# Solving Quadratic Equations by Factoring

**Quadratic** equations are equations (not just
polynomials) of the second-degree, that is, they have

an x^{2} term in them. Examples are: x^{2} + x = 5 and x^{2} – 2x + 35 = 0.

The standard form of a quadratic equation is: ax^{2} + bx +
c = 0.

To solve the quadratic equation x^{2} + 7x + 6 = 0, we must
find the values of x that make this

equation true. We can do this by factoring.

Factoring a polynomial gives us a product of factors:

x^{2} + 7x + 6 = (x + 6)(x + 1)

If a product of two numbers equals zero, for example, ab =
0, then either a must equal zero or b

must equal zero (because anything multiplied by zero equals zero), or both may
equal zero. This

is called the principle of zero products.

So, if (x + 6)(x + 1) = 0, then either (x + 6) = 0 or (x +
1) = 0. To solve the original equation, we

solve for x in both of these equations:

The solutions for x^{2} + 7x + 6 = 0 are –6 and –1.

If the polynomial turns out to be a perfect square, you
only get one solution (or really the same

solution twice):

The solution is x = -3. In this case, both factors will equal zero when x = -3.

**Solving a Quadratic Equation**

Step 1: Make sure all terms are on

one side of the equation and zero is

on the other side.

a. x^{2} – 7x = -12 → x^{2} – 7x + 12 = 0

b. 5x^{2} = 15x → 5x^{2} – 15x = 0

c. x(x - 6) = 27 → x(x - 6) – 27 = 0

d. (2x + 5)(x + 9) = 15 → (2x + 5)(x + 9) – 15 = 0

**Step 2:** If some terms were written

in factored form originally, do all the

multiplication and combine like terms.

c. x(x - 6) – 27 = 0 → x^{2} – 6x – 27 = 0

d. (2x + 5)(x + 9) – 15 = 0

→ 2x^{2} + 23x + 45 – 15 = 0

→ 2x^{2} + 23x + 30 = 0

**Step 3: **Factor and solve by setting each factor equal to zero:

a. x^{2} – 7x + 12 = 0

The solutions are x = 3 or x = 4.

b. 5x^{2} – 15x = 0

The solutions are x = 0 or x = 3.

c. x2 – 6x – 27 = 0

The solutions are x = 9 or x = -3.

d. 2x^{2} + 23x + 30 = 0

The solutions are x = -3/2 or x = -10.

**So why are we doing this?**

This section is an introduction to topics in Math 102 – problem solving using
quadratics and

graphing quadratic functions. Quadratic functions are useful when we are trying
to find the

maximum or minimum of a function, such as maximizing area or minimizing cost.
These topics

will be covered later.

The graph of a quadratic function is called a parabola and looks like these:

With what you know now, you can find the x-intercepts of these graphs. The
x-intercepts are

where the graph crosses the x-axis. At these points, y = 0. (In Section 3.2, you
looked at the

y-intercept of straight-line functions - where the line crossed the y-axis. This
concept is similar.)

When you solve a quadratic equation, you are finding the x-intercepts of the
quadratic function

(called finding the “zeros” or “roots”).

For the graphs above:

y = x^{2} – 3x – 4

y = -2x^{2} + x + 6

Set y = 0:

0 = x^{2} – 3x – 4

0 = -2x^{2} + x + 6

Factor and solve:

Using these values of x, and remembering y = 0, we can state these values as
ordered pairs of

coordinates on a graph in the form (x, y). These represent the points on the
graphs which are the

x-intercepts:

x = -1, y = 0 and x = 4, y = 0

x-intercepts: (-1, 0) and (4, 0)

x = -3/2, y = 0 and x = 2, y = 0

x-intercepts: (-3/2, 0) and (2, 0)

And you can see above, these points are where the graphs cross their x-axes.

**Exercises:
**Solve, using the principle of zero products:

1. (x + 3)(x + 11) = 0

2.

3. (x – 7)(x + 75)(x – 43) = 0

4. 0.35x (0.1x + 0.4)(0.05x – 15) = 0

Solve by factoring:

5. 9x^{2} – 16 = 0

6. x^{2} + 7x = 18

7. 3x^{2} – 10x – 8 = 0

8. 36x^{2} +84x + 49 = 0

9. 15x^{2} +10x = 0

10. (5x + 3)(x – 1) = 13

Find the x-intercepts:

11. y = x^{2} – 6x + 9

12. y = -3x^{2} +5x + 2

Answers:

1. x = -3 or –11

2. x = 0 or ¾

3. x = 7 or –75 or 43

4. x = 0 or –4 or 300

5. x = 4/3 or –4/3

6. x = -9 or 2

7. x = -2/3 or 4

8. x = -7/6

9. x = 0 or –2/3

10. x = -8/5 or 2

11. (3, 0)

12. (-1/3, 0), (2, 0)