Welcome! Today we're exploring The Remainder Theorem: Substitution Instead of Division — a powerful shortcut that will save you tons of time.

Imagine you need to find the remainder when a cubic is divided by a linear polynomial.

Long division takes half a page.

The Remainder Theorem gives the same answer in one substitution.

Method	Effort
Long Division	Half a page
Remainder Theorem	One step

In this lesson, you will learn:

Why the theorem works
How to apply it
How to find unknown coefficients from remainder conditions

1. Applying the remainder theorem

📚 The Remainder Theorem: Substitution Instead of Division

Imagine you need to find the remainder when a cubic polynomial is divided by a linear polynomial. Long division takes half a page of work!

But here's the beautiful shortcut:

The Remainder Theorem converts a division problem into a simple substitution problem.

The Remainder Theorem:

When $p (x)$ is divided by $(x - a)$ , the remainder is simply $p(a)$ .

One evaluation replaces an entire long division. No messy calculations needed!

✍️ Question

Let's see if you can apply this! 🎯

What is the remainder when $p (x) = x^{3} - 4 x^{2} + 5 x - 2$ is divided by $(x - 2)$ ?

Think about what value you need to substitute.

The Remainder TheoremYour shortcut for finding remainders:

When dividing a polynomial $p (x)$ by $(x - a)$ , the remainder is simply $p(a)$ Substitute and evaluate instead of dividing.

What this means:

Instead of performing long division, just substituteThe divisor tells you what value to plug in $x = a$ into the polynomial
The value you get is the remainderOne substitution gives you the answer!

In our problem: To find the remainder when $p (x) = x^{3} - 4 x^{2} + 5 x - 2$ is divided by $(x - 2)$ a=2This tells you to use x equals 2, we simply calculate $p (2)$ .

✍️ MCQ

Choose one

To find the remainder when

p(x) = x^3 - 4x^2 + 5x - 2

is divided by

(x - 2)

, what value do we substitute for

x

?

✓

For $p (x) = x^{3} - 4 x^{2} + 5 x - 2$ , divided by $(x - 2)$ The divisor tells you what value to substitute:

By the Remainder Theorem, we find $p(2)$ keyImmediately know to substitute 2:

Step 1: Substitute $x = 2$ into $p (x)$ :

p(2) = (2)^3 - 4(2)^2 + 5(2) - 2

✍️ MCQ

Choose one

What is

(2)^3

?

✓

Step 2: Calculate each term:

$(2)^3 = 8$
$4(2)^2 = 4 \times 4 = 16$
$5(2) = 10$

✍️ FIB

Fill in the blank

What is

8 - 16 + 10 - 2

?

✓

Step 3: Combine all terms:

$p (2) = 8 - 16 + 10 - 2 = 0$

\therefore \text{Remainder} = 0

(Your signal that the divisor is actually a factor)

The remainder is $0$ .

✍️ MCQ

Choose one

If we had divided

p(x) = x^3 - 4x^2 + 5x - 2

by

(x - 3)

instead, what would we substitute to find the remainder?

✓

When the remainder is $0$ , $(x - 2)$ is a factor of $p (x)$ .

This is exactly the Factor TheoremThe key theorem for checking factors on exams: it's the special case of the Remainder TheoremSame idea, just a special case where the remainder equals $0$ .

Factor Theorem: If $p(a) = 0$ key conditionThis is the test for checking factors, then $(x - a)$ is a factor of $p (x)$ .

✍️ MCQ

Choose one

Since

p(2) = 0

for

p(x) = x^3 - 4x^2 + 5x - 2

, which of the following is true?

✓

Why it works: The division algorithmThe starting point that makes everything work says $p (x) = q (x) (x - a) + R$ .

This is the foundation — any polynomial division can be written as: dividend equals quotient times divisor plus remainderThe structure that makes the Remainder Theorem possible.

Substitute $x = a$ :

p(a) = q(a)(a - a) + R = q(a) \cdot 0 + R = 0 + R = R

(Makes the divisor term become zero and vanish)

So $p(a) = R$ key resultThe value you get by substituting IS the remainder. That's the entire proof — one clever substitution reveals the remainderRemember this logic for proofs in exams!

✍️ MCQ

Choose one

In the proof, why does the term

q(a)(a - a)

become zero?

✓

2. Finding unknowns from remainder conditions

Setting the Scene 📐

When a problem says "the remainder when $p (x)$ is divided by $(x - a)$ is $R$ ", the Remainder Theorem translates this directly into:

p(a) = R

This gives us an equation for unknown coefficients — no long division needed.

📋 Given Info

Key Point ⚠️

If the remainder when $p (x)$ is divided by $(x - 3)$ is 21, then:

p(3) = 21

Key Result

Not $p(3) = 0$ — that would mean $(x - 3)$ is a factor, which is the special case when remainder is zero.

✍️ Question

Your Turn ✏️

The remainder when $x^3 + 2x^2 + kx + 3$ is divided by $(x - 3)$ is 21given.

Find $k$ .

The Remainder TheoremYour shortcut for finding remainders states:

When a polynomial $p (x)$ is divided by $(x - a)$ , the remainder equals $p(a)$ Substitute a to get the remainder.

Key Insight:Important understanding about remainders The remainder doesn't have to be zero!

When the remainder is 21givenThe given remainder value (not 0), we simply set:

p(3) = 21

(Set up this equation from the remainder)

This gives us an equation we can solve for the unknown $k$ Solve this equation for k.

✍️ MCQ

Choose one

For the polynomial

p(x) = x^3 + 2x^2 + kx + 3

, what expression do we get when we substitute

x = 3

?

✓

Let's work through it:

p(3) = (3)^3 + 2(3)^2 + k(3) + 3 = 21

(Plug in the value instead of doing long division)

27

+

18

+

3k

+ 3 =

21

remainder

48

+ 3k = 21

✍️ MCQ

Choose one

To isolate

k

from

48 + 3k = 21

, what should we do next?

✓

$3 k = 21 - 48 =$ $- 27$

k = -9

(Setting up the equation with remainder, then solving)

Note: This is NOT the factor theorem(Factor theorem only works when remainder is zero) (which requires remainder = 0Factor theorem needs zero remainder).

The remainder is 21Our specific remainder value, so $(x - 3)$ is NOT a factorNon-zero remainder means not a factor.

The remainder theorem is more generalWorks no matter what remainder you get — it works for any remainder valueAny remainder value works.

3. Connecting remainder theorem and factor theorem

The factor theorem is actually a special case of the remainder theorem — it applies when the remainder $R = 0$ .

Understanding this connection clarifies when each theorem applies and prevents confusion between the two.

Quick recap:

Remainder Theorem: The remainder when dividing $p (x)$ by $(x - a)$ is $p(a)$ .
Factor Theorem: $(x - a)$ is a factor of $p (x)$ if and only if $p(a) = 0$ .

✍️ Question

Question 🤔

For $k = - 9$ , is $(x - 3)$ a factor of $x^{3} + 2 x^{2} - 9 x + 3$ ?

Justify your answer using the appropriate theorem.

Let's compute $p(3)$ for $x^3 + 2x^2 - 9x + 3$ :

p(3) = (3)^3 + 2(3)^2 - 9(3) + 3

(Just substitute — no long division needed)

$p (3) = 27 + 18 - 27 + 3 =$ $21$ result

✍️ Yes/No

Yes or No?

Since

p(3) = 21

, is

(x - 3)

a factor of

x^3 + 2x^2 - 9x + 3

?

✓

The remainder when dividing by $(x - 3)$ is 21Check if remainder equals zero.

Since $21 \neq 0$ Since 21 is not zero, not a factor, by the Factor Theorem(Factor Theorem connects remainder to factors), $(x - 3)$ is NOTRemainder non-zero means not a factor a factor of $x^{3} + 2 x^{2} - 9 x + 3$ .

Here's the connection between the two theorems:

Remainder TheoremWorks for any remainder value: remainder $= p (a)$ . This works for any remainder(Any remainder value works).
Factor TheoremOnly applies when remainder is zero: $(x - a)$ is a factor when $p (a) = 0$ . This is the special caseApplies when remainder is specifically zero of the remainder theorem where the remainder is specifically 0.

So: the remainder theorem is the general toolFor finding any remainder. The factor theorem is a specific applicationWhen you want to check if something divides evenly when remainder $= 0$ .

Theorem	What it tells you	When to use
Remainder Theorem	$p (a)$ = remainder when dividing by $(x - a)$	Finding any remainderUse for any remainder
Factor Theorem	$(x - a)$ is a factor $\Leftrightarrow p (a) = 0$	Checking if something divides evenlyWhen remainder equals zero

✍️ MCQ

Choose one

If

p(2) = 0

for some polynomial

p(x)

, which statement is correct?

✓