Factoring and Solving Quadratic Equations: College Algebra Lesson 1.3 Study Notes

Solving Quadratic Equations by Factoring

Introduction to Factoring Quadratic Equations

Factoring is a fundamental method for solving quadratic equations in College Algebra. Quadratic equations are equations of the form ax2 + bx + c = 0. The process involves rewriting the equation so that one side equals zero, factoring the quadratic expression, and then solving for the variable using the zero product property.

• Quadratic Equation: An equation where the highest power of the variable is 2.

• Zero Product Property: If a product of factors equals zero, then at least one of the factors must be zero.

• Factoring: Expressing a polynomial as a product of its factors.

Steps to Solve by Factoring:

1. Get 0 on one side of the equation.

2. Factor the nonzero side completely.

3. Set each factor equal to 0 and solve for the variable.

Example: Solve x2 - 25x = 0

• Step 1: Already set to zero.

• Step 2: Factor: $x^2 - 25x = x(x - 25)$

• Step 3: Set each factor to zero: $x = 0$ or $x - 25 = 0 \Rightarrow x = 25$

Solving Quadratic Equations with Mixed Terms

Factoring When Rearrangement is Needed

Sometimes, quadratic equations are not initially set to zero and require rearrangement. After moving all terms to one side, factor and solve as before.

• Example: Solve 3x2 - 2 = 5x

• Step 1: Move all terms to one side: $3x^2 - 5x - 2 = 0$

• Step 2: Factor: $(3x + 1)(x - 2) = 0$

• Step 3: Set each factor to zero: $3x + 1 = 0 \Rightarrow x = -\frac{1}{3}$, $x - 2 = 0 \Rightarrow x = 2$

Solving Rational Equations by Factoring

Combining Rational Expressions and Factoring

Rational equations involve fractions with polynomials in the numerator and denominator. To solve, clear denominators by multiplying both sides by the least common denominator (LCD), then factor and solve the resulting quadratic equation.

• Example: Solve $\frac{x}{x+2} + \frac{4}{x^2 - 4} = \frac{1}{x-2}$

• Step 1: Find LCD: $(x+2)(x-2)$

• Step 2: Multiply both sides by LCD and simplify:

• $x(x-2) + 4 = x+2$

• Step 3: Rearrange: $x^2 - 2x + 4 - x - 2 = 0 \Rightarrow x^2 - 3x + 2 = 0$

• Step 4: Factor: $(x-2)(x-1) = 0$

• Step 5: Solutions: $x = 2$, $x = 1$

Note: Always check for extraneous solutions that may make the original denominators zero.

Summary Table: Factoring Steps and Solutions

Equation	Factored Form	Solutions
$x^2 - 25x = 0$	$x(x - 25)$	$x = 0$, $x = 25$
$3x^2 - 2 = 5x$	$(3x + 1)(x - 2)$	$x = -\frac{1}{3}$, $x = 2$
$\frac{x}{x+2} + \frac{4}{x^2 - 4} = \frac{1}{x-2}$	$(x-2)(x-1)$	$x = 2$, $x = 1$

Additional info: These examples reinforce the process of solving quadratic and rational equations by factoring, a key skill in College Algebra.