Lesson 2: Solving Quadratic Equations by Factoring

Property

If $a \cdot b = 0$, then either $a = 0$ or $b = 0$ or both.

Examples

• To solve $(x-3)(x+5) = 0$, we set each factor to zero: $x-3=0$ or $x+5=0$. The solutions are $x=3$ and $x=-5$.
• For $7y(2y-1)=0$, we set $7y=0$ or $2y-1=0$. This gives the solutions $y=0$ and $y=\frac{1}{2}$.
• If $(z+4)^2 = 0$, it means $(z+4)(z+4)=0$. Both factors give the same solution, $z=-4$. This is called a double root.

Explanation

This property is the secret to solving factored equations. If a product of several things equals zero, at least one of those things must be zero. This lets us break a complicated product into simpler, separate equations.

Property

To use the Zero Product Property, the quadratic equation must be factored, with zero on one side. We must start with the quadratic equation in standard form, $ax^2 + bx + c = 0$.
HOW TO Solve a Quadratic Equation by Factoring.
Step 1. Write the quadratic equation in standard form, $ax^2 + bx + c = 0$.
Step 2. Factor the quadratic expression.
Step 3. Use the Zero Product Property.
Step 4. Solve the linear equations.
Step 5. Check.

Examples

• To solve $x^2 - x - 12 = 0$, we first factor it into $(x-4)(x+3)=0$. Using the Zero Product Property, we get $x=4$ and $x=-3$.
• To solve $3y^2 - 7y = -2$, first write it in standard form: $3y^2 - 7y + 2 = 0$. Factoring gives $(3y-1)(y-2)=0$, so the solutions are $y=\frac{1}{3}$ and $y=2$.
• To solve $k^2 = 8k$, rewrite it as $k^2 - 8k = 0$. Factor out the common factor $k(k-8)=0$. The solutions are $k=0$ and $k=8$.

Explanation

Factoring transforms a single, complex quadratic problem into a product of simple linear factors that equals zero. This allows you to apply the Zero Product Property and solve for the variable by handling each factor separately.