Solving Quadratic Equations by Factoring

By the end of this lesson, you will be able to use the zero product property, get an equation to zero before factoring, find both solutions, check them, and reject the ones a word problem rules out.

Why Zero Is Special

Everything in this lesson rests on one fact, and the fact is about zero and nothing else.

$$\text{if } A \cdot B = 0 \text{ then } A = 0 \text{ or } B = 0$$

Two things multiplying to zero means at least one of them is zero. There is no other possibility, because any two non-zero numbers have a non-zero product.

Compare it with any other number. If $A \cdot B = 12$, then $A$ could be $3$, or $2$, or $\tfrac{1}{2}$, or $100$. Knowing the product tells you almost nothing about the factors. Zero is the only product that pins one of them down, and that is the entire reason for factoring quadratics.

$$(x + 2)(x - 5) = 0$$

So $x + 2 = 0$ or $x - 5 = 0$, giving $x = -2$ or $x = 5$. A quadratic becomes two one-step equations.

Get It to Zero First

The property is about zero, so the equation has to say zero.

$$x^2 + 3x = 10$$

You cannot split this. $x(x + 3) = 10$ does not mean $x = 10$ or $x+3 = 10$, and checking shows why: $x = 10$ gives $130$, not $10$. Move everything to one side first.

$$x^2 + 3x - 10 = 0$$

$$(x + 5)(x - 2) = 0$$

$$x = -5 \quad \text{or} \quad x = 2$$

This is the single most common error in the topic, and it comes from applying a rule about zero to a number that is not zero. The order is always: zero on one side, then factor, then split.

$$\textbf{standard form} \rightarrow \textbf{factor} \rightarrow \textbf{split} \rightarrow \textbf{solve each} \rightarrow \textbf{check}$$

Example 1: The Whole Method

Here is every step on one equation, worked all the way through. $$x^2 + 3x = 10$$

• Move the 10 across. Write the equation with zero on the right x^2 + 3x - 10 = 0
• Two numbers multiply to -10 and add to 3. What is the negative one? -2
• Write the factored equation (x - 2)(x + 5) = 0
• Set the first factor to zero and solve. What is x? 2
• Set the second factor to zero. What is x? -5

Two Solutions, Usually

A quadratic has two solutions, and forgetting the second one throws away half the answer.

$$x^2 - 9 = 0 \quad \Rightarrow \quad (x+3)(x-3) = 0 \quad \Rightarrow \quad x = \pm 3$$

Both are genuine. On a graph they are the two places the parabola crosses the $x$-axis, which is a useful picture: solutions are $x$-intercepts.

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