Lesson 2: Intercepts, Solutions, and Factors

New Concept

This lesson connects a quadratic's factors, its solutions, and its graph's x-intercepts. We'll use the Zero-Factor Principle, which states that if $ab = 0$, then $a = 0$ or $b = 0$, to solve equations by factoring.

What’s next

Now, let's apply this principle. You'll work through examples of solving equations by factoring, followed by interactive practice problems to master the skill.

Property

The product of two factors equals zero if and only if one or both of the factors equals zero. In symbols,

Examples

• To solve $(x+4)(x-2)=0$, set each factor to zero. $x+4=0$ gives the solution $x=-4$, and $x-2=0$ gives the solution $x=2$.

Property

The $x$-intercepts of the graph of

are the solutions of the equation

Property

To Solve a Quadratic Equation by Factoring:

1. Write the equation in standard form, $ax^2 + bx + c = 0$.
2. Factor the left side of the equation.
3. Apply the zero-factor principle: Set each factor equal to zero.
4. Solve each of the resulting linear equations.

Examples

• To solve $x^2 - 2x = 15$, first write it in standard form: $x^2 - 2x - 15 = 0$. Factoring gives $(x-5)(x+3)=0$, so the solutions are $x=5$ and $x=-3$.

• To solve $3x^2 = 12x$, rearrange to $3x^2 - 12x = 0$. Factor out the common term $3x$ to get $3x(x-4)=0$. The solutions are $x=0$ and $x=4$.

Property

The solutions of the quadratic equation $a(x - r_1)(x - r_2) = 0$ are $r_1$ and $r_2$. This is called the factored form of the quadratic equation. If you know the two solutions of a quadratic equation, you can work backwards to reconstruct the equation.

Examples

• A quadratic equation has solutions $x=3$ and $x=-6$. The factors are $(x-3)$ and $(x-(-6))$, or $(x+6)$. The equation is $(x-3)(x+6)=0$, which expands to $x^2+3x-18=0$.

• To find an equation with solutions $x=2$ and $x=\frac{1}{4}$, start with factors $(x-2)$ and $(x-\frac{1}{4})$. For integer coefficients, use $(x-2)$ and $(4x-1)$. The equation is $(x-2)(4x-1)=0$, or $4x^2-9x+2=0$.