Dividing a polynomial by a polynomial

Dividing polynomials, also called algebraic division, is not a trivial operation, and it is used fairly often when simplifying polynomials and expressions.

First example

Dividing polynomials is quite a hard task, at least compared with the earlier operations. However, the method is very similar to ordinary long division. We will show the whole algorithm through examples. Let’s start with a simple one:

$(4 x^{3} + 8 x^{2}) / (2 x^{2})$

We assume that both polynomials are written in descending order and fully simplified – that is, we have already collected any like terms. This is not essential for the algorithm, but it makes the working easier. In the first step we divide the first term of the first polynomial by the first term of the second polynomial (here the second polynomial has only one term, so its first term is the whole polynomial). That is, we work out

$4 x^{3} / (2 x^{2}) = \frac{4}{2} x^{3 - 2} = 2 x$

Dividing these basic terms is the reverse of multiplying them, so there should be no surprises. Since this is the first step, we write it into the original expression that we want to work out:

$(4 x^{3} + 8 x^{2}) / (2 x^{2}) = 2 x$

and carry on with the second step. We multiply our partial result by the second polynomial, the one we are dividing by.

$2 x \cdot (2 x^{2}) = 4 x^{3}$

We subtract this from the whole first polynomial, writing it under the first polynomial like this:

$\begin{array}{rc} (4 x^{3} + 8 x^{2}) & / (2 x^{2}) = 2 x \\ \underset{―}{- (4 x^{3})} \\ 8 x^{2} \end{array}$

Now what is left under the line is a new expression, and we apply the same method to it as to the original polynomial. We divide it by the first term of the second polynomial.

$8 x^{2} / (2 x^{2}) = 4$

We add this to the previous partial result:

$(4 x^{3} + 8 x^{2}) / (2 x^{2}) = 2 x + 4$

Then we multiply the whole second polynomial by 4 again

$(2 x^{2}) \cdot 4 = 8 x^{2}$

and subtract it from what was left under the line.

$\begin{array}{rc} (4 x^{3} + 8 x^{2}) & / (2 x^{2}) = 2 x + 4 \\ \underset{―}{- (4 x^{3})} \\ 8 x^{2} \\ \underset{―}{- (8 x^{2})} \\ 0 \end{array}$

After subtracting we are left with zero, so the algorithm ends here. The right-hand side is the result of the division. To check it, we multiply the resulting polynomial by the second polynomial, and we must get the first polynomial.

$(2 x^{2}) \cdot (2 x + 4) = 4 x^{3} + 8 x^{2}$

Second example

Let’s change the question only slightly and work out

$(4 x^{3} + 8 x^{2} + 7) / (2 x^{2}) .$

The working goes exactly as in the previous example, except that after each subtraction the result has an extra 7. In brief, it goes like this:

$\begin{array}{rc} (4 x^{3} + 8 x^{2} + 7) & / (2 x^{2}) = 2 x + 4 \\ \underset{―}{- (4 x^{3})} \\ 8 x^{2} + 7 \\ \underset{―}{- (8 x^{2})} \\ 7 \end{array}$

The method has not changed, there is just an extra 7. At this point we cannot divide the new remainder (the 7) by the second polynomial to get a nice result. We simply write the result as 7 over the polynomial and add it to the partial result.

$\begin{array}{rc} (4 x^{3} + 8 x^{2} + 7) & / (2 x^{2}) = 2 x + 4 + \frac{7}{2 x^{2}} \\ \underset{―}{- (4 x^{3})} \\ 8 x^{2} + 7 \\ \underset{―}{- (8 x^{2})} \\ 7 \end{array}$

When we multiply back by the second polynomial, we get 7

$(2 x^{2}) \cdot \frac{7}{2 x^{2}} = 7$

and so the subtraction gives zero.

$\begin{array}{rc} (4 x^{3} + 8 x^{2} + 7) & / (2 x^{2}) = 2 x + 4 + \frac{7}{2 x^{2}} \\ \underset{―}{- (4 x^{3})} \\ 8 x^{2} + 7 \\ \underset{―}{- (8 x^{2})} \\ 7 \\ \underset{―}{- 7} \\ 0 \end{array}$

The algorithm ends here, as we have reached zero.

Third example

Let’s add another term to the second polynomial.

$(6 x^{7} + 19 x^{4} + 7) / (3 x^{3} + 5)$

In the first step we divide the first term of the first polynomial by the first term of the second, as in the previous examples.

$6 x^{7} / 3 x^{3} = 2 x^{4}$

In the next step we multiply this partial result by the whole second polynomial and subtract it from the first.

$(3 x^{3} + 5) \cdot 2 x^{4} = 6 x^{7} + 10 x^{4}$

We write it out neatly under the whole question.

$\begin{array}{rc} (6 x^{7} + 19 x^{4} + 7) & / (3 x^{3} + 5) = 2 x^{4} \\ \underset{―}{- (6 x^{7} + 10 x^{4})} \\ 9 x^{4} + 7 \end{array}$

We continue by dividing the first term of the new polynomial by the first term of the second polynomial.

$9 x^{4} / 3 x^{3} = 3 x$

We multiply by the whole second polynomial:

$(3 x^{3} + 5) \cdot 3 x = 9 x^{4} + 15 x$

We write it out and subtract:

$\begin{array}{rc} (6 x^{7} + 19 x^{4} + 7) & / (3 x^{3} + 5) = 2 x^{4} + 3 x \\ \underset{―}{- (6 x^{7} + 10 x^{4})} \\ 9 x^{4} + 7 \\ \underset{―}{- (9 x^{4} + 15 x)} \\ - 15 x + 7 \end{array}$

This cannot sensibly be divided any further, so we simply add the fraction made from the remainder and the second polynomial.

$(6 x^{7} + 19 x^{4} + 7) / (3 x^{3} + 5) = 2 x^{4} + 3 x + \frac{- 15 x + 7}{3 x^{3} + 5}$

If we multiplied back and subtracted once more, we would get zero straight away, as in the last example. Let’s check the answer. Before a full check, we can try a quick test: substitute a simple value, such as zero, into the expressions. Substituting zero into the result of the division, we get

$0 + 0 + \frac{0 + 7}{0 + 5} = \frac{7}{5} .$

Substituting zero into the expression we are dividing, we get

$(0 + 0 + 7) / (0 + 5) = \frac{7}{5} .$

If our working is correct, both values must give the same result, and as we can see, they are equal. This does not guarantee that the answer is fully correct, as we would have to test every possible value of x, but it can quickly reveal a mistake. Let’s try the same with 1. Substituting into the result gives:

$2 + 3 + \frac{- 15 + 7}{3 + 5} = 5 - \frac{8}{8} = 4.$

Substituting into the original expression gives

$(6 + 19 + 7) / (3 + 5) = \frac{32}{8} = 4.$

That looks promising, but for final confirmation we need to carry out the whole multiplication.

$\begin{array}{rcl} (3 x^{3} + 5) \cdot (2 x^{4} + 3 x + \frac{- 15 x + 7}{3 x^{3} + 5}) & = & (3 x^{3} + 5) \cdot 2 x^{4} + (3 x^{3} + 5) \cdot 3 x + (- 15 x + 7) \\ = & 6 x^{7} + 10 x^{4} + 9 x^{4} + 15 x - 15 x + 7 \\ = & 6 x^{7} + 19 x^{4} + 7 \end{array}$

Chapters: Polynomials, Dividing a polynomial by a polynomial, Root of a polynomial, Factorising polynomials

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