Linear Factors of a Polynomial

Try this

There are only $8$ candidates: $\pm 1, \pm 2, \pm 3, \pm 6$ . How are they related to the constant term $6$ ? Why is there no need to try $4$ or $5$ ?
Tap $1$ : the remainder is $0$ , and the graph crosses the $x$ -axis exactly at $x = 1$ . Now tap $2$ : the remainder is $- 4$ . Look at the graph: how high is the curve at $x = 2$ ?
When the remainder is $0$ , the first few numbers in the bottom row of the table are the coefficients of the quotient.
Once you have found all three roots, multiply the three linear factors together and compare the result with $f (x)$ .

Two theorems are at work here:

The Remainder Theorem: when $f (x)$ is divided by $x - c$ , the remainder is $f (c)$ .
The Factor Theorem: if $f (c) = 0$ , then $x - c$ is a factor of $f (x)$ , and the converse is true too.

So “find a linear factor” becomes “find a root”. The table on the workbench is called synthetic division: while it works out $f (c)$ , it also gives you the quotient for free.

Predict first

Without dividing: what is the remainder when $f (x) = x^{3} - 2 x^{2} - 5 x + 6$ is divided by $x - 2$ ?

Choose an answer first; then you can see the result.

Why

Why are only a few rational numbers worth trying? Suppose the fraction $\frac{p}{q}$ in lowest terms is a root of the integer polynomial $a_{n} x^{n} + \dots + a_{1} x + a_{0}$ . Substitute it and multiply both sides by $q^{n}$ :

a_{n} p^{n} + a_{n - 1} p^{n - 1} q + \dots + a_{1} p q^{n - 1} + a_{0} q^{n} = 0.

Every term except the last contains $p$ , so $p$ divides $a_{0} q^{n}$ ; and $p$ and $q$ are coprime, so $p ∣ a_{0}$ . In the same way, every term except the first contains $q$ , so $q ∣ a_{n}$ .

Example: find all the linear factors of 2x³ − 3x² − 3x + 2

Step 1 / 7

List the candidates
The numerator $p$ is a divisor of the constant term $2$ : $1, 2$ . The denominator $q$ is a divisor of the leading coefficient $2$ : $1, 2$ . The candidates are $\pm 1, \pm 2, \pm \frac{1}{2}$ : $6$ in all.

$\pm \frac{p}{q} : \pm 1, \pm 2, \pm \frac{1}{2}$
Try 1
Try −1: bring down the leading coefficient
Multiply once, add once
The remainder is 0
Do not miss the fractions
Finish

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: It must be on the candidate list $\pm \frac{p}{q}$

All three roots of $x^{3} - 7 x + 6$ are rational, and all of them are on the list. Sift them out.

Practice

Exercise 1Basic

What is the remainder when $f (x) = x^{3} + 2 x^{2} - 5 x - 6$ is divided by $x - 1$ ?

Exercise 2Basic

Find all the rational roots of $x^{3} - 7 x + 6$ . (Separate them with commas.)

Exercise 3Advanced

Factor $x^{3} - 4 x^{2} + x + 6$ .

Exercise 4Challenge

Factor $2 x^{3} + x^{2} - 7 x - 6$ .

Try this

Why

Example: find all the linear factors of 2x³ − 3x² − 3x + 2

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Substituting with the wrong sign in the Remainder Theorem

Trying only integers when the leading coefficient is not 1