Factor Theorem Practice Test

Use polynomial values to test factors, find remainders and parameters, and build or factor polynomials.

Factor Theorem Practice Test

20 factor-theorem and remainder-theorem questions with worked solutions.

Instant feedback · Worked explanations

Factor verification lab

Test the value. Prove the factor.

The Factor Theorem connects four ideas at once: polynomial evaluation, roots, division, and factorization. A candidate linear factor is confirmed by one decisive test—substitute its corresponding root into the polynomial. If the result is zero, the factor is valid; if not, the same value becomes the remainder.

Anchor 01Factor

x - a

corresponds to test value

a

Anchor 02Zero polynomial value proves the factor.

Anchor 03Nonzero polynomial value is the remainder.

Anchor 04After confirmation, synthetic division can produce the quotient.

1. The Factor Theorem is a zero-value test

A candidate linear factor does not become a factor because it looks plausible. It becomes a factor only when its corresponding input makes the polynomial equal zero.

P (a) = 0 \Leftrightarrow (x - a) is a factor

Read candidate factor

→

Solve it equal to zero

→

Evaluate the polynomial

→

Accept factor only if result is zero

Root connection

Factor and zero are paired

If $x - a$ is a factor, then $x = a$ is a root.

Division connection

Zero remainder

An exact factor produces remainder $0$ when the polynomial is divided by that linear factor.

Evaluation connection

One number settles the question

$P (a)$ is enough to confirm or reject the candidate factor.

2. Decode the factor sign before evaluating

The visible sign in the factor is opposite the corresponding root sign. Rewrite the candidate factor as an equation and solve it equal to zero before substituting.

Factor sign decoder

The factor itself tells you which input must be tested.

Subtraction form

x - a = 0 \Rightarrow x = a

Addition form

x + a = 0 \Rightarrow x = - a

Verification ruleSubstitute the solved value into the polynomial and check whether the result is zero.

Factor with subtraction

x - 4

uses test value

4

Factor with addition

x + 3

uses test value

- 3

Quick checkThe chosen input should make the candidate factor itself equal

0

Why sign errors spreadA wrong input changes the value of every odd-powered term.

3. Evaluate carefully, especially at negative values

After the correct test value is identified, the task becomes polynomial evaluation. Parentheses protect the sign of negative inputs under powers.

Positive test value

For $P (x) = x^{3} - 5 x + 4$ and candidate factor $x - 2$ , test $P (2)$ .

{(2)}^{3} - 5 (2) + 4 = 2

The nonzero result rejects the candidate factor.

Negative test value

For factor $x + 2$ , use $- 2$ and keep the input grouped.

{(- 2)}^{3} + 3 {(- 2)}^{2} - 4

Even vs odd powers

${(- 3)}^{2} = 9$ , while ${(- 3)}^{3} = - 27$ . The parentheses are essential.

4. Zero is the factor verdict

The entire theorem turns on one distinction: zero versus nonzero evaluation.

Zero value

P (a) = 0

The candidate linear factor is valid.

Nonzero value

P (a) \neq 0

The candidate is not a factor.

Remainder meaningThe same nonzero value is the remainder on division by

x - a

5. Factor conditions can determine unknown coefficients

If a polynomial contains a parameter and a linear factor is guaranteed, substitute the corresponding root and set the polynomial value equal to zero.

Illustrative parameter equation

Suppose $P (x) = x^{3} + k x^{2} - 4 x + 4$ has factor $x - 2$ . Then:

{(2)}^{3} + k {(2)}^{2} - 4 (2) + 4 = 0

Do not force zero without a factor condition

If the problem states a nonzero remainder instead, set $P (a)$ equal to that stated remainder. The equation equals zero only when exact divisibility or a factor is specified.

P (a) = R

6. Once a factor is confirmed, synthetic division finds the quotient

The Factor Theorem answers the yes-or-no question. Synthetic division then uses the same root value to remove that linear factor and expose the quotient polynomial.

value

1

- 3

- 4

12

3

1

0

- 4

0

Interpretation

Zero final entry

The last entry $0$ confirms that the chosen divisor is a factor.

Quotient

Earlier entries become coefficients

The preceding bottom-row entries form the quotient polynomial in descending powers.

7. Factor Theorem and root language are two views of the same fact

A root, a zero polynomial value, a zero remainder, and a linear factor all encode the same condition.

Root

Polynomial value is zero

$P (a) = 0$ means $a$ is a zero of the polynomial.

Factor

Linear factor appears

The corresponding factor is $x - a$ .

Division

Remainder disappears

Division by that factor produces remainder $0$ .

8. Build a polynomial from known zeros

Known zeros can be converted directly into linear factors. Multiplying those factors constructs a polynomial with the required roots.

Root-to-factor pipeline

Zeros

2, - 3

→

Factors

(x - 2) (x + 3)

→

Polynomial

x^{2} + x - 6

Scaling keeps the same zeros

Multiplying the entire polynomial by any nonzero constant changes coefficients but preserves the same root set.

P (x) = k (x - 2) (x + 3), k \neq 0

9. The Remainder Theorem handles rejected factors too

If the evaluation is nonzero, the test is still useful. The candidate factor is rejected, and the same polynomial value tells you the exact remainder.

P (a) = R

Zero result

Factor confirmed

$R = 0$

Nonzero result

Factor rejected

$R \neq 0$

Division identity

Why the value is the remainder

P (x) = (x - a) Q (x) + R

10. Error analysis: factor tests fail for predictable reasons

Most distractors come from the wrong test value, careless negative substitution, or declaring a factor before the polynomial value has actually been checked.

Factor sign copied directly

Factor $x + 4$ requires test value $- 4$ .

Input confused with remainder

The test input identifies where to evaluate; the resulting polynomial value is the remainder.

Negative powers evaluated incorrectly

Use parentheses around a negative input before applying odd or even powers.

Candidate assumed valid

A factor must be verified by showing $P (a) = 0$ .

Parameter equation set to wrong value

Factor conditions use zero; stated-remainder conditions use the stated remainder.

Stopped after proving the factor

If the quotient is also requested, continue with synthetic or polynomial division.

11. Worked mini-set: one factor decision at a time

These examples are illustrative teaching examples, not questions copied from the test.

Example A

Test a minus-form factor

Candidate factor $x - 5$ is tested with input $5$ .

Example B

Test a plus-form factor

Candidate factor $x + 2$ is tested with input $- 2$ .

Example C

Confirm a factor

If $P (3) = 0$ , then $x - 3$ is a factor.

Example D

Reject a factor

If $P (4) = 7$ , then $x - 4$ is not a factor and the remainder is $7$ .

Example E

Parameter from factor condition

If the evaluation simplifies to $k - 6$ and the divisor is a factor, solve $k - 6 = 0$ .

Example F

Build from roots

Zeros $1$ and $- 4$ produce factors $(x - 1) (x + 4)$ .

Final Factor Theorem checklist

Before accepting a factor, make sure the test value came from solving the factor equation and that the evaluated polynomial actually equals zero.

Did I solve the candidate factor equal to zero?This gives the correct test input.

Did I substitute carefully?Parenthesize negative values before applying powers.

Is the polynomial value exactly zero?Only zero confirms a factor.

If the value is nonzero, did I interpret it as the remainder?The candidate is rejected, but the evaluation still answers a division question.

For a parameter, did I use the correct condition?Factor means zero remainder; a stated remainder means set the value equal to that number.

If a quotient is requested, did I divide after verification?Synthetic division is often the fastest next step.