Before you read

Answer first — the lesson below explains it.

A projectile's height satisfies $x^{2} - 7 x + 12 = 0$ at two different times. What are the solutions?

A
$x = 3$ and $x = 4$
B
$x = - 3$ and $x = - 4$
C
$x = 2$ and $x = 6$
D
$x = 1$ and $x = 12$

Check your answer

Answer:

Before you read

Answer first — the lesson below explains it.

A ball's trajectory satisfies $x^{2} + 2 x - 15 = 0$ . What are the solutions?

A
$x = 3$ or $x = - 5$
B
$x = - 3$ or $x = 5$
C
$x = 1$ or $x = - 15$
D
$x = 15$ or $x = - 1$

Check your answer

Answer:

Theory

The Zero Product Property

If $A \cdot B = 0$ , then $A = 0$ or $B = 0$ (or both).

This is the foundation for solving quadratics by factoring:

1. Get everything on one side (set equation $= 0$ )
2. Factor the quadratic
3. Set each factor $= 0$
4. Solve each resulting equation

The solutions are also called roots, zeros, or x-intercepts of the corresponding function.

x² - 2x - 8 = 0: roots at x = -2 and x = 4

Quick check

You just read it. Can you apply it?

An acceleration equation gives $x^{2} - 36 = 0$ . What are the solutions?

A
$x = 6$ only
B
$x = - 6$ only
C
$x = 6$ or $x = - 6$
D
$x = 18$ or $x = - 18$

Check your answer

Answer:

$x^{2} - 36 = (x + 6) (x - 6) = 0$ .
$x = 6$ or $x = - 6$ .

Theory

Solving Step by Step

The method works for any factorable quadratic.

Example 1

Solve $x^2 - 5x + 6 = 0$

Factor: $(x - 2)(x - 3) = 0$

Set each factor to zero:

$x - 2 = 0 \quad \Rightarrow \quad x = 2$

$x - 3 = 0 \quad \Rightarrow \quad x = 3$

x = 2

x = 3

Example 2

Solve $2x^2 + 8x = 0$

Factor out GCF: $2x(x + 4) = 0$

$2x = 0 \quad \Rightarrow \quad x = 0$

$x + 4 = 0 \quad \Rightarrow \quad x = -4$

x = 0

x = -4

Example 3

Solve $x^2 = 9$

Rearrange: $x^2 - 9 = 0$

Factor: $(x + 3)(x - 3) = 0$

$x = 3$ or $x = -3$

x = \pm 3

Example 4

Solve $3x^2 - 7x + 2 = 0$

Factor (AC method): $(3x - 1)(x - 2) = 0$

$3x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{3}$

$x - 2 = 0 \quad \Rightarrow \quad x = 2$

x = \frac{1}{3}

x = 2

Quick check

You just read it. Can you apply it?

A velocity equation is $3 x^{2} - 12 x = 0$ . What are all solutions?

A
$x = 4$ only
B
$x = 0$ or $x = 4$
C
$x = 0$ or $x = - 4$
D
$x = 12$ or $x = 0$

Check your answer

Answer:

Factor: $3 x (x - 4) = 0$ .
$x = 0$ or $x = 4$ .

Common Mistakes

Common Mistakes to Avoid

Dividing both sides by

x

(losing a solution)

$x^2 = 5x$ $\to$ dividing by $x$ : $x = 5$ (lost $x = 0$ !)

Move everything to one side:

x^2 - 5x = 0

\to

x(x-5) = 0

\to

x = 0

x = 5

Forgetting to set equal to zero first

$x^2 - 3x = 10$ $\to$ factoring $x(x-3) = 10$ $\to$ $x = 10$ or $x - 3 = 10$

The zero product property only works when one side is

0

. Rearrange first:

x^2 - 3x - 10 = 0

Forgetting there can be two solutions

Finding $x = 3$ from $x^2 - 9 = 0$ but forgetting $x = -3$

Every quadratic can have up to two solutions. Always check both factors.

Quick check

You just read it. Can you apply it?

If $(x - a) (x - b) = 0$ where $a$ and $b$ are positive constants, which describes the graph of $y = (x - a) (x - b)$ ?

A
Parabola crossing x-axis at two positive values
B
Parabola crossing x-axis at two negative values
C
Parabola crossing x-axis at one positive and one negative value
D
Parabola that doesn't cross the x-axis

Check your answer

Answer:

The zeros are $x = a$ and $x = b$ , both positive. The parabola opens upward (leading coefficient $1 > 0$ ) and crosses the x-axis at two positive values.

Strategy

SAT Strategy Tip

Solving by factoring is the fastest method on the SAT when the quadratic factors nicely. If you can't find the factors within 30 seconds, switch to the quadratic formula (next lesson). Also, the SAT sometimes gives you the factored form directly: "If $(x - 3)(x + 7) = 0$ , what is a possible value of $x$ ?" — just read off the answers: $x = 3$ or $x = -7$ .

After reading

Same question as before. What do you say now?

A projectile's height satisfies $x^{2} - 7 x + 12 = 0$ at two different times. What are the solutions?

A
$x = 3$ and $x = 4$
B
$x = - 3$ and $x = - 4$
C
$x = 2$ and $x = 6$
D
$x = 1$ and $x = 12$

Check your answer

Answer:

Factor: $(x - 3) (x - 4) = 0$ .
$x = 3$ or $x = 4$ .

After reading

Same question as before. What do you say now?

A ball's trajectory satisfies $x^{2} + 2 x - 15 = 0$ . What are the solutions?

A
$x = 3$ or $x = - 5$
B
$x = - 3$ or $x = 5$
C
$x = 1$ or $x = - 15$
D
$x = 15$ or $x = - 1$

Check your answer

Answer:

Factor: $(x + 5) (x - 3) = 0$ .
$x = - 5$ or $x = 3$ .

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