Solving Quadratic Equations by Factoring, SAT math lesson

If two numbers multiply to give $0$ , at least one of them has to be $0$ . No other product has that property, since two numbers can multiply to $6$ in endless ways. That one fact turns a hard question, which numbers make $x^2 - 5x + 6$ equal to zero, into two easy ones, because $x^2 - 5x + 6 = (x - 2)(x - 3)$ and each factor is a simple linear expression.

See also Expanding and Factoring for a Purpose and Solving Equations and Checking Solutions in Foundations for Calculus.

Key terms

Quadratic equation

Definition A quadratic equation in $x$ is an equation that can be rewritten in the standard form

ax^2 + bx + c = 0,

where $a$ , $b$ and $c$ are real numbers and $a \ne 0$ . If $a = 0$ the equation is linear, not quadratic. A quadratic equation has at most two real solutions.

In plain words Gather every term on one side so the other side is zero, and if the highest power left is $x^2$ , the equation is quadratic.

Example The equation $2x^2 + 3 = 7x$ is quadratic. Subtracting $7x$ from both sides puts it in standard form,

2x^2 - 7x + 3 = 0,

with $a = 2$ , $b = -7$ and $c = 3$ . The equation $x^2 + 4x = x^2 - 8$ is not quadratic, because the $x^2$ terms cancel and leave $4x = -8$ .

Zero-product property

Definition For all real numbers $A$ and $B$ ,

\begin{gathered} AB = 0 \quad \text{if and only if} \\[0.4em] A = 0 \ \text{or} \ B = 0. \end{gathered}

The same holds for a product of more than two factors. The property works only because the right side is $0$ . An equation such as $AB = 2$ cannot be split into cases this way, because many pairs of numbers have product $2$ .

In plain words Once one side is zero and the other side is a product, you may split into separate cases, one for each factor. If the right side is any other number, splitting is not allowed.

Example The equation $(x - 2)(x - 3) = 2$ is a trap. Setting $x - 2 = 2$ and $x - 3 = 2$ gives $4$ and $5$ . At $x = 5$ the product is $3 \cdot 2 = 6$ , not $2$ , and $x = 4$ works only by luck while the solution $x = 1$ is missed. The correct way first moves everything to one side,

\begin{aligned} x^2 - 5x + 6 - 2 &= 0 \\[0.4em] x^2 - 5x + 4 &= 0 \\[0.4em] (x - 1)(x - 4) &= 0. \end{aligned}

So $x = 1$ or $x = 4$ . Both check, since $(-1)(-2) = 2$ and $(2)(1) = 2$ .

Factored form

Definition A quadratic expression $ax^2 + bx + c$ with $a \ne 0$ is in factored form when it is written as

a(x - r)(x - s)

for some real numbers $r$ and $s$ . Then the equation $ax^2 + bx + c = 0$ has solutions exactly $x = r$ and $x = s$ . When $r = s$ the form is $a(x - r)^2$ and the equation has the single solution $r$ , called a double root.

In plain words In factored form the solutions can be read off directly, but watch the signs. The factor $x - r$ gives the solution $r$ , and $x + 4$ gives $-4$ .

Example The expression $3(x + 4)(x - 1)$ is in factored form with $r = -4$ and $s = 1$ , so the equation $3(x + 4)(x - 1) = 0$ has solutions $-4$ and $1$ . The constant factor $3$ is never zero, so it contributes no solution. The equation $x^2 - 6x + 9 = 0$ factors as $(x - 3)^2 = 0$ , with only the double root $3$ .

Solution

Definition A solution of an equation in $x$ is a value of $x$ that makes the equation a true statement. The solution set is the set of all solutions. Solving means finding every solution and no extra values.

Dividing both sides by an expression that contains $x$ is not a safe step, because that expression could be zero for one of the solutions. Factor instead.

In plain words A solution is a number that works when you plug it back in. Report all of them, and check each one in the original equation.

Example Solve $3x^2 = 12x$ . Dividing by $3x$ gives $x = 4$ and silently loses a solution. Instead, move everything to one side and factor out the common factor,

\begin{aligned} 3x^2 - 12x &= 0 \\[0.4em] 3x(x - 4) &= 0. \end{aligned}

So $x = 0$ or $x = 4$ . Both work, since $3 \cdot 0 = 12 \cdot 0$ and $3 \cdot 16 = 48 = 12 \cdot 4$ . The lost solution was exactly the value that made $3x$ zero.

Root

Definition A root of the equation $p(x) = 0$ , also called a zero of the function $p$ , is a real number $r$ with $p(r) = 0$ . The real zeros of $p$ are exactly the $x$ -coordinates of the points where the graph of $y = p(x)$ meets the $x$ -axis, its $x$ -intercepts.

In plain words Roots are the solutions of the equation set to zero, and on the graph they are where the curve crosses or touches the $x$ -axis.

Example The function $f(x) = x^2 - 5x + 6 = (x - 2)(x - 3)$ has roots $2$ and $3$ , so its graph crosses the $x$ -axis at $(2, 0)$ and $(3, 0)$ . A quick table confirms it.

$x$	$1$	$2$	$2.5$	$3$	$4$
$f(x)$	$2$	$0$	$-0.25$	$0$	$2$

In the Desmos graphing calculator on the SAT you can type $y = x^2 - 5x + 6$ and click the curve. Its $x$ -intercepts appear as gray points, and clicking a point shows its coordinates. That is a good way to check a factoring answer.

Solving by factoring

Definition Solving a quadratic equation by factoring means rewriting it as $p(x) = 0$ , writing $p(x)$ as a product of linear factors (taking out any greatest common factor first), setting each factor equal to $0$ by the zero-product property, and reporting every value found, after checking each one in the original equation. The method works only when $p(x)$ can be factored, and many quadratics have no factorization with integer coefficients.

In plain words Zero on one side, factor, split, solve and check. Skipping the first step is the most common way to go wrong.

Example Solve $2x^2 + 3 = 7x$ . Following the steps,

\begin{aligned} 2x^2 - 7x + 3 &= 0 \\[0.4em] (2x - 1)(x - 3) &= 0. \end{aligned}

So $2x - 1 = 0$ or $x - 3 = 0$ , which gives $x = \frac{1}{2}$ or $x = 3$ . A check at $x = \frac{1}{2}$ gives $2 \cdot \frac{1}{4} + 3 = 3.5$ and $7 \cdot \frac{1}{2} = 3.5$ , and at $x = 3$ both sides equal $21$ .

Check yourself

What are all the solutions of $x^2 + 2x = 15$ ? (A) $-3$ and $5$ (B) $-5$ and $3$ (C) $3$ only (D) $13$ and $15$

Show the answer

The answer is (B). Rewriting gives $x^2 + 2x - 15 = 0$ , which factors as $(x + 5)(x - 3) = 0$ . Choice (A) reads the factors with the wrong signs. Choice (C) misses the negative solution, although $(-5)^2 + 2(-5) = 15$ works. Choice (D) comes from writing $x(x + 2) = 15$ and setting each factor equal to $15$ , which the zero-product property does not allow.

Which of the following gives all solutions of $x^2 = 7x$ ? (A) $7$ only (B) $0$ and $7$ (C) $-7$ and $0$ (D) $\sqrt{7}$ only

Show the answer

The answer is (B). Moving everything to one side gives $x^2 - 7x = 0$ , so $x(x - 7) = 0$ and $x = 0$ or $x = 7$ . Choice (A) is what you get by dividing both sides by $x$ , which throws away $x = 0$ . Choice (C) has the sign of $7$ wrong. Choice (D) treats the equation as $x^2 = 7$ and keeps only the positive root.

The graph of $f(x) = 2x^2 - 5x - 3$ crosses the $x$ -axis at $(a, 0)$ and $(b, 0)$ , where $a < b$ . What is the value of $b - a$ ?

Show the answer

The answer is $\frac{7}{2}$ , which can also be entered as $3.5$ . Factoring gives $2x^2 - 5x - 3 = (2x + 1)(x - 3)$ , so the roots are $a = -\frac{1}{2}$ and $b = 3$ , and

b - a = 3 - \left(-\frac{1}{2}\right) = \frac{7}{2}.

Solving Quadratic Equations by Factoring

Theory, part 47

Practice, part 48

Key terms

Quadratic equation

Zero-product property

Factored form

Solution

Root

Solving by factoring

Check yourself

How this lesson connects

Builds on

Used later in