Lesson 3: Solving Quadratic Equations by Factoring

New Concept

This lesson combines factoring with the Zero-Factor Principle. You'll learn to rewrite quadratic equations like $x^2 + bx + c = 0$ into a factored form, which makes finding the solutions as simple as setting each factor to zero.

What’s next

Next, you'll work through interactive examples to master factoring trinomials and then use these skills to solve equations in a series of practice cards.

Property

If the product of two numbers is zero, then one (or both) of the numbers must be zero. Using symbols,

If $AB = 0$, then either $A = 0$ or $B = 0$.

Examples

• To solve $(x - 7)(x + 2) = 0$, we set each factor to zero. This gives us $x - 7 = 0$ or $x + 2 = 0$, so the solutions are $x = 7$ and $x = -2$.
• If $y(y - 10) = 0$, then either $y = 0$ or $y - 10 = 0$. The two solutions for the equation are $y = 0$ and $y = 10$.
• Given $(2a + 1)(a - 5) = 0$, we solve $2a + 1 = 0$ to get $a = -\frac{1}{2}$, and we solve $a - 5 = 0$ to get $a = 5$.

Property

To factor $x^2 + bx + c$, we look for two numbers $p$ and $q$ so that

$pq = c$ and $p + q = b$

When we expand the factored form $(x + p)(x + q)$, we get $x^2 + (p + q)x + pq$.

Property

Assume that $b$, $c$, $p$, and $q$ are positive integers.

1. $x^2 + bx + c = (x + p)(x + q)$. If all the coefficients of the trinomial are positive, then both $p$ and $q$ are positive.
2. $x^2 - bx + c = (x - p)(x - q)$. If the linear term of the trinomial is negative and the other two terms positive, then $p$ and $q$ are both negative.
3. $x^2 \pm bx - c = (x + p)(x - q)$. If the constant term of the trinomial is negative, then $p$ and $q$ have opposite signs.

Examples

• To factor $x^2 - 7x + 10$, the constant term is positive and the middle term is negative. So, we need two negative numbers that multiply to 10 and add to -7. The factors are $(x - 2)(x - 5)$.
• In $y^2 + 4y - 12$, the constant term is negative, so the factors have opposite signs. We need numbers that multiply to -12 and add to 4. The factorization is $(y + 6)(y - 2)$.
• For $z^2 - z - 30$, the negative constant term means opposite signs. We need numbers that multiply to -30 and add to -1. The correct pair is -6 and 5, so we get $(z - 6)(z + 5)$.

Explanation

The signs in the trinomial give you huge clues! A positive last term means the signs in your factors are the same. A negative last term means the signs are different. Use this to factor faster!

Property

To Solve a Quadratic Equation by Factoring:

1. Write the equation in standard form, $ax^2 + bx + c = 0$ ($a \neq 0$)
2. Factor the left side of the equation.
3. Apply the Zero-Factor Principle; that is, set each factor equal to zero.
4. Solve each equation to obtain two solutions.

Examples

• To solve $x^2 - 5x = -6$, first write it as $x^2 - 5x + 6 = 0$. Factoring gives $(x - 2)(x - 3) = 0$. The solutions are $x = 2$ and $x = 3$.
• Solve $y^2 = 7y$. First, rearrange to $y^2 - 7y = 0$. Factor out $y$ to get $y(y - 7) = 0$. The solutions are $y = 0$ and $y = 7$.
• Solve $2a^2 + 14a + 24 = 0$. First, factor out the common factor of 2 to get $2(a^2 + 7a + 12) = 0$. This simplifies to $(a + 3)(a + 4) = 0$. Solutions are $a = -3$ and $a = -4$.

Explanation

First, get everything on one side so the equation equals zero. Then, factor the expression. Finally, use the Zero-Factor Principle to find the two values of $x$ that make the equation true.

Property

Before factoring and applying the Zero-Factor Principle, we must write the equation in standard form, so that one side of the equation is zero. For an equation like $x(x + 3) = 18$, we cannot apply the Zero-Factor principle.

Examples

• A common mistake in solving $x(x - 2) = 8$ is to set $x=8$ or $x-2=8$. This is incorrect because the product on the right side is not zero.
• To correctly solve $x(x - 2) = 8$, you must first distribute and rearrange: $x^2 - 2x = 8$, which becomes $x^2 - 2x - 8 = 0$.
• After rearranging to $x^2 - 2x - 8 = 0$, we can factor it into $(x - 4)(x + 2) = 0$. Now we can use the Zero-Factor Principle to find the correct solutions, $x = 4$ and $x = -2$.

Explanation

The Zero-Factor Principle only works with zero! If a product equals 18, the factors could be 2 and 9, or 3 and 6. There are too many options. Only a product of zero guarantees a factor is zero.