Math · College algebra · Concept

Factoring quadratics and finding zeros

Factoring rewrites a polynomial as a product, such as $x^{2} - 5 x + 6$ $= (x - 2) (x - 3)$ . Set the product equal to zero and each factor gives one small equation, whose solutions are where the graph meets the x-axis.

A concrete case

Multiply, then run it backward

Multiply $(x + 2) (x + 5)$ . Each term of the first bracket multiplies each term of the second, which makes four products, $x^{2}$ , $5 x$ , $2 x$ and $10$ . The two middle products combine, so $(x + 2) (x + 5)$ $= x^{2} + 7 x + 10$ .

Try it with a number to be sure. At $x = 10$ the brackets are $12$ and $15$ , and $12 \times 15 = 180$ . The expanded form gives $100 + 70 + 10 = 180$ as well.

Look at where $7$ and $10$ came from. The $7$ is $2 + 5$ , the sum of the two numbers in the brackets. The $10$ is $2 \times 5$ , their product. Factoring runs this backward. Given $x^{2} + 7 x + 10$ , you look for two numbers whose sum is $7$ and whose product is $10$ , find $2$ and $5$ , and write $(x + 2) (x + 5)$ .

(x + 2) (x + 5) = x^{2} + 5 x + 2 x + 10 = x^{2} + 7 x + 10

\begin{aligned} (x + 2) (x + 5) & = x^{2} + 5 x + 2 x + 10 \\ = x^{2} + 7 x + 10 \end{aligned}

The four products of $(x + 2) (x + 5)$ , as the areas of a rectangle $x + 2$ tall and $x + 5$ wide
	$x$	$+ 5$
$x$	$x^{2}$	$5 x$
$+ 2$	$2 x$	$10$

The rule

The product and sum rule

A polynomial is a sum of terms like $3 x^{2}$ , $- 5 x$ and $6$ , with whole-number powers of $x$ . To factor it is to write it as a product of simpler polynomials, called its factors.

For a quadratic whose $x^{2}$ coefficient is $1$ , multiplying $(x + p) (x + q)$ gives $x^{2} + (p + q) x + p q$ . To factor $x^{2} + b x + c$ , find numbers $p$ and $q$ with $p + q = b$ and $p q = c$ . For $x^{2} - 5 x + 6$ they need a product of $6$ and a sum of $- 5$ . The pair $- 2$ and $- 3$ does both, so $x^{2} - 5 x + 6$ $= (x - 2) (x - 3)$ .

The zero-product property turns a factored equation into small ones. If a product of numbers is $0$ , at least one of the factors is $0$ . From $(x - 2) (x - 3) = 0$ , either $x - 2 = 0$ or $x - 3 = 0$ , which gives $x = 2$ or $x = 3$ .

(x + p) (x + q) = x^{2} + (p + q) x + p q

Factor pairs of $6$ , looking for a sum of $- 5$
$p$	$q$	Product	Sum
$1$	$6$	$6$	$7$
$2$	$3$	$6$	$5$
$- 1$	$- 6$	$6$	$- 7$
$- 2$	$- 3$	$6$	$- 5$

Why it works

Why a zero product forces a zero factor

Suppose $A \times B = 0$ and $A$ is not zero. Then you may divide both sides by $A$ , which leaves $B = 0$ . Either $A$ is zero, or $B$ has to be. No other way out exists.

The number 0 is special here. A product of $6$ can come from $2 \times 3$ , $1 \times 6$ , $12 \times \frac{1}{2}$ and endlessly many other pairs, so knowing the product is $6$ tells you nothing about either factor alone. Only a product of $0$ pins a factor down.

The sign of each factor also explains the graph. For $y = (x - 2) (x - 3)$ , both factors are negative when $x < 2$ , so $y$ is positive. Between $2$ and $3$ the factors have opposite signs and $y$ is negative. Past $3$ both are positive. The curve dips below the axis between its two zeros.

A B = 0 ⟹ A = 0 or B = 0

The method

The order of moves

Factoring goes fastest when the moves come in a fixed order. The table runs them on $2 x^{3} - 18 x$ .

The last step matters most. Multiply your factors back together and compare with the start, middle term included.

2 x^{3} - 18 x = 2 x (x^{2} - 9) = 2 x (x - 3) (x + 3)

\begin{aligned} 2 x^{3} - 18 x & = 2 x (x^{2} - 9) \\ = 2 x (x - 3) (x + 3) \end{aligned}

Factoring in order, on $2 x^{3} - 18 x$
Step	What to do	For $2 x^{3} - 18 x$
1	Take out the greatest common factor of every term	$2 x (x^{2} - 9)$
2	Count the terms that are left	Two terms, $x^{2}$ and $- 9$
3	Two terms: a difference of squares, $a^{2} - b^{2}$ $= (a - b) (a + b)$	$x^{2} - 9$ $= (x - 3) (x + 3)$
4	Three terms: the product and sum rule, or split the middle term	Not needed here
5	Factor each factor again until none will go further	$2 x (x - 3) (x + 3)$
6	Multiply back to check	$2 x (x^{2} - 9)$ $= 2 x^{3} - 18 x$

When the $x^{2}$ coefficient is not $1$

For $a x^{2} + b x + c$ with a not $1$ , split the middle term. Multiply $a$ by $c$ . Find two numbers whose product is $a c$ and whose sum is $b$ . Write $b x$ as the sum of those two terms, then factor the four terms in pairs.

For $6 x^{2} + x - 2$ , $a c$ is $6 \times (- 2) = - 12$ . The pair $4$ and $- 3$ has product $- 12$ and sum $1$ . Writing $x$ as $4 x - 3 x$ gives $6 x^{2} + 4 x - 3 x - 2$ . The first pair is $2 x (3 x + 2)$ and the second is $- 1 (3 x + 2)$ . The shared bracket comes out, leaving $(3 x + 2) (2 x - 1)$ .

If the two pairs do not share a bracket, check the signs first. Taking $- 1$ out of $- 3 x - 2$ gives $- (3 x + 2)$ , with both signs inside changed.

6 x^{2} + x - 2 = 6 x^{2} + 4 x - 3 x - 2 = (3 x + 2) (2 x - 1)

\begin{aligned} 6 x^{2} + x - 2 & = 6 x^{2} + 4 x - 3 x - 2 \\ = (3 x + 2) (2 x - 1) \end{aligned}

Solving an equation by factoring

Factoring an expression and solving an equation are different jobs. $x^{2} - 5 x + 6$ is an expression, and factoring it changes its form, not its value. $x^{2} - 5 x + 6 = 0$ is an equation, and solving it finds the values of $x$ that make it true.

To solve by factoring, move every term to one side so the other side is $0$ . Factor that side. Set each factor equal to $0$ and solve each small equation. Then substitute each answer into the original equation.

x^{2} - 5 x + 6 = 0 ⟹ (x - 2) (x - 3) = 0 ⟹ x = 2 or x = 3

\begin{array}{c} x^{2} - 5 x + 6 = 0 \\ (x - 2) (x - 3) = 0 \\ x = 2 or x = 3 \end{array}

Where it breaks

The zero-product property needs $0$ on one side. $(x - 2) (x - 3) = 6$ does not mean $x - 2 = 6$ or $x - 3 = 6$ . Try $x = 8$ from the first of those. It gives $6 \times 5 = 30$ , not $6$ . Expand instead, so $x^{2} - 5 x + 6 = 6$ becomes $x^{2} - 5 x = 0$ . That factors as $x (x - 5) = 0$ , and $x = 0$ or $x = 5$ . Both check, since $(- 2) (- 3) = 6$ and $(3) (2) = 6$ .

Not every quadratic factors over the integers. $x^{2} + x - 1$ has no integer pair with product $- 1$ and sum $1$ , yet the equation $x^{2} + x - 1 = 0$ has two real solutions. Failing to find a pair is not proof there is no solution. The quadratic formula finds them.

A sum of squares is the other trap. $x^{2} - 9$ factors as $(x - 3) (x + 3)$ , but $x^{2} + 9$ does not factor at all with real numbers, because $x^{2} + 9$ is never $0$ .

(x - 2) (x - 3) = 6 ⟹ x^{2} - 5 x = 0 ⟹ x (x - 5) = 0

\begin{array}{c} (x - 2) (x - 3) = 6 \\ x^{2} - 5 x = 0 \\ x (x - 5) = 0 \end{array}

Check your answer

A factorization is true for every $x$ , so any number you choose can test it.

Multiply back. Expand your factors and compare with the start, term by term. The middle term is where sign slips hide. $(x - 1) (x - 6)$ has the right first and last terms for $x^{2} - 5 x + 6$ , but its middle term is $- 7 x$ .

Put $x = 10$ into the original and into your factors. For $x^{2} - 5 x + 6$ the original gives $56$ , and $(8) (7) = 56$ . Two different polynomials rarely agree at a number you picked yourself.

For an equation, substitute each answer into the original equation and look at the graph. The real solutions are the x-intercepts.

To have Chalk check the work, write the expression on the first row of a Math box and its factored form on the row below. The Checker marks a correct factoring ✓. $(x - 1) (x - 6)$ under $x^{2} - 5 x + 6$ reads ✗ “This factoring doesn’t multiply back to row 1.” and names a value of $x$ where the two disagree. For an equation, put the equation first, the factored equation next and the solutions last, joined by “or”. Leaving out a solution reads ✗ “This step loses a solution of row 2.”, unless the one left out is the negative root of a square or quadratic. Then keeping only the positive root reads ✓ with a note that names the other, such as “Keeps the positive root. x = −2 also solves row 2.”

Zeros and factors are the same information

The Factor Theorem says that a number $c$ is a zero of a polynomial exactly when $x - c$ is one of its factors. One direction is the zero-product property. If $p (x) = (x - c) q (x)$ , then $p (c) = 0 \times q (c) = 0$ . The other direction needs polynomial division, and it is why finding one zero of a cubic lets you divide it down to a quadratic.

There is a test for whether a quadratic factors with whole numbers. $a x^{2} + b x + c$ with integer coefficients factors into integer brackets exactly when $b^{2} - 4 a c$ is a perfect square. For $x^{2} - 5 x + 6$ it is $25 - 24 = 1$ , a perfect square. For $x^{2} + x - 1$ it is $1 + 4 = 5$ , which is not. That quantity is the discriminant of the quadratic formula.

Four terms can sometimes be factored in pairs, called grouping. $x^{3} + 2 x^{2} - 9 x - 18$ is $x^{2} (x + 2) - 9 (x + 2)$ , which is $(x + 2) (x - 3) (x + 3)$ , with zeros $- 2$ , $3$ and $- 3$ .

Common mistakes

Checking only the first and last terms. $(x - 1) (x - 6)$ gives $x^{2}$ and $+ 6$ , but its middle term is $- 7 x$ , so it is not $x^{2} - 5 x + 6$ . Multiply all four products back before you trust a factoring.
Dropping one root after factoring. $(x - 2) (x - 3) = 0$ has two solutions, $x = 2$ and $x = 3$ , because either factor can be the one that equals zero.
Using the zero-product property on a product that is not zero. $(x - 2) (x - 3) = 6$ does not give $x - 2 = 6$ . Expand, subtract $6$ to get $x^{2} - 5 x = 0$ , and factor that to find $x = 0$ or $x = 5$ .
Forgetting the common factor. $2 x^{2} - 18$ is $2 (x - 3) (x + 3)$ , not $(x - 3) (x + 3)$ , because the $2$ still multiplies everything. Take the greatest common factor out first.
Factoring a sum of squares like a difference. $(x + 3) (x - 3)$ multiplies to $x^{2} - 9$ , not $x^{2} + 9$ . With real numbers, $x^{2} + 9$ does not factor at all.
Dividing both sides by $x$ to solve $x^{2} = 3 x$ . That leaves only $x = 3$ and loses $x = 0$ . Subtract $3 x$ instead, so $x^{2} - 3 x = 0$ , and factor to $x (x - 3) = 0$ .
Assuming every quadratic factors over the integers. $x^{2} + x - 1$ has no integer pair with product $- 1$ and sum $1$ , yet $x^{2} + x - 1 = 0$ has the solutions $(- 1 \pm \sqrt{5}) / 2$ .

Key terms

Polynomial: An expression made by adding terms, each a number times whole-number powers of the variables, such as $3 x^{2} - 5 x + 2$ . An expression with a variable in a denominator, under a root or inside a function like $sin$ is not a polynomial.
Factorization: Writing an expression as a product of simpler pieces, such as $x^{2} - 9 = (x - 3) (x + 3)$ . How far it splits depends on the numbers allowed: $x^{2} + 1$ has no real factors but factors with complex numbers.
Zero-product property: If a product equals zero, at least one of its factors is zero. That is why you move everything to one side, leaving $0$ , before factoring to solve.
Difference of squares: The identity $a^{2} - b^{2} = (a - b) (a + b)$ . Both terms must be squares and the operation between them must be subtraction.
Factoring by grouping: A method that groups terms, factors each group, and then factors a common expression that appears in the group results. Regrouping does not change the original sum.
x-intercept: A point where a graph crosses or touches the x-axis, so its y-value is $0$ . For $y = f (x)$ , its x-value is a zero (root) of $f$ .

Work through an example

Solve $x^{2} - 5 x + 6 = 0$ , then connect the answer to a graph.

Solve a quadratic equation by factoring →

Factor a quadratic by splitting the middle term →

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