# Remainder and Factor Theorems

## Key Questions

• You have only learned tools to factor quadratics; for higher degree polynomials, you can use long or synthetic division.

However, long/synthetic division is slow. If you guess the wrong divisor/factor, you will be wasting even more time.

For example: P(x)=2x^2−x−1. (I am using a quadratic for the sake of brevity).

Using the remainder theorem, we can quickly get the remainder with P(2)=2⋅2^2−2−1=8−2−1=5 and P(0)=2⋅0^2−0−1=−1. So, $x - 2$ and $x - 0$ are not factors of $P \left(x\right)$. But consider that $P \left(0\right)$ is below the x-axis and $P \left(2\right)$ is above the x-axis; this means the $P \left(x\right)$ must cross the x-axis somewhere between 0 and 2.

This should lead you to try $P \left(1\right)$ as a factor. P(1)=2⋅1^2−1−1=0, so x−1 is a factor of $P \left(x\right)$ by the factor theorem.

Once you have a factor, you can proceed with long/synthetic division to get your quotient. Then you will need to repeat and factor the quotient. Again, if it is a quadratic, switch to your quadratic factoring methods.

• The remainder theorem states that if you want to find f(x) of any function, you can synthetically divide by whatever "x" is, get the remainder and you will have the corresponding "y" value. Lets go through an example: (I have to assume you know synthetic division)

Say you had the function $f \left(x\right) = 2 {x}^{2} + 3 x + 7$ and wanted to find f(3), rather than plugging in 3, you could SYNTHETICALLY DIVIDE by 3 to find the answer.

To find f(3) you would set up synthetic division so that your "x" value (3 in this case) is in a box on the left and you write out all the coefficients of the function on the right! (Don't forget to add place holders if necessary!) Just as a quick review for synthetic division, you bring the first term down, multiply by number on the left, write your answer in the next column, then add and so on!

After the synthetic division, you notice that the remainder is 34...

If I were to find f(3) by substitution I would get:

$f \left(3\right) = 2 {\left(3\right)}^{2} + 3 \left(3\right) + 7$
$= 18 + 9 + 7$
$= \ast 34 \ast$

Hopefully you notice that the remainder is the same as the answer you get when using substitution! THIS WILL ALWAYS BE THE CASE IF YOU DO THE SYNTHETIC DIVISION CORRECTLY! Hopefully you've understood this! :)

• Your question isn't phrased quite correctly. The remainder theorem is a short cut to find the remainder of polynomial long division or synthetic division.

The remainder theorem only applies if your divisor is a monic linear binomial, that is, $x - a$. If you have a polynomial $P \left(x\right)$ and divide it by $x - a$, then the remainder is $P \left(a\right)$. Note that the remainder theorem doesn't give you the quotient, so you can't use it for questions that are looking for the quotient and remainder.

For example: $P \left(x\right) = 2 {x}^{2} - x - 1$ divided by $x - 2$. If we do long or synthetic division, we get $Q \left(x\right) = 2 x + 3$ and $R \left(x\right) = 5$. But using the remainder theorem, we can quickly get the remainder with $P \left(2\right) = 2 \cdot {2}^{2} - 2 - 1 = 8 - 2 - 1 = 5$.

When we combine the remainder theorem with the factor theorem, we can use it to find/verify the factors of the polynomial. So, $x - 2$ is not a factor of $P \left(x\right)$. But $P \left(1\right) = 2 \cdot {1}^{2} - 1 - 1 = 0$, so $x - 1$ is a factor of $P \left(x\right)$.

If instead, we tried $P \left(0\right) = 2 \cdot {0}^{2} - 0 - 1 = - 1$, so $x - 0$ is not a factor. But consider that $P \left(0\right)$ is below the x-axis and $P \left(2\right)$ is above the x-axis; this means the $P \left(x\right)$ must cross the x-axis somewhere between 0 and 2. This should lead you to try $P \left(1\right)$ as a factor.

See explanation

#### Explanation:

Suppose you have an equation. For example: $y = {x}^{2} - x - 12$

In this case if we set $y = 0$ and substitute $4$ for $x$ then we have: $y = {\left(4\right)}^{2} - \left(4\right) - 12 = 16 - 4 - 12 = 0$

So if the equation equals 0 $\to f \left(x\right) = 0$

and by substituting $x = 4 \to f \left(4\right)$ we get the answer $f \left(4\right) = 0$

then by using $\left(x - 4\right) \left(\text{something else") ->0("something else}\right) = 0$
So $\left(x - 4\right)$ is a factor of $f \left(x\right)$