# Remainder Theorem

and Factor Theorem

*Or: how to avoid Polynomial Long Division when finding factors*

Do you remember doing division in Arithmetic?

* "7 divided by 2 equals 3 with a remainder of 1"*

Each part of the division has names:

Which can be **rewritten** as a sum like this:

## Polynomials

Well, we can also divide polynomials.

f(x) ÷ g(x) = q(x) with a remainder of r(x)

But it is better to write it as a sum like this:

Like in this example using Polynomial Long Division:

### Example: 2x^{2}-5x-1 divided by x-3

- f(x) is 2x
^{2}-5x-1 - g(x) is x-3

After dividing we get the answer 2x+1, but there is a remainder of 2.

- q(x) is 2x+1
- r(x) is 2

In the style f(x) = g(x)·q(x) + r(x) we can write:

2x^{2}−5x−1 = (x−3)(2x+1) + 2

But you need to know one more thing:

When we divide by a polynomial of **degree 1** (such as "x−3") the remainder will have **degree 0** (in other words a constant, like "4").

And we will use that idea in the "Remainder Theorem":

## The Remainder Theorem

When we divide a polynomial f(x) by x-c we get:

f(x) = (x−c)·q(x) + r(x)

But r(x) is simply the constant r *(remember? when we divide by (x-c) the remainder is a constant)* .... so we get this:

f(x) = (x−c)·q(x) + r

Now see what happens when we have x equal to c:

f(c) = (c−c)·q(c) + r

f(c) = (0)·q(c) + r

f(c) = r

So we get this:

**The Remainder Theorem:**

When we divide a polynomial f(x) by x-c the remainder r equals f(c)

So when we want to know the remainder after dividing by x-c we don't need to do any division:

Just calculate f(c).

Let us see that in practice:

### Example: 2x^{2}−5x−1 divided by x-3

(Continuing our example from above)

We don't need to divide by **(x−3)** ... just calculate **f(3)**:

2(3)^{2}−5(3)−1 = 2x9−5x3−1 = 18−15−1 = **2**

And that is the remainder we got from our calculations above.

We didn't need to do Long Division at all!

### Example: Dividing by x−4

(Continuing our example)

What is the remainder when we divide by "x−4" ?

"c" is 4, so let us check f(4):

2(4)^{2}−5(4)−1 = 2x16−5x4−1 = 32−20−1 = **11**

Once again ... We didn't need to do Long Division to find it.

## The Factor Theorem

Now ...

What if we calculate **f(c)** and it is **0**?

... that means the **remainder is 0**, and ...

... **(x−c) must be a factor** of the polynomial!

### Example: x^{2}−3x−4

f(4) = (4)^{2}−3(4)−4 = 16−12−4 = 0

so (x−4) must be a factor of x^{2}−3x−4

And so we have:

**The Factor Theorem:**

When f(c)=0 then x−c is a factor of the polynomial

*And the other way around, too:*

When x−c is a factor of the polynomial then f(c)=0

## Why Is This Useful?

Knowing that x−c is a factor is the same as knowing that c is a root (and vice versa).

The **factor "x−c"** and the **root "c"** are the same thing

Know one and we know the other

For one thing, it means that we can quickly check if (x-c) is a factor of the polynomial.

### Example: 2x^{3}−x^{2}−7x+2

The polynomial is degree 3, and could be difficult to solve. So let us plot it first:

The curve crosses the x-axis at three points, and one of them **might be at 2**. We can check easily:

**f(2)** = 2(2)^{3}−(2)^{2}−7(2)+2

= 16−4−14+2

= **0**

Yes! **f(2)=0**, so we have found a root **and** a factor.

So (x−2) must be a factor of 2x^{3}−x^{2}−7x+2

How about where it crosses near **−1.8**?

**f(−1.8)** = 2(−1.8)^{3}−(−1.8)^{2}−7(−1.8)+2

= −11.664−3.24+12.6+2

= **−0.304**

No, (x+1.8) is not a factor. But we could try some other values close by.

## Summary

**The Remainder Theorem:**

- When we divide a polynomial f(x) by x−c the remainder equals f(c)

**The Factor Theorem:**

- When f(c)=0 then x−c is a factor of the polynomial
- When x−c is a factor of the polynomial then f(c)=0