Factoring Polynomials

To factor a polynomial means to rewrite it as a product of simpler polynomials that cannot be factored further over the real numbers. For example,

Below you will find the main factoring formulas, worked examples with full solutions, and practice questions with answers. A factor polynomial calculator is also available to check your work.

Factoring Polynomials: Key Formulas

1. Common factor (reverse distributive property) \[ ax + ay = a(x + y) \]
2. Factoring by grouping \[ ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y) \]
3. Difference of two squares \[ x^2 - y^2 = (x + y)(x - y) \]
4. Difference of two squares (extended) \[ (x + y)^2 - z^2 = (x + y + z)(x + y - z) \]
5. Sum of two cubes \[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]
6. Difference of two cubes \[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
7. Difference of fourth powers \[ x^4 - y^4 = (x + y)(x - y)(x^2 + y^2) \]
8. Perfect square trinomials \[ x^2 + 2xy + y^2 = (x + y)^2 \] \[ x^2 - 2xy + y^2 = (x - y)^2 \]
9. Perfect cubes \[ x^3 + 3x^2y + 3xy^2 + y^3 = (x + y)^3 \] \[ x^3 - 3x^2y + 3xy^2 - y^3 = (x - y)^3 \]

Worked Examples

Example 1: Factoring Out a Common Factor

Factor the binomial

Solution

The greatest common factor of 12 and 8 is 4. Rewrite each term using this factor:

Factor out 4:

You can check the result by expanding \(4(3x - 2)\).

Example 2: Difference of Two Squares

Factor the binomial

Solution

Rewrite the expression as a difference of squares:

Apply the difference of squares formula:

Example 3: Factoring a Trinomial

Factor the trinomial

Solution

Assume a factorization of the form

Expanding gives

Match coefficients with \(9x^2 + 3x - 2\):

Choosing \(a = 3\), \(b = 3\), \(m = 2\), and \(n = -1\) satisfies all conditions. Therefore,

Example 4: Factoring by Grouping

Factor the polynomial

Solution

Group terms with common factors:

Factor each group:

Factor out \(x + 2\):

Apply the difference of squares:

Practice Questions

Part A: Factor the polynomials

1. \((x + 1)^2 - 4\)
2. \(x^2 + 5x + 6\)
3. \(x^3 - 1\)
4. \(x^3 - x^2 - 25x + 25\)
5. \(27 + 8x^3\)

Part B

Prove that

Solutions

Part A

1. \[(x + 1)^2 - 4 = (x - 1)(x + 3)\]
2. \[x^2 + 5x + 6 = (x + 2)(x + 3)\]
3. \[x^3 - 1 = (x - 1)(x^2 + x + 1)\]
4. \[x^3 - x^2 - 25x + 25 = (x - 1)(x + 5)(x - 5)\]
5. \[27 + 8x^3 = (3 + 2x)(4x^2 - 6x + 9)\]

   Note: The quadratic \(4x^2 - 6x + 9\) cannot be factored over the real numbers.

Part B

Grouping and factoring step by step:

More References on Polynomials

• Multiplicity of Zeros and Graphs of Polynomials
• Finding Zeros of Polynomial Functions
• Polynomial Functions: Zeros, Factors, and Intercepts