(x) \)

c)
check if \( x - 3 \) is a factor of \( P(x) \).
\( x - 3 = x - k \) gives \( k = 3 \) \[ P(3) = (3)^4-4(3)^3-31(3)^2+70(3) = -96\] Hence \( x - 3\) IS NOT a factor of \( P(x) \)

d)
check if \( x + 5 \) is a factor of \( P(x) \).
\( x + 5= x - k \) gives \( k = - 5 \) \[ P(-5) = (-5)^4-4(-5)^3-31(-5)^2+70(-5) = 0\] Hence \( x + 5\) IS a factor of \( P(x) \)

Part B
a)
For the polynomial \( P_1(x) = x^3 + r x^2 - x - 4 \) to have \( x - 1 \) as a factor and according to the factor theorem we need to have \( P_1(1) = 0 \)
Hence the equation \[ (1)^3 + r (1)^2 - (1) - 4 = 0 \] Simplify
\( r - 4 = 0 \)
\( r = 4 \)

b)
For the polynomial \( P_2(x) = x^5- r x^4-3x^3+3x + s \) to have \( (x + 1) \) and \( x - 2 \) as factors and according to the factor theorem we need to have \( P_2(-1) = 0 \) and \( P_2(2) = 0 \).
Hence the system of two equations \[ (-1)^5- r (-1)^4-3(-1)^3+3(-1) + s = 0 \quad \text{and} \quad (2)^5- r (2)^4-3(2)^3+3(2) + s = 0 \] Simplify \[ - r + s - 1 = 0 \quad \text{and} \quad - 16 r + s + 14 = 0 \] Solve the above system of linear equations to obtain: \( r = 1\) and \( s = 2\)

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

Solution of synthetic division applications question a Part C
The result of the division may be written as a multiplication, hence factoring \( P_1(x) \) as follows \[ P_1(x) = (x^2-x-6)(x+4)\] Factor the quadratic term \( x^2-x-6 \) and hence completely factor \( P_1(x) \) as follows \[ P_1(x) = (x^2-x-6)(x+4) = (x-3)(x+2)(x+4)\] b)
Use synthetic division to divide \( P_2(x) \) by \( (x + 2) \)

Solution of synthetic division applications question b1 Part C

The result of the division may be written as a multiplication, hence factoring \( P_2(x) \) as follows \[ P_2(x) = (x^3-3x^2-16x+48)(x+2) \quad \quad (I)\] Use synthetic division to divide the quotient in the above division \( \quad x^3-3x^2-16x+48 \quad \) by \( (x - 3) \)
Solution of synthetic division applications question b2 Part C

The result of the division may be written as a multiplication, hence factoring \( x^3-3x^2-16x+48 \) as follows \[ x^3-3x^2-16x+48 = (x^2-16)(x-3) \] factor completely the above and write as \[ x^3-3x^2-16x+48 = (x-4)(x+4)(x-3) \] Substitute the above in \( (I) \) and factor \( P_2(x) \) completely as follows \[ P_2(x) = (x-4)(x+4)(x-3)(x+2) \]



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

Solution of synthetic division applications question a Part D
Factor the numerator, using the result of the above synthetic division, and denominator, which is a quadratic expression, of \( F_1(x) \). \[ F_1(x) = \dfrac{(x^2-7)(x+5)}{(x -1)(x+5)} \] Divide numerator and the denominator by \( (x + 5) \) and simplify \[ F_1(x) = \dfrac{x^2-7}{x -1} \] It is easy to check that \( x - 1 \) is not a factor of \( x^2-7 \) and therefore we cannot further simplify \( F_1(x) \).

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

Solution of synthetic division applications question b1 Part D
hence factor the numerator \[ F_2(x) = \dfrac{(x^3+3x^2+2x+6)(x-6)}{x^2-3x-18} \] Use synthetic division to divide \( x^3+3x^2+2x+6 \) in the numerator above by \( (x + 6) \)
Solution of synthetic division applications question b2 Part D
hence factor the numerator \[ F_2(x) = \dfrac{(x^2+2)(x+3)(x-6)}{x^2-3x-18} \] Factor the quadratic expression \( x^2-3x-18\) \[ F_2(x) = \dfrac{(x^2+2)(x+3)(x-6)}{(x+3)(x-6)} \] Divide numerator and the denominator by \( (x+3)(x-6) \) and simplify \[ F_2(x) = x^2+2 \]

More references and links to polynomial functions