Factor, root, and remainder tests

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Connect factors, roots, and remainders to test polynomial claims quickly in Desmos and choose when direct substitution or algebra is stronger.

Difficulty: Advanced
Estimated time: 50 minutes
Domains: Advanced Math

What you’ll learn

See a linear factor, its root and a zero remainder as one fact.
Turn $x-c$ , $x+c$ and $ax+b$ into the right input to test.
Use graph roots to find possible factors.
Use function evaluation to confirm a factor or find a remainder.
Find every root that fits, then apply words like positive, smallest and largest.
Tell when a repeated or nonlinear factor needs more proof than one root test.
Choose between plugging in, graphing and algebra, whichever gets you to a sure answer fastest.

Prerequisites

Why this matters on the SAT

Turn factor questions into one input

The SAT asks about factors in a few different ways. Is the polynomial divisible by $x-3$ ? Which choice is a factor? What's the remainder? They sound like different questions, but each one comes down to a single number: the input that makes the factor equal zero.

SAT example

One of the factors of the polynomial

x^4-27x^2+14x+120

is $x+b$ , where $b$ is a positive constant. Which choice gives the smallest possible value of $b$ ?

Fast Desmos solution

Graph the polynomial and click its $x$ -intercepts. They're at

x=-5,-2,3,\text{ and }4.

The factor $x+b$ equals $0$ when $x=-b$ , so a positive $b$ comes from a negative root. The negative roots give $b=5$ and $b=2$ , so the smallest possible value is

\boxed{2}.

The answer is D.

One more step before you submit. Desmos finds intercepts numerically, so treat each one as a strong guess until you check it. Plug the root back in and make sure you get exactly $0$ :

(-2)^4-27(-2)^2+14(-2)+120=0.

Calculator loads as you approach

The polynomial and x + 2 cross the x-axis at the same point, x = -2.

When should you use these tests?

You’ll know these questions by their wording: “ $x-c$ is a factor,” “divisible by $x-c$ ,” “ $c$ is a zero,” “the remainder when divided by $x-c$ ,” or “can be rewritten with no remainder.” Then pick the quickest way in.

Use the calculator when…

The question hands you the divisor, like “Is $x-3$ a factor?” or “What’s the remainder when you divide by $x-3$ ?” Plug in its root.
You have to find the roots first, like when $x+b$ is a factor for some positive $b$ . Graph the polynomial.
The answer choices list several linear factors. The graph’s roots tell you which ones match.

Use algebra when…

Grouping, a special product or a common factor finishes the job in a line or two, like $2x^3-5x^2-18x+45=(2x-5)(x^2-9)$ .
The factor is squared, like $(x-3)^2$ .
The factor has a power of $x$ above $1$ , like $x^2+9$ .
An unknown constant leaves a question one graph can’t settle, like when $(x-3)^2$ has to be a factor of $x^3+ax^2-9x-9a$ .

A method that works isn’t always the fastest one, so pick the one that gets you to an exact answer with the least setup.

Check your understanding:

A question asks for the remainder when $P(x)$ , a fourth-degree polynomial, is divided by $x-3$ . Would you graph $P$ to find its roots, or evaluate $P(3)$ ?

Connect factors, roots, and remainders

Here's where all of this comes from. When you divide a polynomial $P(x)$ by $x-c$ , you get a quotient $Q(x)$ and a remainder $r$ , which is a constant:

P(x)=(x-c)Q(x)+r.

Now plug in $x=c$ . The first term disappears, because $c-c=0$ :

\begin{aligned} P(c) &=(c-c)Q(c)+r\\[1.4em] &=0+r\\[1.4em] &=r. \end{aligned}

So you don't have to do the division to find the remainder. This is the Remainder Theorem:

\boxed{\text{The remainder after division by }x-c\text{ is }P(c).}

And when the remainder is $0$ , the equation becomes $P(x)=(x-c)Q(x)$ , which says $x-c$ is a factor. This is the Factor Theorem:

\boxed{x-c\text{ is a factor of }P(x)\text{ exactly when }P(c)=0.}

"Exactly when" works in both directions. If $P(c)=0$ , then $x-c$ is a factor. And if $x-c$ is a factor, then $P(c)=0$ .

So these three sentences all say the same thing:

$x-c$ is a factor of $P(x)$ .
$c$ is a root or zero of $P$ .
Dividing $P(x)$ by $x-c$ leaves remainder $0$ .

A root and a zero are two names for the same thing: an input $c$ that makes $P(c)=0$ . When $c$ is a real number, you'll see it on the graph as the $x$ -intercept $(c,0)$ . Complex roots don't show up as $x$ -intercepts, because the graph only shows real numbers.

Here's a quick example. For $P(x)=x^2-5x+6$ ,

P(2)=2^2-5(2)+6=0,

so three things are true at once: $2$ is a root, $x-2$ is a factor, and dividing by $x-2$ leaves remainder $0$ .

Check your understanding:

Suppose $H(-4)=0$ . What’s the root, what’s the linear factor, and what can you say about the remainder?

Common mistake:

It’s easy to assume a factor and its root have the same sign. They don’t: $x+4$ equals $0$ at $x=-4$ , so the factor $x+4$ goes with the root $-4$ .

Translate the whole linear factor

Every linear factor test starts with one question: where does the whole factor equal zero? That value is the input you test.

For $x-c$ , test $x=c$ .
For $x+c$ , test $x=-c$ .
For $ax+b$ , where $a\ne0$ , test $x=-\frac{b}{a}$ .

The number in front of $x$ is the part that's easy to miss. Take $2x-5$ . Setting $2x-5=0$ gives $x=\frac52$ , not $5$ . So the factor test for $2x-5$ is $P\left(\frac52\right)=0$ .

Remainders work the same way. When you divide a polynomial by $ax+b$ , the remainder is $P\left(-\frac{b}{a}\right)$ .

Check your understanding:

What input would you test for each divisor: (1) $x-7$ , (2) $x+7$ , and (3) $3x-7$ ?

Common mistake:

Plugging in a number you can see in the factor instead of solving for its zero. In $4x+1$ , neither $4$ nor $1$ is the input. Solve $4x+1=0$ and you get $x=-\frac14$ .

Graph to find possible factors

When the factor has an unknown constant, or the choices list several linear factors, let the graph find the real roots for you.

For a possible factor $x-c$ :

Graph $y=P(x)$ .
Click each $x$ -intercept that could matter.
Read its $x$ -coordinate. That's a possible value of $c$ .
Apply every condition the question gives, such as positive, negative, smallest or largest.
Plug that value into $P$ and check that $P(c)=0$ exactly.

For a factor like $x+b$ , the root is $-b$ , so a negative root gives a positive $b$ . That's why the opening example kept only the roots $-5$ and $-2$ , and then smallest picked $b=2$ .

You can also graph the factor itself to see the connection. With $b=2$ , the line $y=x+2$ crosses the $x$ -axis at $(-2,0)$ , the same point as the polynomial. For a linear factor, that shared root is exactly the input the Factor Theorem tests.

Check your understanding:

A polynomial has real roots $-6$ , $-1$ , $2$ , and $5$ . One factor is $x+b$ , where $b$ is positive. What are the possible values of $b$ , and which is the largest?

Try it yourself:

Change the second line from y=x+2 to y=x+5, and watch the shared intercept move from $x=-2$ to $x=-5$ . Then reset the example.

Calculator loads as you approach

The polynomial and x + 2 cross the x-axis at the same point, x = -2.

Evaluate once to find a remainder

When the question already gives you the divisor, you usually don't need to graph the whole polynomial. Define it, then plug in the divisor's root.

Say $p(x)=3x^3-2x^2+5x-7$ , and you want the remainder when you divide by $x-2$ . That's

\begin{aligned} p(2) &=3(2)^3-2(2)^2+5(2)-7\\[1.4em] &=24-8+10-7\\[1.4em] &=19. \end{aligned}

In Desmos, type p(x)=3x^3-2x^2+5x-7 on one line and p(2) on the next, and Desmos shows $19$ .

The calculator also has a third line, p(-1). It gives the remainder for a different divisor, $x+1$ , whose root is $-1$ , and it comes out to $-17$ . So once $p$ is defined, each new remainder costs you one short line.

If the question asks only for the remainder, you can skip the quotient entirely.

Check your understanding:

If $F(x)=2x^3+x-4$ , what’s the remainder when $F(x)$ is divided by $x+2$ ?

Common mistake:

Dividing by $x+2$ but plugging in $2$ . It’s the same sign flip as before: $x+2$ equals $0$ at $x=-2$ , so that’s your input.

Calculator loads as you approach

p(2) is the remainder for x - 2, and p(-1) is the remainder for x + 1.

Example: Test a factor without factoring first

Worked example

Which choice is a factor of

2x^3-5x^2-18x+45?

A
$2x+5$
B
$x^2+9$
C
$2x-5$
D
$x+5$

Step 1

Find where each choice equals zero

Set each linear choice equal to $0$ :

\begin{aligned} 2x+5=0&\Rightarrow x=-\frac52,\\[1.4em] 2x-5=0&\Rightarrow x=\frac52,\\[1.4em] x+5=0&\Rightarrow x=-5. \end{aligned}

Choice B, $x^2+9$ , is quadratic, so a single input can't test it. Start with the linear choices.

Step 2

Plug in each root

Define

p(x)=2x^3-5x^2-18x+45.

Then evaluate $p$ at each of those roots. Desmos shows

\begin{aligned} p\left(-\frac52\right)&=27.5,\\[1.4em] p\left(\frac52\right)&=0,\\[1.4em] p(-5)&=-240. \end{aligned}

Only $\frac52$ gives zero, so the factor is $2x-5$ .

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p(5/2) = 0, and y = 2x - 5 crosses the x-axis at the same point as the polynomial.

Step 3

Choose the factor

Since $p\left(\frac52\right)=0$ , the Factor Theorem says $x-\frac52$ is a factor. Multiply it by $2$ and you get $2x-5$ , the same factor without the fraction.

The answer is C.

Step 4

Check it by factoring

Grouping shows how the whole polynomial breaks apart:

\begin{aligned} 2x^3-5x^2-18x+45 &=x^2(2x-5)-9(2x-5)\\[1.4em] &=(2x-5)(x^2-9)\\[1.4em] &=(2x-5)(x-3)(x+3). \end{aligned}

So plugging in found the zero, the graph showed where it sits, and the factoring confirms every piece.

Common mistake:

Testing $p(5)$ because you see a $5$ in $2x-5$ . You’d get $p(5)=80$ and wrongly rule out C. The root of $2x-5$ is $\frac52$ .

Know what one root test proves

You now have a fast, reliable test. For a linear factor, one exact zero settles it: if $P(c)=0$ , then $x-c$ is a factor.

The tricky part is knowing where that stops. It's tempting to ask one zero to prove more than it can, so here are the two places it falls short.

A repeated factor needs more than one zero

Say a question claims $(x-c)^2$ is a factor. Checking $P(c)=0$ tells you there's at least one copy of $x-c$ . It doesn't tell you there are two. A short way to remember it: one zero, one copy.

Sometimes that check can't help at all. For

p(x)=x^3+ax^2-9x-9a,

plugging in $3$ gives $p(3)=0$ for every value of $a$ . So it can't tell you when $(x-3)^2$ is a factor.

Group the terms instead:

\begin{aligned} p(x) &=x^2(x+a)-9(x+a)\\[1.4em] &=(x^2-9)(x+a)\\[1.4em] &=(x-3)(x+3)(x+a). \end{aligned}

One copy of $x-3$ is always there. A second copy shows up only when

x+a=x-3,

so $a=-3$ .

A nonlinear factor needs algebraic proof

Now say the possible factor is a quadratic, or has any power of $x$ above $1$ . Sharing one real root with the polynomial doesn't prove it's a factor. A quadratic can have two roots, one repeated root, or no real roots at all, and one matching root doesn't mean the others match too. Choice B in the last example, $x^2+9$ , has no real roots, so there isn't even an intercept to check.

To prove a factor like that, use an exact method, such as factoring, polynomial division or matching coefficients. A graph can hint at the answer, but it shouldn't be your only evidence.

Check your understanding:

Why doesn’t $P(4)=0$ prove that $(x-4)^2$ is a factor of $P(x)$ ?

Common mistake:

Seeing a graph touch the $x$ -axis and taking the picture as proof of a squared factor. The touch is a good clue, but confirm the repeated factor with algebra.

Finish the solution

Here's the Factor Theorem working on an unknown constant. The polynomial and the line $y=x-3$ are already graphed, with $k$ set to $0$ . Find the $k$ that makes the remainder $0$ , then submit it.

Make the stated factor work

Finish the solution

Let $k$ be a constant such that the polynomial $p$ defined by

p(x)=2x^3+kx^2-18x+9

is divisible by $x-3$ . What is the value of $k$ ?

First steps

Dividing evenly by $x-3$ means $p(3)=0$ .
Right now $k=0$ , so $p(3)$ isn't $0$ yet.
Change $k$ until $p(3)=0$ and the polynomial crosses the $x$ -axis at $3$ , right where $x-3$ does.

Finish it

Your answer

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Change k until p(3) = 0 and both graphs cross the x-axis at x = 3.

Practice problems

Now it's your turn. These three problems each lean on a different idea, so before you type anything, decide what each one needs: one evaluation, a graph, or some algebra. If you get stuck, go back to the question that started it all: where does the factor equal zero?

Find and check a factor

Practice problem

One of the factors of the polynomial

x^4-3x^3-24x^2+28x+48

is $x+b$ , where $b$ is a positive constant. Which choice gives the largest possible value of $b$ ?

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Graph it, read the roots, then check your pick exactly.

Test the second copy of a factor

Practice problem

Which choice gives the value of $a$ such that $(x+2)^2$ is a factor of

p(x)=x^3+(a+2)x^2+(2a+6)x+12?

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Check p(-2), then find when the quotient gives a second factor of x + 2.

Combine known roots with a remainder

Practice problem

The function $f$ is defined by

f(x)=x^3+ax^2+bx+c,

where $a$ , $b$ , and $c$ are real constants. The zeros of the function include $x=2$ and $x=-1$ . When $f(x)$ is divided by $x-3$ , the remainder is $20$ . What is the value of $f(0)$ ?

Your answer

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Build the cubic from its roots, use f(3) = 20, then find f(0).

Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Ask where the whole factor equals zero. That's the input you test.
$x-c$ is a factor exactly when $c$ is a root, which is exactly when dividing by $x-c$ leaves remainder $0$ .
To find possible roots of a polynomial with no unknown constants, graph it, then apply conditions like positive or smallest.
The remainder after dividing by $x-c$ is $P(c)$ , so one evaluation answers it.
Graph intercepts are strong guesses until you plug them in and get exactly $0$ .
One zero, one copy. A zero proves one linear factor, not a repeated or higher-degree one.
When grouping or a common factor settles the question quickly, do it by hand.

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