Quotient of polynomials

Now we will explain a method to divide polynomials of one variable. We will use an example to illustrate the procedure:

Example

Let's consider,

$p (x) = x^{5} - 3 x^{3} + 2 x - 1$

$q (x) = x^{2} - 1 - 2 x$

Calculate this quotient $\frac{p (x)}{q (x)}$ .

1) Complete and put in order both polynomials.

In our case,

$p (x) = x^{5} + 0 x^{4} - 3 x^{3} + 0 x^{2} + 2 x - 1$

$q (x) = x^{2} - 2 x - 1$

2) Write both polynomials as if we wanted to solve a traditional division of two numbers (the dividend into the left, the divisor into the right). Let's consider that every monomial is a number.

Here we will use the following table:

x^{5}

0

- 3 x^{3}

0

2 x

- 1

x^{2} - 2 x - 1

3) Divide the first monomial of the dividend by the first monomial of the divisor.

In our case: $\frac{x^{5}}{x^{2}} = x^{3}$

4) Multiply the result by every monomial of the dividing polynomial and subtract the result from the polynomial dividend.

The result of the product is $x^{3} \cdot q (x) = x^{3} (x^{2} - 2 x - 1) = x^{5} - 2 x^{4} - x^{3}$

And we subract it by the dividend. Then, we schematize it:

$x^{5}$	$0$	$- 3 x^{3}$	$0$	$2 x$	$- 1$	$x^{2} - 2 x - 1$
$- x^{5}$	$+ 2 x^{4}$	$+ x^{3}$	$0$	$0$	$0$	$x^{3}$
$0$	$+ 2 x^{4}$	$- 2 x^{3}$	$0$	$2 x$	$- 1$

The result of the subtraction appears in the third line. We take note of the result of the division of monomials placed just under the divisor: this will be our quotient.

Let's focus on the box of the degree of the polynomial that we have divided. In this case, we find a $0$ . This must happen in each one of the steps that we make.

5) Repeat steps $3$ and $4$ until the degree of the polynomial by which we need to divide is lower than the degree of the dividing polynomial.

Let's see how we continue: $\frac{2 x^{4}}{x^{2}} = 2 x^{2}$

$2 x^{2} (x^{2} - 2 x - 1) = 2 x^{4} - 4 x^{3} - 2 x^{2}$

$x^{5}$	$0$	$- 3 x^{3}$	$0$	$2 x$	$- 1$	$x^{2} - 2 x - 1$
$- x^{5}$	$+ 2 x^{4}$	$+ x^{3}$	$0$	$0$	$0$	$x^{3} + 2 x^{2}$
$0$	$+ 2 x^{4}$	$- 2 x^{3}$	$0$	$2 x$	$- 1$
	$- 2 x^{4}$	$4 x^{3}$	$2 x^{2}$	$0$	$0$
	$0$	$2 x^{3}$	$2 x^{2}$	$2 x$	$- 1$

Now, we have a $0$ in the degree $4$ monomial. Let's continue:

$\frac{2 x^{3}}{x^{2}} = 2 x$

$2 x (x^{2} - 2 x - 1) = 2 x^{3} - 4 x^{2} - 2 x$

$x^{5}$	$0$	$- 3 x^{3}$	$0$	$2 x$	$- 1$	$x^{2} - 2 x - 1$
$- x^{5}$	$+ 2 x^{4}$	$+ x^{3}$	$0$	$0$	$0$	$x^{3} + 2 x^{2} + 2 x$
$0$	$+ 2 x^{4}$	$- 2 x^{3}$	$0$	$2 x$	$- 1$
	$- 2 x^{4}$	$4 x^{3}$	$2 x^{2}$	$0$	$0$
	$0$	$2 x^{3}$	$2 x^{2}$	$2 x$	$- 1$
		$- 2 x^{3}$	$+ 4 x^{2}$	$+ 2 x$	$0$
		$0$	$6 x^{2}$	$4 x$	$- 1$

We can see, again, that we find a $0$ in the degree $3$ monomial. We repeat the operation:

$\frac{6 x^{2}}{x^{2}} = 6$

$6 (x^{2} - 2 x - 1) = 6 x^{2} - 12 x - 6$

$x^{5}$	$0$	$- 3 x^{3}$	$0$	$2 x$	$- 1$	$x^{2} - 2 x - 1$
$- x^{5}$	$+ 2 x^{4}$	$+ x^{3}$	$0$	$0$	$0$	$x^{3} + 2 x^{2} + 2 x + 6$
$0$	$+ 2 x^{4}$	$- 2 x^{3}$	$0$	$2 x$	$- 1$
	$- 2 x^{4}$	$+ 4 x^{3}$	$+ 2 x^{2}$	$0$	$0$
	$0$	$2 x^{3}$	$2 x^{2}$	$2 x$	$- 1$
		$- 2 x^{3}$	$+ 4 x^{2}$	$+ 2 x$	$0$
		$0$	$6 x^{2}$	$4 x$	$- 1$
			$- 6 x^{2}$	$+ 12 x$	$+ 6$
			$0$	$16 x$	$+ 5$

Again, a $0$ appears in the monomial of second degree. Now, the polynomial that we want to divide has degree $1$ , which is less than the degree of the divisor (degree $2$ ). At this point, the division is finished. Then:

The quotient will be the polynomial placed just under the divisor: $x^{3} + 2 x^{2} + 2 x + 6$
The remainder will be the polynomial located at the end, which degree will be always lower than the one of the divisor: $16 x + 5$

VERIFICATION

To verify that we have done the division correctly, we will calculate: $quotient \times divisor + remainder$ The result, if we have done the operation correctly, should be the dividend.

So, in our example: $(x^{3} + 2 x^{2} + 2 x + 6) \cdot (x^{2} - 2 x - 1) + (16 x + 5)$

We calcule the multiplication:

$x^{3} \cdot (x^{2} - 2 x - 1) = x^{5} - 2 x^{4} - x^{3}$

$2 x^{2} \cdot (x^{2} - 2 x - 1) = 2 x^{4} - 4 x^{3} - 2 x^{2}$

$2 x \cdot (x^{2} - 2 x - 1) = 2 x^{3} - 4 x^{2} - 2 x$

$6 \cdot (x^{2} - 2 x - 1) = 6 x^{2} - 12 x - 6$

$(x^{5} - 2 x^{4} - x^{3}) + (2 x^{4} - 4 x^{3} - 2 x^{2}) + (2 x^{3} - 4 x^{2} - 2 x) +$

$+ (6 x^{2} - 12 x - 6) = x^{5} - 3 x^{3} - 14 x - 6$

Then, we add the remainder:

$(x^{5} - 3 x^{3} - 14 x - 6) + (16 x + 5) = x^{5} - 3 x^{3} + 2 x - 1$

As we can see, the result coincides with our dividend.

We can also verify that:

degree(quotient)=degree(dividend)-degree(divisor)

degree(remainder)

Example

Calculate the quotient $3$ where $x^{3} + 2 x^{2} + 2 x + 6$ and $x^{5} - 3 x^{3} + 2 x - 1$ .

We complete and put in order

$x^{2} - 2 x - 1$

$5 - 2 = 3$

We define the initial table

16 x + 5

1 < 2

x^{2} - 2 x - 1

\frac{p (x)}{q (x)}

p (x) = 1 - x^{3}

Continuing with the operation: $q (x) = x + 2$ $p (x) = - x^{3} + 0 x^{2} + 0 x + 1$

$q (x) = x + 2$	$- x^{3}$	$0$	$0$	$1$
$x + 2$	$\frac{- x^{3}}{x} = - x^{2}$	$- x^{2} (x + 2) = - x^{3} - 2 x^{2}$	$- x^{3}$	$0$
$0$	$1$	$x + 2$	$+ x^{3}$

And then, the next step: $+ 2 x^{2}$ $0$

$0$	$- x^{2}$	$0$	$+ 2 x^{2}$	$0$
$1$	$\frac{2 x^{2}}{x} = 2 x$	$2 x (x + 2) = 2 x^{2} + 4 x$	$- x^{3}$	$0$
$0$	$1$	$x + 2$	$+ x^{3}$
	$+ 2 x^{2}$	$0$	$0$
	$- x^{2} + 2 x$	$0$	$+ 2 x^{2}$

Third step: $0$ $1$

$- 2 x^{2}$	$- 4 x$	$0$	$0$	$- 4 x$
$1$	$\frac{- 4 x}{x} = - 4$	$- 4 (x + 2) = - 4 x - 8$	$- x^{3}$	$0$
$0$	$1$	$x + 2$	$+ x^{3}$
	$+ 2 x^{2}$	$0$	$0$
	$- x^{2} + 2 x - 4$	$0$	$+ 2 x^{2}$
		$0$	$1$
		$- 2 x^{2}$	$- 4 x$

Now, we can see that

degree $0$ degree $0$

With this operation, the process is finished. We verify the process:

$- 4 x$

We do the multiplication:

$1$

$+ 4 x$

$+ 8$

$0$

And, adding the remainder, we obtain the dividend:

$9$

Concerning the degrees, we can see that:

$(9) = 0 < 1 =$

degree $(x + 2)$ degree $(- x^{2} + 2 x - 4) (x + 2) + (9)$