Division of polynomials. Dividing a polynomial by a polynomial with a remainder Dividing a polynomial by a binomial examples

A proof is given that an improper fraction composed of polynomials can be represented as the sum of a polynomial and a proper fraction. Examples of dividing polynomials with a corner and multiplying with a column are analyzed in detail.

Content

Theorem

Let P k (x),Qn (x)- polynomials in the variable x of degrees k and n, respectively, with k ≥ n. Then the polynomial P k (x) can be represented in the only way in the following form:
(1) Pk (x) = S k-n (x) Q n (x) + U n-1 (x),
where S k-n (x)- polynomial of degree k-n, U n- 1 (x)- polynomial of degree not higher than n- 1 , or zero.

Proof

By definition of a polynomial:
;
;
;
,
where p i, q i are known coefficients, s i, u i are unknown coefficients.

Let us introduce the notation:
.
Let's substitute in (1) :
;
(2) .
The first term on the right hand side is a polynomial of degree k. The sum of the second and third terms is a polynomial of degree no higher than k - 1 . Let us equate the coefficients for x k:
p k = s k-n q n .
Hence s k-n = p k / q n.

Let's transform the equation (2) :
.
Let us introduce the notation: .
Since s k-n = p k / q n, then the coefficient for x k is equal to zero. Therefore - this is a polynomial of degree no higher than k - 1 , . Then the previous equation can be rewritten as:
(3) .

This equation has the same form as the equation (1) , only the value of k became 1 less. Repeating this procedure k-n times, we obtain the equation:
,
from which we determine the coefficients of the polynomial U n- 1 (x).

So, we have determined all the unknown coefficients s i, ul. Moreover, s k-n ≠ 0 . The lemma is proven.

Division of polynomials

Dividing both sides of the equation (1) on Qn (x), we get:
(4) .
By analogy with decimal numbers, S k-n (x) called the integer part of the fraction or quotient, U n- 1 (x)- the remainder of the division. A fraction of polynomials in which the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator is called a proper fraction. A fraction of polynomials in which the degree of the polynomial in the numerator is greater than or equal to the degree of the polynomial in the denominator is called an improper fraction.

The equation (4) shows that any improper fraction of polynomials can be simplified by representing it as the sum of the integer part and the proper fraction.

At their core, decimal integers are polynomials in which the variable is equal to the number 10 . For example, take the number 265847. It can be represented as:
.
That is, this is a fifth-degree polynomial in 10 . The numbers 2, 6, 5, 8, 4, 7 are the coefficients of the expansion of the number into powers of 10.

Therefore, the rule of division (sometimes called long division) that applies to dividing numbers can be applied to polynomials. The only difference is that when dividing polynomials, you do not need to convert numbers greater than nine to the highest digits. Let's consider the process of dividing polynomials with a corner using specific examples.

An example of dividing polynomials with a corner

Here the numerator contains a polynomial of the fourth degree. The denominator is a polynomial of the second degree. Because the 4 ≥ 2 , then the fraction is incorrect. Let’s select the whole part by separating the polynomials with a corner (in a column):

Let's give detailed description division process. We write the original polynomials in the left and right columns. Under the denominator polynomial, in the right column, draw a horizontal line (corner). Below this line, under the corner, there will be a whole part of the fraction.

1.1 We find the first term of the whole part (under the corner). To do this, divide the leading term of the numerator by the leading term of the denominator: .

1.2 Multiply 2 x 2 by x 2 - 3 x + 5:
. We write the result in the left column:

1.3 We take the difference of the polynomials in the left column:

.

So, we got an intermediate result:
.

The fraction on the right side is improper because the degree of the polynomial in the numerator ( 3 ) is greater than or equal to the degree of the polynomial in the denominator ( 2 ). We repeat the calculations. Only now the numerator of the fraction is in the last line of the left column.
2.1 Let's divide the leading term of the numerator by the leading term of the denominator: ;

2.2 Multiply by the denominator: ;

2.3 And subtract from the last line of the left column: ;

Intermediate result:
.

We repeat the calculations again, since there is an improper fraction on the right side.
3.1 ;
3.2 ;
3.3 ;

So we got:
.
The degree of the polynomial in the numerator of the right fraction is less than the degree of the polynomial in the denominator, 1 < 2 . Therefore the fraction is correct.

;
2 x 2 - 4 x + 1- this is a whole part;
x- 8 - remainder of the division.

Example 2

Select the whole part of the fraction and find the remainder of the division:
.

We perform the same actions as in the previous example:

Here the remainder of the division is zero:
.

Multiplying polynomials by column

You can also multiply polynomials in a column, similar to multiplying integers. Let's look at specific examples.

An example of multiplying polynomials by a column

Find the product of polynomials:
.

2.1
.

2.2
.

2.3
.
We write the result in a column, leveling the degrees x.

3
;
;
;
.

Note that only the coefficients could be written, and the powers of the variable x could be omitted. Then multiplying with a column of polynomials will look like this:

Example 2

Find the product of polynomials in a column:
.

When multiplying polynomials in a column, it is important to write the same powers of the variable x one below the other. If some powers of x are missing, then they should be written explicitly, multiplied by zero, or left blanks.

In this example, some degrees are missing. Therefore, we write them explicitly, multiplied by zero:
.
Multiplying polynomials in a column.

1 We write the original polynomials one below the other in a column and draw a line.

2.1 Multiply the lowest term of the second polynomial by the first polynomial:
.
We write the result in a column.

2.2 The next term of the second polynomial is zero. Therefore, its product by the first polynomial is also zero. The zero line may not be written.

2.3 Multiply the next term of the second polynomial by the first polynomial:
.
We write the result in a column, leveling the degrees x.

2.3 We multiply the next (highest) term of the second polynomial by the first polynomial:
.
We write the result in a column, leveling the degrees x.

3 After all terms of the second polynomial have been multiplied by the first, draw a line and add the terms with the same powers x:
.

General form monomial

f(x)=ax n, Where:

-a- coefficient that can belong to any of the sets N, Z, Q, R, C

-x- variable

-n exponent that belongs to a set N

Two monomials are similar if they have the same variable and the same exponent.

Examples: 3x2 And -5x2; ½x 4 And 2√3x4

The sum of monomials that are not similar to each other is called a polynomial (or polynomial). In this case, the monomials are terms of the polynomial. A polynomial containing two terms is called a binomial (or binomial).
Example: p(x)=3x 2 -5; h(x)=5x-1
A polynomial containing three terms is called a trinomial.

General view of a polynomial with one variable

Where:

a n ,a n-1 ,a n-2 ,...,a 1 ,a 0- polynomial coefficients. They can be natural, integer, rational, real or complex numbers.
a n- coefficient of the term with the largest exponent (leading coefficient)
a 0- coefficient of the term with the smallest exponent (free term, or constant)
n- degree of polynomial

Example 1
p(x)=5x 3 -2x 2 +7x-1

third degree polynomial with coefficients 5, -2, 7 And -1
5 - leading coefficient
-1 - free member
x- variable

Example 2
h(x)=-2√3x 4 +½x-4

fourth degree polynomial with coefficients -2√3.½ And -4
-2√3 - leading coefficient
-4 - free member
x- variable

Division of polynomials

p(x) And q(x)- two polynomials:
p(x)=a n x n +a n-1 x n-1 +...+a 1 x 1 +a 0
q(x)=a p x p +a p-1 x p-1 +...+a 1 x 1 +a 0

To find the quotient and remainder of division p(x) on q(x), you need to use the following algorithm:

Degree p(x) must be greater than or equal to q(x).
We must write both polynomials in decreasing order of degree. If in p(x) there is no term with any degree, it must be added with a coefficient of 0.
Lead Member p(x) divided by the leading term q(x), and the result is written under the dividing line (in the denominator).
Multiply the result by all terms q(x) and write the result with opposite signs under the terms p(x) with relevant degrees.
Add terms with the same powers term by term.
We assign the remaining terms to the result p(x).
Divide the leading term of the resulting polynomial by the first term of the polynomial q(x) and repeat steps 3-6.
This procedure is repeated until the newly obtained polynomial has a degree less than q(x). This polynomial will be the remainder of the division.
The polynomial written below the dividing line is the result of division (quotient).

Example 1
Step 1 and 2) $p(x)=x^5-3x^4+2x^3+7x^2-3x+5 \\ q(x)=x^2-x+1$

3) x 5 -3x 4 +2x 3 +7x 2 -3x+5

4) x 5 -3x 4 +2x 3 +7x 2 -3x+5

5) x 5 -3x 4 +2x 3 +7x 2 -3x+5

6) x 5 -3x 4 +2x 3 +7x 2 -3x+5

/ -2x 4 -x 3 +7x 2 -3x+5

7) x 5 -3x 4 +2x 3 +7x 2 -3x+5

/ -2x 4 +x 3 +7x 2 -3x+5

2x 4 -2x 3 +2x 2

/ -x 3 +9x 2 -3x+5

8) x 5 -3x 4 +2x 3 +7x 2 -3x+5

/ -2x 4 -x 3 +7x 2 -3x+5

2x 4 -2x 3 +2x 2

/ -x 3 +9x 2 -3x+5

/ 6x-3 STOP

x 3 -2x 2 -x+8 --> C(x) Private

Answer: p(x) = x 5 - 3x 4 + 2x 3 + 7x 2 - 3x + 5 = (x 2 - x + 1)(x 3 - 2x 2 - x + 8) + 6x - 3

Example 2
p(x)=x 4 +3x 2 +2x-8
q(x)=x 2 -3x

X 4 +0x 3 +3x 2 +2x-8

/ 3x 3 +3x 2 +2x-8

/ 38x-8 r(x) STOP

x 2 +3x+12 --> C(x) Quotient

Answer: x 4 + 3x 2 + 2x - 8 = (x 2 - 3x)(x 2 + 3x + 12) + 38x - 8

Division by a polynomial of the first degree

This division can be done using the above algorithm or even faster using Horner's method.
If f(x)=a n x n +a n-1 x n-1 +...+a 1 x+a 0, the polynomial can be rewritten as f(x)=a 0 +x(a 1 +x(a 2 +...+x(a n-1 +a n x)...))

q(x)- polynomial of the first degree ⇒ q(x)=mx+n
Then the polynomial in the quotient will have degree n-1.

According to Horner's method, $x_0=-\frac(n)(m)$.
b n-1 =a n
b n-2 =x 0 .b n-1 +a n-1
b n-3 =x 0 .b n-2 +a n-2
...
b 1 =x 0 .b 2 +a 2
b 0 =x 0 .b 1 +a 1
r=x 0 .b 0 +a 0
Where b n-1 x n-1 +b n-2 x n-2 +...+b 1 x+b 0- private. The remainder will be a polynomial of degree zero, since the degree of the polynomial in the remainder must be less than the degree of the divisor.
Division with remainder ⇒ p(x)=q(x).c(x)+r ⇒ p(x)=(mx+n).c(x)+r if $x_0=-\frac(n)(m)$
Note that p(x 0)=0.c(x 0)+r ⇒ p(x 0)=r

Example 3
p(x)=5x 4 -2x 3 +4x 2 -6x-7
q(x)=x-3
p(x)=-7+x(-6+x(4+x(-2+5x)))
x 0 =3

b 3 =5
b 2 =3.5-2=13
b 1 =3.13+4=43 ⇒ c(x)=5x 3 +13x 2 +43x+123; r=362
b 0 =3.43-6=123
r=3.123-7=362
5x 4 -2x 3 +4x 2 -6x-7=(x-3)(5x 3 +13x 2 +43x+123)+362

Example 4
p(x)=-2x 5 +3x 4 +x 2 -4x+1
q(x)=x+2
p(x)=-2x 5 +3x 4 +0x 3 +x 2 -4x+1
q(x)=x+2
x 0 =-2
p(x)=1+x(-4+x(1+x(0+x(3-2x))))

b 4 =-2 b 1 =(-2).(-14)+1=29
b 3 =(-2).(-2)+3=7 b 0 =(-2).29-4=-62
b 2 =(-2).7+0=-14 r=(-2).(-62)+1=125
⇒ c(x)=-2x 4 +7x 3 -14x 2 +29x-62; r=125
-2x 5 +3x 4 +x 2 -4x+1=(x+2)(-2x 4 +7x 3 -14x 2 +29x-62)+125

Example 5
p(x)=3x 3 -5x 2 +2x+3
q(x)=2x-1
$x_0=\frac(1)(2)$
p(x)=3+x(2+x(-5+3x))
b 2 =3
$b_1=\frac(1)(2)\cdot 3-5=-\frac(7)(2)$
$b_0=\frac(1)(2)\cdot \left(-\frac(7)(2)\right)+2=-\frac(7)(4)+2=\frac(1)(4 )$
$r=\frac(1)(2)\cdot \frac(1)(4)+3=\frac(1)(8)+3=\frac(25)(8) \Rightarrow c(x)= 3x^2-\frac(7)(2)x+\frac(1)(4)$
$\Rightarrow 3x^3-5x^2+2x+3=(2x-1)(3x^2--\frac(7)(2)x+\frac(1)(4))+\frac(25) (8)$
Conclusion
If we divide by a polynomial of degree higher than one, we need to use the algorithm to find the quotient and remainder 1-9 .
If we divide by a polynomial of first degree mx+n, then to find the quotient and remainder you need to use Horner’s method with $x_0=-\frac(n)(m)$.
If we are only interested in the remainder of the division, it is enough to find p(x 0).
Example 6
p(x)=-4x 4 +3x 3 +5x 2 -x+2
q(x)=x-1
x 0 =1
r=p(1)=-4.1+3.1+5.1-1+2=5
r=5

Today we will learn how to divide polynomials by each other, and we will do the division with a corner, by analogy with ordinary numbers. This is a very useful technique that, unfortunately, is not taught in most schools. Therefore, listen carefully to this video lesson. There is nothing complicated in such a division.

First, let's divide two numbers by each other:

How can this be done? First of all, we cut off so many bits so that the resulting numeric value was more than what we divide by. If we cut off one digit, we get five. Obviously, seventeen can't fit into five, so it's not enough. We take two digits - we get 59 - it is already more than seventeen, so we can perform the operation. So how many times does seventeen fit into 59? Let's take three. We multiply and write the result under 59. In total, we get 51. Subtract and we get “eight.” Now we take down the next digit - five. Divide 85 by seventeen. Let's take five. Multiply seventeen by five and we get 85. Subtract and we get zero.

Solving real examples

Task No. 1

Now let's perform the same steps, but not with numbers, but with polynomials. For example, let's take this:

\[\frac(((x)^(2))+8x+15)(x+5)=x+3\]

Please note that if when dividing numbers by each other we assumed that the dividend is always greater than the divisor, then in the case of dividing polynomials by a corner, it is necessary that the degree of the dividend be greater than the divisor. In our case, everything is in order - we are working with constructions of the second and first degrees.

So, the first step: compare the first elements. Question: What should you multiply $x$ by to get $((x)^(2))$? Obviously for another $x$. Multiply $x+5$ by the number $x$ we just found. We have $((x)^(2))+5$, which we subtract from the dividend. That leaves $3x$. Now we take down the next term - fifteen. Let's look at the first elements again: $3x$ and $x$. What should $x$ be multiplied by to get $3x$? Obviously, three. We multiply $x+5$ term by three. When we subtract, we get zero.

As you can see, the entire operation of dividing by a corner has been reduced to comparing the highest coefficients of the dividend and the divisor. It's even easier than when you divide numbers. There is no need to select a certain number of digits - we simply compare the highest elements at each step. That's the whole algorithm.

Problem No. 2

Let's try again:

\[\frac(((x)^(2))+x-2)(x-1)=x+2\]

First step: look at the leading odds. How much do you need to multiply $x$ to write $((x)^(2))$? We multiply term by term. Please note that when subtracting, we get exactly $2x$, because

We remove -2 and again compare the first coefficient obtained with the highest element of the divisor. In total, we came up with a “beautiful” answer.

Let's move on to the second example:

\[\frac(((x)^(3))+2((x)^(2))-9x-18)(x+3)=((x)^(2))-x-6\ ]

This time the dividend is a third-degree polynomial. Let's compare the first elements with each other. In order to get $((x)^(3))$, it is necessary to multiply $x$ by $((x)^(2))$. After subtraction we take away $9x$. Multiply the divisor by $-x$ and subtract. As a result, our expression was completely divided. We write down the answer.

Problem No. 3

Let's move on to the last task:

\[\frac(((x)^(3))+3((x)^(2))+50)(x+5)=((x)^(2))-2x+10\]

Let's compare $((x)^(3))$ and $x$. Obviously, you need to multiply by $((x)^(2))$. As a result, we see that we received a very “beautiful” answer. Let's write it down.

That's the whole algorithm. There are two key points here:

Always compare the first power of the dividend and the divisor - we repeat this at every step;
If any degrees are missing in the original expression, they must be added when dividing by a corner, but with zero coefficients, otherwise the answer will be incorrect.

There are no more wisdom and tricks in this division.

The material from today’s lesson is never found in its “pure” form anywhere. It is rarely taught in schools. However, the ability to divide polynomials by each other will greatly help you when solving equations of higher degrees, as well as all kinds of “increased difficulty” problems. Without this technique, you will have to factor polynomials, select coefficients - and the result is by no means guaranteed. However, polynomials can also be divided by a corner - just like ordinary numbers! Unfortunately, this technique is not taught in schools. Many teachers believe that dividing polynomials by a corner is something incredibly complicated, from the field of higher mathematics. I hasten to assure you: this is not so. In fact, dividing polynomials is even easier than dividing regular numbers! Watch the lesson and see for yourself. :) In general, be sure to adopt this technique. The ability to divide polynomials by each other will be very useful to you when solving equations of higher degrees and in other non-standard problems.

I hope this video will help those who work with polynomials, especially higher degrees. This applies to both high school and university students. And that's all for me. See you!

Let's start with some definitions. Polynomial nth degree(or nth order) we will call an expression of the form $P_n(x)=\sum\limits_(i=0)^(n)a_(i)x^(n-i)=a_(0)x^(n)+ a_(1)x^(n-1)+a_(2)x^(n-2)+\ldots+a_(n-1)x+a_n$. For example, the expression $4x^(14)+87x^2+4x-11$ is a polynomial whose degree is $14$. It can be denoted as follows: $P_(14)(x)=4x^(14)+87x^2+4x-11$.

The coefficient $a_0$ is called the leading coefficient of the polynomial $P_n(x)$. For example, for the polynomial $4x^(14)+87x^2+4x-11$ the leading coefficient is $4$ (the number before $x^(14)$). The number $a_n$ is called the free term of the polynomial $P_n(x)$. For example, for $4x^(14)+87x^2+4x-11$ the free term is $(-11)$. Now let us turn to the theorem on which, in fact, the presentation of the material on this page will be based.

For any two polynomials $P_n(x)$ and $G_m(x)$, one can find polynomials $Q_p(x)$ and $R_k(x)$ such that the equality

\begin(equation) P_n(x)=G_m(x)\cdot Q_p(x)+R_k(x) \end(equation)

and $k< m$.

The phrase “divide the polynomial $P_n(x)$ by the polynomial $G_m(x)$” means “represent the polynomial $P_n(x)$ in the form (1)”. We will call the polynomial $P_n(x)$ divisible, the polynomial $G_m(x)$ a divisor, the polynomial $Q_p(x)$ the quotient of division of $P_n(x)$ by $G_m(x)$, and the polynomial $ R_k(x)$ - remainders from division of $P_n(x)$ by $G_m(x)$. For example, for the polynomials $P_6(x)=12x^6+3x^5+16x^4+6x^3+8x^2+2x+1$ and $G_4(x)=3x^4+4x^2+2 $ you can get the following equality:

$$ 12x^6+3x^5+16x^4+6x^3+8x^2+2x+1=(3x^4+4x^2+2)(4x^2+x)+2x^3+1 $$

Here the polynomial $P_6(x)$ is divisible, the polynomial $G_4(x)$ is a divisor, the polynomial $Q_2(x)=4x^2+x$ is the quotient of $P_6(x)$ divided by $G_4(x) $, and the polynomial $R_3(x)=2x^3+1$ is the remainder of division of $P_6(x)$ by $G_4(x)$. Note that the degree of the remainder (i.e. 3) is less than the degree of the divisor (i.e. 4), therefore the equality condition is met.

If $R_k(x)\equiv 0$, then the polynomial $P_n(x)$ is said to be divisible by the polynomial $G_m(x)$ without remainder. For example, the polynomial $21x^6+6x^5+105x^2+30x$ is divisible by the polynomial $3x^4+15$ without remainder, since the equality is satisfied:

$$ 21x^6+6x^5+105x^2+30x=(3x^4+15)\cdot(7x^2+2x) $$

Here the polynomial $P_6(x)=21x^6+6x^5+105x^2+30x$ is divisible; polynomial $G_4(x)=3x^4+15$ - divisor; and the polynomial $Q_2(x)=7x^2+2x$ is the quotient of $P_6(x)$ divided by $G_4(x)$. The remainder is zero.

To divide a polynomial into a polynomial, division by a “column” or, as it is also called, “corner” is often used. Let's look at the implementation of this method using examples.

Before moving on to examples, I will introduce one more term. He not generally accepted, and we will use it solely for the convenience of presenting the material. For the rest of this page, we will call the highest element of the polynomial $P_n(x)$ the expression $a_(0)x^(n)$. For example, for the polynomial $4x^(14)+87x^2+4x-11$ the leading element will be $4x^(14)$.

Example No. 1

Divide $10x^5+3x^4-12x^3+25x^2-2x+5$ by $5x^2-x+2$ using long division.

So we have two polynomials, $P_5(x)=10x^5+3x^4-12x^3+25x^2-2x+5$ and $G_2(x)=5x^2-x+2$. The degree of the first is $5$, and the degree of the second is $2$. The polynomial $P_5(x)$ is the dividend, and the polynomial $G_2(x)$ is the divisor. Our task is to find the quotient and remainder. We will solve the problem step by step. We will use the same notation as for dividing numbers:

First step

Let's divide the highest element of the polynomial $P_5(x)$ (i.e. $10x^5$) by the highest element of the polynomial $Q_2(x)$ (i.e. $5x^2$):

$$ \frac(10x^5)(5x^2)=2x^(5-2)=2x^3. $$

The resulting expression $2x^3$ is the first element of the quotient:

Multiply the polynomial $5x^2-x+2$ by $2x^3$, obtaining:

$$ 2x^3\cdot (5x^2-x+2)=10x^5-2x^4+4x^3 $$

Let's write down the result:

Now subtract from the polynomial $10x^5+3x^4-12x^3+25x^2-2x+5$ the polynomial $10x^5-2x^4+4x^3$:

$$ 10x^5+3x^4-12x^3+25x^2-2x+5-(10x^5-2x^4+4x^3)=5x^4-16x^3+25x^2-2x+ 5 $$

This concludes the first step. The result we got can be written in expanded form:

$$ 10x^5+3x^4-12x^3+25x^2-2x+5=(5x^2-x+2)\cdot 2x^3+5x^4-16x^3+25x^2-2x +5 $$

Since the degree of the polynomial $5x^4-16x^3+25x^2-2x+5$ (i.e. 4) is greater than the degree of the polynomial $5x^2-x+2$ (i.e. 2), then the process divisions must be continued. Let's move on to the second step.

Second step

Now we will work with the polynomials $5x^4-16x^3+25x^2-2x+5$ and $5x^2-x+2$. In exactly the same way as in the first step, we divide the highest element of the first polynomial (i.e. $5x^4$) by the highest element of the second polynomial (i.e. $5x^2$):

$$ \frac(5x^4)(5x^2)=x^(4-2)=x^2. $$

The resulting expression $x^2$ is the second element of the quotient. Let's add $x^2$ to the quotient

Multiply the polynomial $5x^2-x+2$ by $x^2$, obtaining:

$$ x^2\cdot (5x^2-x+2)=5x^4-x^3+2x^2 $$

Let's write down the result:

Now subtract the polynomial $5x^4-x^3+2x^2$ from the polynomial $5x^4-16x^3+25x^2-2x+5$:

$$ 5x^4-16x^3+25x^2-2x+5-(5x^4-x^3+2x^2)=-15x^3+23x^2-2x+5 $$

Let's add this polynomial below the line:

This ends the second step. The result obtained can be written in expanded form:

$$ 10x^5+3x^4-12x^3+25x^2-2x+5=(5x^2-x+2)\cdot (2x^3+x^2)-15x^3+23x^2 -2x+5 $$

Since the degree of the polynomial $-15x^3+23x^2-2x+5$ (i.e. 3) is greater than the degree of the polynomial $5x^2-x+2$ (i.e. 2), we continue the division process. Let's move on to the third step.

Third step

Now we will work with the polynomials $-15x^3+23x^2-2x+5$ and $5x^2-x+2$. In exactly the same way as in the previous steps, we divide the highest element of the first polynomial (i.e. $-15x^3$) by the highest element of the second polynomial (i.e. $5x^2$):

$$ \frac(-15x^3)(5x^2)=-3x^(2-1)=-3x^1=-3x. $$

The resulting expression $(-3x)$ is the third element of the quotient. Let's add $-3x$ to the quotient

Multiply the polynomial $5x^2-x+2$ by $(-3x)$, obtaining:

$$ -3x\cdot (5x^2-x+2)=-15x^3+3x^2-6x $$

Let's write down the result:

Now subtract the polynomial $-15x^3+3x^2-6x$ from the polynomial $-15x^3+23x^2-2x+5$:

$$ -15x^3+23x^2-2x+5-(-15x^3+3x^2-6x)=20x^2+4x+5 $$

Let's add this polynomial below the line:

This ends the third step. The result obtained can be written in expanded form:

$$ 10x^5+3x^4-12x^3+25x^2-2x+5=(5x^2-x+2)\cdot (2x^3+x^2-3x)+20x^2+4x +5 $$

Since the degree of the polynomial $20x^2+4x+5$ (i.e. 2) is equal to the degree of the polynomial $5x^2-x+2$ (i.e. 2), we continue the division process. Let's move on to the fourth step.

Fourth step

Now we will work with the polynomials $20x^2+4x+5$ and $5x^2-x+2$. In exactly the same way as in the previous steps, we divide the highest element of the first polynomial (i.e. $20x^2$) by the highest element of the second polynomial (i.e. $5x^2$):

$$ \frac(20x^2)(5x^2)=4x^(2-2)=4x^0=4. $$

The resulting number $4$ is the fourth element of the quotient. Let's add $4$ to the quotient

Multiply the polynomial $5x^2-x+2$ by $4$, obtaining:

$$ 4\cdot (5x^2-x+2)=20x^2-4x+8 $$

Let's write down the result:

Now let's subtract the polynomial $20x^2-4x+8$ from the polynomial $20x^2+4x+5$.