Dividing polynomials

This chapter explores dividing polynomials. It covers algebraic division of polynomials by (x + a) or (x – a). Before attempting this chapter you must have prior knowledge of long division and knowing that polynomials can be written as a product of factors.

Introduction

We can write polynomials as a product of factors. Below is an example;
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Here is another example;
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Polynomials work very similar to usual numbers, they can also be divided for example we know that;
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…this implies that;
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…and…
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We can take the same idea above and apply it in polynomials, we know;
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…that must also mean that;
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…it also means that;
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In this lession we shall cover how we divide a polynomial by a linear function for example suppose we had to work out the following;
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Long division with numbers

First let us look at division of standard numbers. Suppose we had to divide 8476 by 13. We can rearrange the question like the following;
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First we find out how many times 13 goes into 84. We find that 13 goes into 84, 6 times as shown below;
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Multiplying 13 by 6 gives us 78 so there was a remainder, we write;
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We write the remainder below as a subtraction and bring the next digit down as shown below.
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Now we have to work out how many times 13 goes into 67. 13 goes into 67, 5 times as shown below;
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Multiplying 5 by 13 gives 67 so there was a remainder again. We bring 65 below as a subtraction as shown below.
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The remainder was a 2 so we write it below as a subtraction as shown above. Now we can bring down the next digit which is a 6.
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No we have to find out how many times 13 goes into 26. 13 goes into 26, 2 times as shown below.
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Multiplying 2 by 13 gives 26 which means there was no remainder so we continue to write;
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So we can conclude that 8476 divided by 13 is 652.

Dividing a quadratic by a linear

Here we shall explore dividing a quadratic by a linear. Suppose we had to work out the following.
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Like we did above we could rewrite it like the following;
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Let us ignore the -3 part. We have to find out how many times x goes into x²
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x-3 goes into x², x times. Multiplying x by x-3 we get;
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We now need to find the remainder. The remainder must be -x–3x = 2x. We write this below as a subtraction as shown below.
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Next we bring the next digit down as we did before. The next digit is -6.
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Next we need to find out how many times x-3 goes into 2x. Again we ignore the -3 part and find out how many times x goes into 2x. x-3 goes into 2x, 2 times. That must mean that;
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So now we have;
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We can see that there is no remainder. We can conclude that;
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Remainder example

Here is another example;
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Let us rearrange it like the following;
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We need to find how many times x – 1 goes into 2x². We ignore the -1 part and simply think how many times x goes into 2x². x-1 goes into 2x², 2x times. That must mean that;
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So we have the following, notice that we have pulled +5 down.
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Now we have to find how many times x – 1 goes into x. Again we ignore the -1 part and simply find how many times x goes into x. x – 1 goes into x once. So we have;
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Notice we have added x-1 below because;
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Now we need to find the remainder. The remainder is 5–1 = 6. So now we have;
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We can conclude that;
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The above can also be rewritten as the following;
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Dividing a cubic by a linear

In this section we shall explore dividing a cubic by a linear. The following is a longer example suppose we had to divide x³ + 7x² + 14x + 6 by (x + 3). We can write such as the following.
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We need to find how many times x+3 goes into x³. Here we ignore the +3 part and simply find how many times x goes into x³. x + 3 goes into x³, x² times. That must mean;
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The remainder is;
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So now we have;
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Notice that above we have written the remainder below as a subtraction. Next we bring the next digit (+14x) down as shown below.
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Now we need to find how many times x + 3 goes into 4x². Again we ignore the +3 part and simply find how many times x goes into 4x². x+3 goes into 4x², 4x times meaning;
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The remainder is;
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We can add this to the work out as shown below;
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Notice above we have written 4x² + 12x as a subtraction to get the remainder. Next we shall bring the next digit down which is a +6.
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Next we need to find how many times x+3 goes into 2x. Again we need to ignore the +3 part and find how many times x goes into 2x. x+3 goes into 2x, 2 times meaning.
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That gives us;
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With remainders

Below is another example to workout. Suppose we had to divide x³ + 2x² – 3x + 10 by (x + 2). We write it like the following.
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Firstly we need to find how many times x+2 goes into x³, we ignore the +2 part and find out how many times x goes into x³.
x+2 goes into x³, x² times meaning…
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…so we write;
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Notice that we have written x³ + 2x² below as a subtraction which resulted in 0. Next we bring the next digit down as usual. The next term is -3x, we also bring +10 down as well since it contains no x. So we get;
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Next we need to find out how many times x+2 goes into -3x. Again we ignore the +2 part and simply find how many times x goes into -3x. x+2 goes into -3x, -3 times, meaning.
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The remainder is;
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So we have;
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That gives us;
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We can also write this like the following;
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