FACTOR THEOREM  ·  A2.A.APR.A.1  ·  p(a) = 0  ·  (x − a) IS A FACTOR  ·  ZEROS OF POLYNOMIALS  ·  REMAINDER THEOREM  ·  SYNTHETIC DIVISION  ·  ROOTS & FACTORS  ·  POLYNOMIAL FUNCTIONS  ·  TN ALGEBRA II    
TN ALGEBRA II · A2.A.APR.A.1

FACTOR
LAB

Master the Factor Theorem — one of the most powerful tools for breaking polynomials apart and finding their zeros.

A2.A.APR.A.1 — Factor Theorem & Polynomial Zeros
THE FACTOR THEOREM
For a polynomial p(x) and a number a:
p(a) = 0  ⟺  (x − a) is a factor of p(x)

What Is the Factor Theorem?

It's a shortcut: instead of doing long division, you can just plug in a number to test if something is a factor.

CLICK ANY CARD TO FLIP IT

THE BIG IDEA

Testing Factors

To test if (x − a) divides evenly into a polynomial, just evaluate p(a). If the result is zero — it's a factor!

Flip for an example

Is (x − 2) a factor of p(x) = x² − 5x + 6?

Plug in a = 2:  p(2) = 4 − 10 + 6 = 0

✓ Yes! Zero means (x − 2) IS a factor.

ZEROS = FACTORS

Zeros & Factors Are Twins

A zero of a polynomial is an x-value that makes it equal zero. Every zero a gives you a factor (x − a).

Flip for more

If p(3) = 0, then:

x = 3 is a zero of p(x)

AND

(x − 3) is a factor of p(x)

These two facts are always linked!

WHY IT MATTERS

Smarter Than Long Division

Instead of grinding through polynomial long division, you can test potential factors in seconds by just evaluating the polynomial at one point.

Flip for context

The Factor Theorem is a special case of the Remainder Theorem. When you divide p(x) by (x − a) and the remainder is zero, you've found a factor — and the Factor Theorem tells you this without doing the division!

ON THE GRAPH

Zeros = x-Intercepts

On a graph, zeros of a polynomial are where the curve crosses the x-axis. The Factor Theorem connects those crossing points to the factored form.

Flip for more

If a parabola crosses the x-axis at x = 2 and x = −3, then:

p(x) = (x − 2)(x + 3)

The Factor Theorem tells you the factors directly from the graph!

REAL WORLD ANALOGY

Like Testing a Lock

The polynomial = a locked door.

A candidate factor (x − a) = a key you want to try.

Plugging in a: if p(a) = 0, the key fits — it's a factor! If not, try a different key.

p(a) = 0  →  key fits!

The Theorem — Both Directions

The Factor Theorem works both ways. This is called a biconditional — it's a two-way street.

DIRECTION 1 →

If p(a) = 0, then (x − a) is a factor

Plug a into the polynomial. If you get zero, you've confirmed (x − a) divides evenly. You found a factor!

p(2) = 0  ⟹  (x − 2) | p(x)
DIRECTION 2 ←

If (x − a) is a factor, then p(a) = 0

If you know (x − a) divides evenly into the polynomial, then a must be a zero — plugging it in always gives zero.

(x − 2) | p(x)  ⟹  p(2) = 0
BICONDITIONAL — "IF AND ONLY IF"
p(a) = 0   ⟺   (x − a) is a factor of p(x)

The double arrow (⟺) means BOTH directions are always true at the same time.

Quick Check Examples

IS IT A FACTOR?

p(x) = x² − 4

Is (x − 2) a factor?

Flip to check

Test a = 2:

p(2) = (2)² − 4 = 4 − 4 = 0

✓ YES — (x − 2) is a factor!

IS IT A FACTOR?

p(x) = x³ − 8

Is (x − 3) a factor?

Flip to check

Test a = 3:

p(3) = (3)³ − 8 = 27 − 8 = 19

✗ NO — remainder is 19, not 0.

FIND THE FACTOR

p(x) = x² − x − 6

If p(3) = 0, what's a factor?

Flip to see

Since p(3) = 9 − 3 − 6 = 0:

(x − 3) is a factor

Divide out to get: (x − 3)(x + 2)

Zeros: x = 3 and x = −2

How To Use It

Here's the systematic approach for applying the Factor Theorem on any problem.

1

Identify the polynomial and the candidate factor

You're given p(x) and asked if (x − a) is a factor. Pull out the value a. Watch the sign — if the factor is (x + 3), then a = −3.

2

Evaluate p(a) — plug in the value

Substitute a into the polynomial and simplify carefully. Use the order of operations. This is the key step — arithmetic errors here will throw off everything.

3

Read the result

If p(a) = 0(x − a) IS a factor. If p(a) ≠ 0 → it is NOT a factor. The answer is always one of these two outcomes.

4

(Optional) Factor the polynomial completely

Once you confirm a factor, use synthetic division or polynomial long division to divide it out and find the remaining factor(s). Then factor those further if possible.

Worked Examples

EXAMPLE 1 — BASIC
Is (x − 1) a factor of p(x) = x³ − 6x² + 11x − 6 ?
Factor is (x − 1), so test a = 1
Evaluate: p(1) = (1)³ − 6(1)² + 11(1) − 6
= 1 − 6 + 11 − 6 = 0
✓ YES — (x − 1) is a factor
EXAMPLE 2 — NEGATIVE VALUE
Is (x + 2) a factor of p(x) = x³ + x² − 4x − 4 ?
Factor is (x + 2) = (x − (−2)), so test a = −2
Evaluate: p(−2) = (−2)³ + (−2)² − 4(−2) − 4
= −8 + 4 + 8 − 4 = 0
✓ YES — (x + 2) is a factor
EXAMPLE 3 — NOT A FACTOR
Is (x − 3) a factor of p(x) = x² + 2x + 1 ?
Factor is (x − 3), so test a = 3
Evaluate: p(3) = (3)² + 2(3) + 1
= 9 + 6 + 1 = 16 ≠ 0
✗ NO — remainder is 16, not a factor
EXAMPLE 4 — FIND ALL ZEROS
Factor completely: p(x) = x³ − 6x² + 11x − 6
Test x = 1: p(1) = 0 ✓ so (x − 1) is a factor
Divide: x³ − 6x² + 11x − 6 ÷ (x − 1) = x² − 5x + 6
Factor the quotient: x² − 5x + 6 = (x − 2)(x − 3)
Complete factorization: (x − 1)(x − 2)(x − 3)
Zeros: x = 1, x = 2, x = 3

Try It Live

Pick a polynomial, enter any value of a, and see instantly whether (x − a) is a factor.

CHOOSE A POLYNOMIAL
SELECTED POLYNOMIAL
p(x) = − 5x + 6
Test value a =
0
PRO TIP

When hunting for rational zeros, try factors of the constant term first (both positive and negative). For example, if the constant is 6, try ±1, ±2, ±3, ±6. This is the Rational Root Theorem working alongside the Factor Theorem.

The Remainder Connection

The Factor Theorem is actually a special case of the Remainder Theorem. Understanding both deepens your mastery.

REMAINDER THEOREM

When p(x) ÷ (x − a)...

The remainder equals p(a). You can find any remainder just by plugging in — no long division needed.

Remainder = p(a)
special case:
remainder = 0
FACTOR THEOREM

When p(a) = 0...

The remainder is zero, meaning (x − a) divides evenly — it's a factor. Zero remainder = perfect division.

p(a) = 0 ⟹ factor
SIDE-BY-SIDE COMPARISON
REMAINDER THEOREM — Example
p(x) = x² + 3x + 1
Divide by (x − 2):
p(2) = 4 + 6 + 1 = 11
Remainder = 11 (not a factor)
FACTOR THEOREM — Example
p(x) = x² − 5x + 6
Divide by (x − 2):
p(2) = 4 − 10 + 6 = 0
Remainder = 0 → it IS a factor!
KEY TAKEAWAY

Think of the Remainder Theorem as the general rule and the Factor Theorem as the special case when everything works out perfectly. Every time you apply the Factor Theorem, you are secretly using the Remainder Theorem — you just already know the remainder will be zero.

Key Vocabulary

Every term you need to know for A2.A.APR.A.1 — searchable and student-friendly.

Quiz Yourself

Test your mastery of A2.A.APR.A.1 — the Factor Theorem!

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