Section 1B

Factoring Polynomials

Learning Objectives

Extract a greatest common factor.

Factor common quadratic patterns and verify by expansion.

Use the zero-product property to solve a factored equation.

Prerequisites: the distributive property, like terms, and exponent rules. Factoring rewrites a sum as a product. It is the reverse of expansion, and it can expose zeros or simplify a quotient.

1. Look for a Common Factor First

A common factor divides every term. For a polynomial with integer coefficients, the greatest common factor includes the greatest shared numerical factor and the smallest power of each shared variable.

Worked Example: Extract a Factor

Factor 6x³ − 9x². The coefficients share 3, and both terms contain x².

Thus 6x³ − 9x² = 3x²(2x − 3). Expanding checks it: 3x²·2x = 6x³ and 3x²·(−3) = −9x².

The factor inside the parentheses contains the remaining part of each original term.

2. Recognize a Difference of Squares

The identity a² − b² = (a − b)(a + b) follows by expansion: the middle terms cancel. Both terms must be squares, and the operation between them must be subtraction. A sum of squares does not use this same real linear-factor pattern.

Example: Two Squares

Factor 9x² − 16 as (3x)² − 4². The result is (3x − 4)(3x + 4).

Check the middle terms: +12x and −12x cancel, leaving 9x² − 16.

3. Find Numbers with the Right Sum and Product

For x² + bx + c, a factorization (x + p)(x + q) needs p + q = b and pq = c. The product determines possible factor pairs; the sum selects the pair with the correct middle coefficient.

Worked Example: A Monic Quadratic

Factor x² + x − 12. We need a product of −12 and a sum of 1.

The pair 4 and −3 works, so x² + x − 12 = (x + 4)(x − 3).

Expanding gives x² − 3x + 4x − 12 = x² + x − 12.

4. A Leading Coefficient May Need Grouping

Worked Example: Split the Middle Term

Factor 2x² + 7x + 3. The product of the leading and constant coefficients is 6. Two numbers with product 6 and sum 7 are 6 and 1.

Rewrite as 2x² + 6x + x + 3. Group: 2x(x + 3) + 1(x + 3).

Factor the shared binomial to get (2x + 1)(x + 3). Expansion confirms the original expression.

5. A Zero Product Gives Equations to Solve

If AB = 0 for real numbers A and B, at least one factor is zero. Therefore (x − 2)(x + 5) = 0 gives x = 2 or x = −5. This rule requires the whole product to equal zero. It does not apply unchanged to a product equal to 7.

Common Mistakes

Extract a common factor before testing more complicated patterns.

Factoring an expression does not itself solve an equation; a zero condition must be supplied.

Check by expansion, especially when signs or a leading coefficient are involved.

6. Practice

1. Factor 4x² − 20x.

2. Factor x² − 7x + 12.

3. Factor 3x² − 12 completely.

4. Solve x² + 2x − 15 = 0 by factoring.

Worked Solutions

1. Extract 4x: 4x(x − 5).

2. The numbers −3 and −4 multiply to 12 and add to −7. The result is (x − 3)(x − 4).

3. First write 3(x² − 4), then factor the difference of squares: 3(x − 2)(x + 2).

4. Factor as (x + 5)(x − 3) = 0. The solutions are x = −5 and x = 3; both make the original quadratic zero.

7. Summary

  • Factoring reverses distribution.
  • Try a greatest common factor, a known pattern, or a sum-product pair.
  • A factored product equal to zero exposes its solutions.
  • Expansion is a direct way to verify a factorization.

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