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If your teacher presented you with the following problem, what would you think? Would you know how to solve it? It certainly does not look like something you would see in a math class. It looks more like a secret code. However, it is the remainder theorem formula.
This remainder theorem formula is used with polynomials. By definition, a polynomial is an expression that contains coefficients and variables.
Simply put, it is an expression with letters and numbers in it. Sometimes, there are more letters than numbers, but the letters represent numbers.
Now, keep in mind that a theorem, as in the remainder theorem, is a proof. This means you are trying to prove the theory and not necessarily get a quick numerical answer.
Before we get into the remainder theorem, let’s find out what exactly a polynomial is. Look at the polynomial below. As you can see in the example below, there are variables (letters) and coefficients (numbers) as well as exponents.
The coefficients are the number (or they could be letters) that are shown first before the variable. The variable is almost always a letter that represents a number.
Now, one other thing you will need to know about the remainder theorem is that a linear factor is involved. So, what is a linear factor? In layman’s terms, a linear factor is just a simple equation like the one shown below. It is also called a degree of 1.
To recap, you have learned about the two parts of a remainder theorem, the polynomial, and the linear factor. Now, it is time to dive into the formula.
Let’s look again at the remainder theorem formula. In this formula, a polynomial is divided by a linear factor. The “p(x)” represents a polynomial with the variable of an unknown amount. With the remainder theorem, you are trying to find the remainder from a division problem.
Let’s review a division problem to understand the terms first before we go any deeper into this polynomial division problem.
FACT: Think of working a polynomial remainder theorem problem like a long division problem!
The polynomial division problem has similar terms. The first part, p(x), means that this is a polynomial with an unknown x-variable. The linear factor is (x-c), and the “c” means it is just an unknown value, and it could be any number.
The “q” is the quotient, and “r” is the remainder. The “x” is again an unknown number.
Now, it is time to work on an example using this formula. Let’s look at an example. The polynomial is below.
The linear factor is $$x\;+2.$$
So, we will divide $$2x^2\;+3x\;+1\;$$ by $$\;x+2.$$
To set it up, you place the polynomial as the dividend, and the linear factor is the divisor. So, $$2x$$ times $$x$$ is equal to $$2x^2$$and $$2x$$ times 2 is equal to $$4x.$$
Now, you subtract. $$2x^2$$ minus $$2x^2$$ is zero. Then you subtract
$$3x$$ and $$4x$$. However, you are subtracting a positive, so this means you subtract. $$3x$$minus $$4x$$is equal to negative $$(-)x.$$
Then you bring down the plus sign and the one. So, you have negative $$(-) x$$ plus 1.
Negative $$(-1)$$ times $$x+2$$ equals negative $$(-) x$$ minus 2. When you subtract negative $$x$$ minus 2 from negative $$x$$ plus 1, you need to realize that you have two negatives that turn into a positive.
Therefore, you have negative $$x$$ plus 1 plus $$x$$ plus two. The negative x plus positive x cancels out. The remainder is 3.
The final answer would be $$2x-1$$ with a remainder of 3.
Like many people, you may wonder where in the world you are going to use this. First, if you are taking an Algebra or other math class, then obviously you will use it there. For the most part, this is not a skill that you will be using every day.
Instead, you will most likely be using it once in a great, great, while. However, when you are trying to find the length of a flat surface, you may be using this theorem.
The surface area will represent the polynomial and the width will represent the linear factor. When you divide those two numbers, then you will get the answer with the remainder just like a polynomial in the remainder theorem.
Let’s look at a word problem using this theorem.
FACT: Remember when you subtract in a long division problem, it might change the sign on the second number. If you look at the problem above, it shows where the subtracting changed to adding.
Linear factor: $$x-2$$
$$2x+8$$ with a remainder of positive 10
To find the remainder, you need to set up the problem correctly first. $$2x$$ times $$x$$ equals your $$2x^2$$ which is what you need since you will be subtracting. Those will cancel out. Then $$2x$$ times negative 2 equals negative $$4x$$. This will turn into a positive when you subtract. Therefore, you will have $$4x$$ plus $$4x$$ which equals $$8x$$. Then bring down the negative 6. Now, 8 will go into $$x-2.$$ Eight times x equals $$8x$$, and positive 8 times a negative 2 equals negative 16. When you subtract, the positive $$8x$$ and the other positive $$8x$$ cancels out. Negative six minus a negative 16 is positive 10 because the minus and negative turns into a positive. Therefore, the answer is $$2x+8$$ with a remainder of positive 10.
Linear factor: $$x+3$$
$$4x-13$$ with a remainder of 40
To find the remainder, you will need to place the polynomial as the dividend and the linear factor as the divisor. Then 4x times x equals to $$4x^2$$, and $$4x$$ times 3 equals to $$12x$$. When you subtract, this will change the positive $$12x$$ to a negative $$12x$$. The $$4x^2$$ will cancel each other out. A negative x plus a negative $$12x$$ will equal negative $$13x$$. Then bring down the plus 1. In the quotient, a negative 13 is needed because negative 13 times x equals negative $$13x$$. Negative 13 times positive 3 equals negative 39. Then subtract. The second negative $$13x$$ will turn into a positive as well as the negative 39. So, the negative $$13x$$ and the positive $$13x$$ will cancel out. The positive 1 and positive 39 equals a positive 40. Therefore, the answer is 4x-13 with a remainder of 40.
A polynomial is an expression that usually has variables, coefficients, and possibly exponents. Variables are usually letters that represent certain values. Many variables are expressed in algebraic expressions “x” and “y.” The coefficients are the numbers that are sometimes listed before the variables. Also, there may be constants in the expression. The constant is merely a number. An exponent shows how many times the number should be multiplied to itself.
A linear factor is when the variable is at the most 1. In this case with this theorem, most linear factors are a degree of 1. That means that they have a variable like x and a plus or minus with a number. An example would be x-1. A linear factor can have a degree 0 instead of a degree of 1. Unlike polynomials, linear factors do not have exponents. They are probably easier to identify than the polymonials.
To solve a polynomial, it is best to understand long division. First, place the polynomial as the dividend. Then place the linear factor as the divisor. Divide the polynomial by the linear factor. If you struggle with this skill, you may want to practice multiplying and dividing variables with coefficients and exponents. Remember that the remainder theorem is a proof, so you are really proving the polynomial.
A common mistake people make is NOT with understanding how to set up the problem. Instead, it is most likely the part in which two negatives become a positive. In a division problem, you must subtract a lot. Each time you subtract, there could be a time when you need to subtract a negative number. This is when it turns into a positive. When something like this turns into a positive, then you add. You may want to practice subtracting equations or expressions in order to be successful with knowing when to subtract and when to add.
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