# 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: 2x2-5x-1 divided by x-3

• f(x) is 2x2-5x-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:

2x2−5x−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: 2x2−5x−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: x2−3x−4

f(4) = (4)2−3(4)−4 = 16−12−4 = 0

so (x−4) must be a factor of x2−3x−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: 2x3−x2−7x+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 2x3−x2−7x+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

Challenging Questions: 1 2 3 4 5 6