*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) ÷ d(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 (the method we want to avoid):

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

- f(x) is 2x
^{2}−5x−1 - d(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) = d(x) q(x) + r(x) we can write:

2x^{2}−5x−1 = (x−3)(2x+1) + 2

And there is a key feature:

The degree of r(x) is always less than d(x)

Say 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").

We will use that idea in the "Remainder Theorem".

## The Remainder Theorem

When we divide f(x) by the simple polynomial x−c we get:

f(x) = (x−c) q(x) + r(x)

x−c is **degree 1**, so r(x) must have **degree 0**, so it is just some constant r:

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 is f(c)

So to find 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: The remainder after 2x^{2}−5x−1 is divided by x−3

(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: The remainder after 2x^{2}−5x−1 is divided by x−5

Similar to our example above but this time we divide by "x−5"

"c" is 5, so let us check f(5):

2(5)^{2}−5(5)−1 = 2x25−5x5−1

= 50−25−1

= **24**

The remainder is **24**

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

## 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!

We see this when dividing whole numbers. For example 60 ÷ 20 = 3 with no remainder. So 20 must be a factor of 60.

### 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 f(x)

*And the other way around, too:*

When x−c is a factor of f(x) 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: Find the factors of 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. We could try some other values near by and maybe get lucky.

But at least we know (x−2) is a factor.

Let us check using Polynomial Long Division:

2x^{2}+3x−1

x−2)2x^{3}− x^{2}−7x+2

2x^{3}−4x^{2}

3x^{2}−7x

3x^{2}−6x

−x+2

−x+2

0

As expected the remainder is zero.

Better still, we are left with the quadratic equation **2x ^{2}+3x−1** which is easy to solve:

It's roots are −1.78... and 0.28..., so the final result is:

2x^{3}−x^{2}−7x+2 = (x−2)(x+1.78...)(x−0.28...)

We were able to solve a difficult polynomial.

## Summary

**The Remainder Theorem:**

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

**The Factor Theorem:**

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

482, 483, 4014, 4015, 484, 485, 4016, 1124, 1125, 4017

Challenging Questions:

96,227,228,229,230,231

Polynomial Long Division Algebra Index