The goal of this post is clear.

Learn how to find the remainder when dividing by $(x-c)$,
and use that result to decide whether $(x-c)$ is a factor.

Start with the main flow.

• When we divide by $(x-c)$, the remainder must be a constant.
• From $P(x)=(x-c)Q(x)+r$, substituting $x=c$ gives $P(c)=r$.
• So the remainder when dividing by $(x-c)$ is $P(c)$.
• In particular, if $P(c)=0$, then $(x-c)$ is a factor of $P(x)$.

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The Remainder Theorem and Factor Theorem

In the previous post, we learned polynomial division. Now an important question appears.

When the divisor is $(x-c)$, why can we know the remainder without finishing the whole long division?

The answer is the remainder theorem, and the next step is the factor theorem.

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1. What is special about dividing by $(x-c)$?

If we divide a polynomial $P(x)$ by $(x-c)$, writing the quotient as $Q(x)$ and the remainder as $r$, then

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

Since the divisor has degree 1, the remainder must have degree less than 1. That means the remainder is just a constant.

Now substitute $x=c$.

$$P(c)=(c-c)Q(c)+r=0+r=r$$

So,

$$P(c)=r$$

This is the key idea of the remainder theorem. It is not just a formula to memorize; it comes directly from the division statement.

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2. The remainder theorem

2-1. Statement

If a polynomial $P(x)$ is divided by $(x-c)$, the remainder is $P(c)$.

2-2. Why is it useful?

Usually we need long division or synthetic division to find a remainder. But if the divisor has the form $(x-c)$, we can simply substitute $x=c$.

So,

• remainder upon division by $(x-2)$ -> $P(2)$
• remainder upon division by $(x+1)$ -> $P(-1)$
• remainder upon division by $(x-3)$ -> $P(3)$

Be careful with signs. $(x+1)$ means $(x-(-1))$, so we substitute $-1$.

2-3. Example 1: direct substitution

Let

$$P(x)=2x^3-3x+1$$

Find the remainder when $P(x)$ is divided by $(x-2)$.

By the remainder theorem,

$$P(2)=2(2^3)-3(2)+1=16-6+1=11$$

So the remainder is $11$.

2-4. Example 2: missing terms do not matter

Let

$$P(x)=x^4-5x^2+4$$

Find the remainder when dividing by $(x+2)$.

Since $(x+2)=(x-(-2))$, substitute $x=-2$.

$$P(-2)=(-2)^4-5(-2)^2+4=16-20+4=0$$

So the remainder is $0$.

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3. The factor theorem

3-1. Statement

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

This follows directly from the remainder theorem.

• remainder when dividing by $(x-c)$ = $P(c)$
• if $P(c)=0$, the remainder is 0
• if the remainder is 0, the polynomial is divisible by $(x-c)$
• therefore $(x-c)$ is a factor

3-2. Example 1: checking a factor

For

$$P(x)=x^3-6x^2+11x-6$$

check whether $(x-1)$ is a factor.

$$P(1)=1-6+11-6=0$$

So $(x-1)$ is a factor.

Also,

$$P(2)=8-24+22-6=0, \quad P(3)=27-54+33-6=0$$

so $(x-2)$ and $(x-3)$ are also factors. Hence

$$x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$$

3-3. Example 2: not a factor

For

$$P(x)=x^2+2x+5$$

check whether $(x+1)$ is a factor.

Since $(x+1)=(x-(-1))$, we substitute $x=-1$.

$$P(-1)=(-1)^2+2(-1)+5=1-2+5=4$$

Since this is not 0, $(x+1)$ is not a factor. At the same time, we also know that the remainder is $4$.

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4. How to use these theorems in problems

4-1. Finding a remainder quickly

Let

$$P(x)=3x^3-2x^2+x+7$$

Find the remainder when dividing by $(x-1)$.

$$P(1)=3-2+1+7=9$$

So the remainder is $9$.

4-2. Deciding whether something is a factor

For $x^3+2x^2-5x-6$, test whether $(x-2)$ is a factor.

$$P(2)=8+8-10-6=0$$

So $(x-2)$ is a factor.

4-3. Determining an unknown coefficient

If $(x-2)$ is a factor of

$$P(x)=x^3+ax+2$$

find $a$.

By the factor theorem, $P(2)=0$.

$$2^3+2a+2=0$$ $$8+2a+2=0$$ $$2a=-10, \quad a=-5$$

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5. Common mistakes

• substituting $3$ into $(x+3)$
  -> use $-3$, because $(x+3)=(x-(-3))$
• memorizing the two theorems separately
  -> the factor theorem comes directly from the remainder theorem
• saying "'$c$ is a factor" when $P(c)=0$
  -> the factor is $(x-c)$, not $c$
• applying the theorem to any divisor
  -> this direct substitution works for divisors of the form $(x-c)$

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6. Key summary

Concept	Core idea
Remainder theorem	The remainder when dividing $P(x)$ by $(x-c)$ is $P(c)$
Factor theorem	If $P(c)=0$, then $(x-c)$ is a factor of $P(x)$
Connection	The factor theorem is a direct consequence of the remainder theorem
Caution	For $(x+a)$, substitute $-a$

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7. Practice problems

Problem 1

For

$$P(x)=2x^2-3x+4$$

find the remainder when dividing by $(x-1)$.

Show answer

By the remainder theorem, the remainder is $P(1)$.

$$P(1)=2-3+4=3$$

So the remainder is $3$.

Problem 2

For

$$P(x)=x^2+x-6$$

determine whether $(x-2)$ is a factor.

Show answer

$$P(2)=4+2-6=0$$

So $(x-2)$ is a factor.

Problem 3

If $(x+1)$ is a factor of

$$P(x)=x^3+2x^2+ax-4$$

find $a$.

Show answer

Since $(x+1)$ is a factor, $P(-1)=0$.

$$(-1)^3+2(-1)^2+a(-1)-4=0$$ $$-1+2-a-4=0$$ $$-3-a=0$$

Therefore,

$$a=-3$$

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8. Next post

In the next post, we move to equality of polynomials and the zero polynomial.

The remainder theorem and factor theorem connect a polynomial's value with its factors. Next, we will study what it means for two polynomial expressions themselves to be equal.