5.3: Factoring Quadratics | Algebra 1 Honors

How it works

For a monic quadratic, seek two numbers whose product is the constant term and whose sum is the linear coefficient. When the leading coefficient is not 1, check both leading and constant products. A factorization is complete only after common factors and special products have been considered.

x^2+(p+q)x+pq=(x+p)(x+q)

Worked examples

Example 1

Factor $\displaystyle x^2+5x+6$ .

The numbers $\displaystyle 2$ and $\displaystyle 3$ multiply to $\displaystyle 6$ and add to $\displaystyle 5$ . Thus $\displaystyle (x+2)(x+3)$ .

Example 2

Factor $\displaystyle 2x^2+7x+3$ .

Try factors of the leading and constant terms: $\displaystyle (2x+1)(x+3)$ expands to $\displaystyle 2x^2+7x+3$ .

Problems

Problem 5.3.1

Factor $\displaystyle x^2-x-12$ .

Show worked solution

The numbers $\displaystyle -4$ and $\displaystyle 3$ multiply to $\displaystyle -12$ and add to $\displaystyle -1$ . The result is $\displaystyle (x-4)(x+3)$ .

Problem 5.3.2

Factor $\displaystyle 3x^2+11x+6$ .

Show worked solution

The factorization is $\displaystyle (3x+2)(x+3)$ , since the middle terms sum to $\displaystyle 11x$ .

Problem 5.3.3

Factor $\displaystyle 2x^2-18$ completely.

Show worked solution

$\displaystyle 2(x^2-9)=2(x-3)(x+3)$ .

Lesson Problems PDF Solutions PDF Unit MCQ

How it works

Worked examples

Example 1

Example 2

Problems

Problem 5.3.1

Problem 5.3.2

Problem 5.3.3

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