Factor, root, and remainder tests

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Difficulty: Advanced
Estimated time: 50 minutes
Domains: Advanced Math
Techniques: FactorsRootsRemainder-theoremEvaluationGraphingMethod-choice

What you’ll learn

Connect a linear factor, its root, and a zero remainder.
Translate linear factors with subtraction, addition, or a leading coefficient into the correct test input.
Use graph roots to discover factor candidates.
Use function evaluation to confirm a factor or calculate a remainder.
Read positive, smallest, and largest conditions after finding all eligible roots.
Recognize when repeated or nonlinear factors need stronger structural evidence.
Choose direct substitution, graphing, or algebra based on the shortest conclusive route.

Prerequisites

Why this matters on the SAT

Turn factor wording into one strategic input

SAT questions may ask whether a polynomial is divisible by a linear expression, which linear expression is a factor, or what remainder results from division. These prompts look different, but they are connected by one input.

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 $b$ ?

A
$3$
B
$4$
C
$5$
D
$2$

Fast Desmos solution

Graph the polynomial and inspect its $x$ -intercepts. They occur at

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

The factor $x+b$ equals $0$ when $x=-b$ . Because $b$ is positive, use the negative roots. They give $b=5$ and $b=2$ , so the smallest possible value is

\boxed{2}.

The answer is D.

The graph finds a candidate quickly. Exact substitution confirms it:

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

Desmos displays numerical graph results, so treat intercepts as candidates and confirm exact zeros by substitution before you submit.

Calculator loads as you approach

The polynomial and x + 2 share the x-intercept x = -2.

When should you use these tests?

Look for wording such as:

“ $x-c$ is a factor”;
“the polynomial is divisible by $x-c$ ”;
“ $c$ is a zero”;
“the remainder when divided by $x-c$ ”; or
“the expression can be rewritten with no remainder.”

Then choose the shortest setup:

Use the calculator when…

Direct evaluation: the divisor is given and you need a factor check or remainder.
Graph the polynomial: you need to discover possible real roots or compare several linear-factor choices.

Use algebra when…

Hand factoring: grouping, a special product, or a visible common factor settles the question in one or two lines.
A stronger algebraic test: the claimed factor is repeated, has degree greater than $1$ (its highest exponent is greater than $1$ ), or contains unknown constants that one numerical graph cannot prove.

Choose the method that reaches an exact conclusion with the least setup. A valid method is not always the fastest method.

Check your understanding:

Which is the shorter opening method: graphing a fourth-degree polynomial to discover its real roots, or evaluating $P(3)$ when the question asks for the remainder after division by $x-3$ ?

1. Connect factors, roots, and remainders

Suppose a polynomial $P(x)$ is divided by $x-c$ . Polynomial division gives

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

where $Q(x)$ is the quotient and $r$ is the constant remainder.

Now substitute $x=c$ :

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

This produces the Remainder Theorem:

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

If the remainder is $0$ , then

P(x)=(x-c)Q(x),

so $x-c$ is a factor. This gives the Factor Theorem:

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

The same statement can be read three ways:

$x-c$ 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 mean the same input: a value $c$ for which $P(c)=0$ . When $c$ is real, that input appears on the graph as the $x$ -intercept $(c,0)$ . Complex roots do not appear as $x$ -intercepts on the real coordinate plane.

For example,

P(x)=x^2-5x+6.

Since

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

$2$ is a root, $x-2$ is a factor, and division by $x-2$ has remainder $0$ .

Check your understanding:

The polynomial $H$ satisfies $H(-4)=0$ . State the corresponding root, linear factor, and remainder statement.

Common mistake:

Thinking that the factor and root have the same sign. The factor $x+4$ produces the root $-4$ because $x+4=0$ at $x=-4$ .

2. Translate the whole linear factor

Before testing a factor, solve the factor equation itself.

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

x=-\frac{b}{a}.

Do not ignore the coefficient of $x$ . For example,

2x-5=0

gives

x=\frac52,

not $5$ .

The factor test for $2x-5$ is therefore

P\left(\frac52\right)=0.

This translation also works for remainders. When a polynomial is divided by $ax+b$ , the constant remainder is

P\left(-\frac{b}{a}\right).

Check your understanding:

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

Common mistake:

Substituting the constant printed in the factor without first solving the factor equation. Always ask, “Where does this entire linear expression equal zero?”

3. Use graph roots to discover factor candidates

If the factor contains an unknown constant or the choices list several possible linear factors, graph the polynomial and inspect its real roots.

For a candidate factor $x-c$ :

Graph $y=P(x)$ .
Click each relevant $x$ -intercept.
Read its $x$ -coordinate as a candidate value of $c$ .
Apply every condition such as positive, negative, smallest, or largest.
Substitute the candidate into $P$ to confirm $P(c)=0$ exactly.

For a candidate factor $x+b$ , remember that its root is $-b$ . A negative polynomial root can therefore produce a positive value of $b$ .

The opening example had roots $-5,-2,3$ , and $4$ . Only the negative roots can equal $-b$ for positive $b$ , so

b=5\quad\text{or}\quad b=2.

The phrase smallest possible selects $b=2$ .

Graphing the candidate linear factor can make the relationship visible. When $b=2$ , the graphs of

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

and

y=x+2

share the $x$ -intercept $(-2,0)$ . For a linear candidate, that shared root is the input used by the Factor Theorem.

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 largest?

Try it yourself:

Change the second line from y=x+2 to y=x+5. Confirm that the shared intercept moves from $x=-2$ to $x=-5$ , then reset the example.

Calculator loads as you approach

The polynomial and x + 2 share the x-intercept x = -2.

4. Evaluate once to find a remainder

When the divisor is already known, graphing the full polynomial is usually unnecessary. Define the polynomial, then evaluate it at the divisor’s root.

Suppose

p(x)=3x^3-2x^2+5x-7.

The remainder after division by $x-2$ is

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

In Desmos, enter:

p(x)=3x^3-2x^2+5x-7

Then enter:

p(2)

The third calculator line shows the sign change for a divisor of $x+1$ :

p(-1)=-17.

You do not need to find the quotient when the question asks only for the remainder.

Check your understanding:

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

Common mistake:

Dividing by $x+2$ but evaluating at $2$ . The correct input is the root of the divisor, so $x+2$ requires $x=-2$ .

Calculator loads as you approach

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

5. 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

Turn each linear factor into an input

Solve each linear candidate equal to $0$ :

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

Choice B is quadratic, so the one-input linear-factor test does not apply to it. Test the linear choices first.

Step 2

Evaluate the polynomial

Define

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

Then evaluate $p$ at the candidate roots. Desmos reports

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

Only $\frac52$ produces zero, so the corresponding linear factor is $2x-5$ .

Calculator loads as you approach

p(5/2) = 0, and y = 2x - 5 shares that x-intercept with the polynomial.

Step 3

Choose the factor

Because

p\left(\frac52\right)=0,

the Factor Theorem confirms that $x-\frac52$ is a factor. Multiplying that factor by $2$ gives the equivalent linear factor $2x-5$ .

The answer is C.

Step 4

Verify the structure

Factoring by grouping confirms the result:

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

Direct evaluation found the zero, the graph displayed the shared intercept, and the factorization verifies the complete structure.

Common mistake:

Testing $p(5)$ because the factor contains the number $5$ . The root of $2x-5$ is $\frac52$ , not $5$ .

6. Know what one root test proves

For a linear candidate factor, one exact zero is conclusive:

P(c)=0\iff x-c\text{ is a factor}.

Do not extend that conclusion too far.

A repeated factor needs repeated-root evidence

If a question claims that $(x-c)^2$ is a factor, checking $P(c)=0$ proves only that at least one copy of $x-c$ is present. It does not prove the square.

For

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

substitution gives $p(3)=0$ for every value of $a$ . That fact cannot determine when $(x-3)^2$ is a factor.

Use the visible algebraic structure instead:

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

One copy of $x-3$ is always present. A second copy appears only when

x+a=x-3,

so $a=-3$ .

A nonlinear candidate needs structural evidence

If a candidate factor has degree greater than $1$ , one shared real root is not enough to prove divisibility. A quadratic candidate may have two roots, a repeated root, or no real roots at all.

Use factoring, polynomial division, coefficient matching, or another exact structural method. A numerical graph can suggest the relationship, but it should not be your only evidence.

Check your understanding:

Why does $P(4)=0$ fail to prove that $(x-4)^2$ is a factor of $P(x)$ ?

Common mistake:

Seeing a graph touch the $x$ -axis and treating the picture as exact proof of a squared factor. The graph is useful evidence, but the repeated factor must be confirmed algebraically.

Finish the solution

The polynomial and the candidate factor are already graphed with $k=0$ . Use the zero-remainder condition to correct $k$ , then submit its value.

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

Divisibility by $x-3$ requires $p(3)=0$ .
The calculator starts with $k=0$ , so $p(3)$ is not yet $0$ .
Edit $k$ until $p(3)=0$ and the polynomial shares the root $3$ with $x-3$ .

Finish it

Your answer

Calculator loads as you approach

Change k until p(3) = 0 and the two graphs share the x-intercept x = 3.

Practice problems

SAT practice problems

Choose the shortest conclusive method. A stated linear divisor usually calls for direct evaluation. Unknown roots may call for a graph. Repeated factors call for stronger algebra.

Discover and confirm 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 $b$ ?

Calculator loads as you approach

Graph the polynomial, inspect its real roots, apply the sign condition, and confirm the selected root 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?

Calculator loads as you approach

Check p(-2), then determine when the quotient supplies 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

Calculator loads as you approach

Build a monic cubic from its roots, use f(3) = 20, and evaluate f(0).

Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Translate: Solve the entire linear factor equal to $0$ to find the test input.
Connect: $x-c$ is a factor exactly when $c$ is a root and division by $x-c$ has remainder $0$ .
Discover: Graph a fixed polynomial when you need to find possible real roots, then apply sign and size conditions.
Evaluate: For division by $x-c$ , calculate $P(c)$ directly. It is the remainder.
Confirm: Treat numerical graph outputs as candidates and verify exact zeros by substitution.
Know the limit: One zero proves one linear factor, not a repeated or higher-degree factor.
Choose: Use hand algebra when visible structure settles the question faster.

Next lesson

Isolate a hidden expression

Rename repeated algebraic chunks, solve a smaller system, and recover the exact expression requested.

Start next lesson

Free Practice

What you’ll learn

Prerequisites

Turn factor wording into one strategic input

Fast Desmos solution

When should you use these tests?

Use the calculator when…

Use algebra when…

1. Connect factors, roots, and remainders

2. Translate the whole linear factor

3. Use graph roots to discover factor candidates

4. Evaluate once to find a remainder

5. Example: Test a factor without factoring first

Turn each linear factor into an input

Evaluate the polynomial

Choose the factor

Verify the structure

6. Know what one root test proves

A repeated factor needs repeated-root evidence

A nonlinear candidate needs structural evidence

Finish the solution

Make the stated factor work

First steps

Finish it

Practice problems

SAT practice problems

Discover and confirm a factor

Test the second copy of a factor

Combine known roots with a remainder

4 practice examples left

Quick recap

Isolate a hidden expression

Free question bank practice problems