Factoring Higher-Degree Polynomials Notes

Published on: September 29, 2026

This Factoring Higher Degree Polynomials Calculator breaks a polynomial of degree three or more into a product of simpler factors, and shows the working line by line. Any rational root has to divide the constant term over the leading coefficient, which turns an endless search into a short list to test. Each root found peels off one linear factor and leaves a smaller polynomial behind. Following the same pattern every time keeps the degree dropping one factor at a time.

Step-by-step method

  1. Set up the problem.
  2. List the possible rational roots.
  3. Test the possible roots until one gives zero.
  4. Divide the polynomial by the factor that root gives.
  5. Factor whatever is left.
  6. Write the polynomial as a product of its factors.
See Example 1 Hide Example 1

Problem

\(x^4 - 1\)

Approach

Step-by-step method

  1. Set up the problem.
  2. List the possible rational roots.
  3. Test the possible roots until one gives zero.
  4. Divide the polynomial by the factor that root gives.
  5. Factor whatever is left.
  6. Write the polynomial as a product of its factors.

Step 1

Step 1 - Set up the problem.

In this problem: The polynomial to factor is \(x^4 - 1\), and factoring means rewriting it as a product of simpler polynomials.

\(x^4 - 1\)

Step 2

Step 2 - List the possible rational roots.

In this problem: Any rational root has to divide the constant term \(−1\) over the leading coefficient \(1\), which leaves \(−1, 1\) to try.

\(-1, 1\)

Step 3

Step 3 - Test the possible roots until one gives zero.

In this problem: Putting \(x = −1\) into the polynomial gives \(0\), so \(x = −1\) is a root.

\(f(-1) = (-1)^4 -1 = 0\)

Step 4

Step 4 - Divide the polynomial by the factor that root gives.

In this problem: A root at \(x = −1\) means \((x + 1)\) divides the polynomial exactly, and each division leaves a smaller one behind.

\(x^4 - 1\)

Step 4a

Step 4a - Divide the polynomial by the factor that root gives.

In this problem: Dividing \(x^4 - 1\) by \((x + 1)\) leaves \(x^3 - x^2 + x - 1\).

\(x^4 - 1 \div (x + 1) = x^3 - x^2 + x - 1\)

Step 4b

Step 4b - Divide the polynomial by the factor that root gives.

In this problem: Dividing \(x^3 - x^2 + x - 1\) by \((x - 1)\) leaves \(x^2 + 1\).

\(x^3 - x^2 + x - 1 \div (x - 1) = x^2 + 1\)

Step 5

Step 5 - Factor whatever is left.

In this problem: Factoring \(x^2 + 1\) gives \(x^2 + 1\).

\(x^2 + 1 = x^2 + 1\)

Step 6

Step 6 - Write the polynomial as a product of its factors.

In this problem: Multiplying the factors back together reproduces \(x^4 - 1\), so the factored form is \((x + 1)(x - 1)(x^2 + 1)\).

\(x^4 - 1 = (x + 1)(x - 1)(x^2 + 1)\)

Final Answer

\((x + 1)(x - 1)(x^2 + 1)\)
See Example 2 Hide Example 2

Problem

\(x^3 - 8\)

Approach

Step-by-step method

  1. Set up the problem.
  2. List the possible rational roots.
  3. Test the possible roots until one gives zero.
  4. Divide the polynomial by the factor that root gives.
  5. Factor whatever is left.
  6. Write the polynomial as a product of its factors.

Step 1

Step 1 - Set up the problem.

In this problem: The polynomial to factor is \(x^3 - 8\), and factoring means rewriting it as a product of simpler polynomials.

\(x^3 - 8\)

Step 2

Step 2 - List the possible rational roots.

In this problem: Any rational root has to divide the constant term \(−8\) over the leading coefficient \(1\), which leaves \(−1, 1, −2, 2, −4, 4, −8, 8\) to try.

\(-1, 1, -2, 2, -4, 4, -8, 8\)

Step 3

Step 3 - Test the possible roots until one gives zero.

In this problem: Each candidate is put into \(x^3 - 8\) in turn, and the first one that returns \(0\) is a root.

\(x^3 - 8\)

Step 3a

Step 3a - Test the possible roots until one gives zero.

In this problem: Putting \(x = −1\) in gives \(−9\), which is not \(0\).

\(f(-1) = (-1)^3 -8 = -9\)

Step 3b

Step 3b - Test the possible roots until one gives zero.

In this problem: Putting \(x = 1\) in gives \(−7\), which is not \(0\).

\(f(1) = (1)^3 -8 = -7\)

Step 3c

Step 3c - Test the possible roots until one gives zero.

In this problem: Putting \(x = −2\) in gives \(−16\), which is not \(0\).

\(f(-2) = (-2)^3 -8 = -16\)

Step 3d

Step 3d - Test the possible roots until one gives zero.

In this problem: Putting \(x = 2\) in gives \(0\), so \(x = 2\) is a root.

\(f(2) = (2)^3 -8 = 0\)

Step 4

Step 4 - Divide the polynomial by the factor that root gives.

In this problem: A root at \(x = 2\) means \((x - 2)\) divides the polynomial exactly, and dividing leaves \(x^2 + 2x + 4\).

\(x^3 - 8 \div (x - 2) = x^2 + 2x + 4\)

Step 5

Step 5 - Factor whatever is left.

In this problem: Factoring \(x^2 + 2x + 4\) gives \(x^2 + 2x + 4\).

\(x^2 + 2x + 4 = x^2 + 2x + 4\)

Step 6

Step 6 - Write the polynomial as a product of its factors.

In this problem: Multiplying the factors back together reproduces \(x^3 - 8\), so the factored form is \((x - 2)(x^2 + 2x + 4)\).

\(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\)

Final Answer

\((x - 2)(x^2 + 2x + 4)\)
See Example 3 Hide Example 3

Problem

\(x^3 + 27\)

Approach

Step-by-step method

  1. Set up the problem.
  2. List the possible rational roots.
  3. Test the possible roots until one gives zero.
  4. Divide the polynomial by the factor that root gives.
  5. Factor whatever is left.
  6. Write the polynomial as a product of its factors.

Step 1

Step 1 - Set up the problem.

In this problem: The polynomial to factor is \(x^3 + 27\), and factoring means rewriting it as a product of simpler polynomials.

\(x^3 + 27\)

Step 2

Step 2 - List the possible rational roots.

In this problem: Any rational root has to divide the constant term \(27\) over the leading coefficient \(1\), which leaves \(−1, 1, −3, 3, −9, 9, −27, 27\) to try.

\(-1, 1, -3, 3, -9, 9, -27, 27\)

Step 3

Step 3 - Test the possible roots until one gives zero.

In this problem: Each candidate is put into \(x^3 + 27\) in turn, and the first one that returns \(0\) is a root.

\(x^3 + 27\)

Step 3a

Step 3a - Test the possible roots until one gives zero.

In this problem: Putting \(x = −1\) in gives \(26\), which is not \(0\).

\(f(-1) = (-1)^3 + 27 = 26\)

Step 3b

Step 3b - Test the possible roots until one gives zero.

In this problem: Putting \(x = 1\) in gives \(28\), which is not \(0\).

\(f(1) = (1)^3 + 27 = 28\)

Step 3c

Step 3c - Test the possible roots until one gives zero.

In this problem: Putting \(x = −3\) in gives \(0\), so \(x = −3\) is a root.

\(f(-3) = (-3)^3 + 27 = 0\)

Step 4

Step 4 - Divide the polynomial by the factor that root gives.

In this problem: A root at \(x = −3\) means \((x + 3)\) divides the polynomial exactly, and dividing leaves \(x^2 - 3x + 9\).

\(x^3 + 27 \div (x + 3) = x^2 - 3x + 9\)

Step 5

Step 5 - Factor whatever is left.

In this problem: Factoring \(x^2 - 3x + 9\) gives \(x^2 - 3x + 9\).

\(x^2 - 3x + 9 = x^2 - 3x + 9\)

Step 6

Step 6 - Write the polynomial as a product of its factors.

In this problem: Multiplying the factors back together reproduces \(x^3 + 27\), so the factored form is \((x + 3)(x^2 - 3x + 9)\).

\(x^3 + 27 = (x + 3)(x^2 - 3x + 9)\)

Final Answer

\((x + 3)(x^2 - 3x + 9)\)
See Example 4 Hide Example 4

Problem

\(x^4 - 16\)

Approach

Step-by-step method

  1. Set up the problem.
  2. List the possible rational roots.
  3. Test the possible roots until one gives zero.
  4. Divide the polynomial by the factor that root gives.
  5. Factor whatever is left.
  6. Write the polynomial as a product of its factors.

Step 1

Step 1 - Set up the problem.

In this problem: The polynomial to factor is \(x^4 - 16\), and factoring means rewriting it as a product of simpler polynomials.

\(x^4 - 16\)

Step 2

Step 2 - List the possible rational roots.

In this problem: Any rational root has to divide the constant term \(−16\) over the leading coefficient \(1\), which leaves \(−1, 1, −2, 2, −4, 4, −8, 8, −16, 16\) to try.

\(-1, 1, -2, 2, -4, 4, -8, 8, -16, 16\)

Step 3

Step 3 - Test the possible roots until one gives zero.

In this problem: Each candidate is put into \(x^4 - 16\) in turn, and the first one that returns \(0\) is a root.

\(x^4 - 16\)

Step 3a

Step 3a - Test the possible roots until one gives zero.

In this problem: Putting \(x = −1\) in gives \(−15\), which is not \(0\).

\(f(-1) = (-1)^4 -16 = -15\)

Step 3b

Step 3b - Test the possible roots until one gives zero.

In this problem: Putting \(x = 1\) in gives \(−15\), which is not \(0\).

\(f(1) = (1)^4 -16 = -15\)

Step 3c

Step 3c - Test the possible roots until one gives zero.

In this problem: Putting \(x = −2\) in gives \(0\), so \(x = −2\) is a root.

\(f(-2) = (-2)^4 -16 = 0\)

Step 4

Step 4 - Divide the polynomial by the factor that root gives.

In this problem: A root at \(x = −2\) means \((x + 2)\) divides the polynomial exactly, and each division leaves a smaller one behind.

\(x^4 - 16\)

Step 4a

Step 4a - Divide the polynomial by the factor that root gives.

In this problem: Dividing \(x^4 - 16\) by \((x + 2)\) leaves \(x^3 - 2x^2 + 4x - 8\).

\(x^4 - 16 \div (x + 2) = x^3 - 2x^2 + 4x - 8\)

Step 4b

Step 4b - Divide the polynomial by the factor that root gives.

In this problem: Dividing \(x^3 - 2x^2 + 4x - 8\) by \((x - 2)\) leaves \(x^2 + 4\).

\(x^3 - 2x^2 + 4x - 8 \div (x - 2) = x^2 + 4\)

Step 5

Step 5 - Factor whatever is left.

In this problem: Factoring \(x^2 + 4\) gives \(x^2 + 4\).

\(x^2 + 4 = x^2 + 4\)

Step 6

Step 6 - Write the polynomial as a product of its factors.

In this problem: Multiplying the factors back together reproduces \(x^4 - 16\), so the factored form is \((x + 2)(x - 2)(x^2 + 4)\).

\(x^4 - 16 = (x + 2)(x - 2)(x^2 + 4)\)

Final Answer

\((x + 2)(x - 2)(x^2 + 4)\)