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:

7 = 2 times 3 + 1


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:

f(x) = g(x) times q(x) + r(x)

Like in this example using Polynomial Long Division:

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

polynomial long division

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


The Remainder Theorem:

The Factor Theorem:

Challenging Questions: 1 2 3 4 5 6