Division of polynomials by "column" ("corner"). Division of polynomials by a corner Divide expression by expression online

Statement

remainder incomplete private.

Comment

For any polynomials $A(x)$ and $B(x)$ (the degree of $B(x)$ is greater than 0) there are unique polynomials $Q(x)$ and $R(x)$ from the condition of the assertion.

The remainder after dividing the polynomial $x^(4) + 3x^(3) +5$ by $x^(2) + 1$ is $3x + 4$:$x^(4) + 3x^(3) +5 = (x^(2) + 3x +1)(x^(2) + 1) +3x + 4.$
The remainder after dividing the polynomial $x^(4) + 3x^(3) +5$ by $x^(4) + 1$ is $3x^(3) + 4$:$x^(4) + 3x^( 3) +5 = 1 \cdot (x^(2) + 1) +3x^(3) + 4.$
The remainder after dividing the polynomial $x^(4) + 3x^(3) +5$ by $x^(6) + 1$ is $x^(4) + 3x^(3) +5$:$x^( 4) + 3x^(3) +5 = 0 \cdot (x^(6) + 1) + x^(4) + 3x^(3) +5.$

Statement

For any two polynomials $A(x)$ and $B(x)$ (where the degree of the polynomial $B(x)$ is non-zero), there exists a polynomial representation $A(x)$ in the form $A(x) = Q (x)B(x) + R(x)$, where $Q(x)$ and $R(x)$ are polynomials and the degree of $R(x)$ is less than the degree of $B(x).$

Proof

We will prove the assertion by induction on the degree of the polynomial $A(x).$ Denote it by $n$. If $n = 0$, the statement is true: $A(x)$ can be represented as $A(x) = 0 \cdot B(x) + A(x).$ Now, let the statement be proved for polynomials of degree $n \ leqm$. Let us prove the assertion for polynomials of degree $k= n+1.$

Let the degree of the polynomial $B(x)$ be equal to $m$. Consider three cases: $k< m$, $k = m$ и $k >m$ and prove the assertion for each of them.

$k< m$
The polynomial $A(x)$ can be represented as
$A(x) = 0 \cdot B(x) + A(x).$
The assertion has been made.
$k = m$
Let the polynomials $A(x)$ and $B(x)$ have the form
$A(x) = a_(n+1)x^(n+1) + a_(n)x^(n) + \dots + a_(1)x + a_(0), \: \mbox(where ) \: a_(n+1) \neq 0;$
$B(x) = b_(n+1)x^(n+1) + b_(n)x^(n) + \dots + b_(1)x + b_(0), \: \mbox(where ) \: b_(n+1) \neq 0.$
Let's represent $A(x)$ as
$A(x) = \dfrac(a_(n+1))(b_(n+1))B(x) - \Big(\dfrac(a_(n+1))(b_(n+1)) B(x) - A(x)\Big).$
Note that the degree of the polynomial $\dfrac(a_(n+1))(b_(n+1))B(x) - A(x)$ is at most $n+1$, then this representation is the desired one and the assertion is satisfied.
$k > m$
We represent the polynomial $A(x)$ in the form
$A(x) = x(a_(n+1)x^(n) + a_(n)x^(n-1) + \dots + a_(1)) + a_(0), \: \mbox (where) \: a_(n+1) \neq 0.$
Consider the polynomial $A"(x) = a_(n+1)x^(n) + a_(n)x^(n-1) + \dots + a_(1).$ can be represented as $A"(x) = Q"(x)B(x) + R"(x)$, where the degree of the polynomial $R"(x)$ is less than $m$, then the representation for $A(x) $ can be rewritten as
$A(x) = x(Q"(x)B(x) + R"(x)) + a_(0) = xQ"(x)B(x) + xR"(x) + a_(0) .$
Note that the degree of the polynomial $xR"(x)$ is less than $m+1$, i.e. less than $k$. Then $xR"(x)$ satisfies the inductive assumption and can be represented as $ xR"(x) = Q""(x)B(x) + R""(x)$, where the degree of the polynomial $R""(x)$ is less than $m$. Rewrite the representation for $A(x)$ how
$A(x) = xQ"(x)B(x) + Q""(x)B(x) + R""(x) + a_(0) =$
$= (xQ"(x)+xQ""(x))B(x) + R""(x) + a_(0).$
The degree of the polynomial $R""(x) + a_(0)$ is less than $m$, so the statement is true.

The assertion has been proven.

In this case, the polynomial $R(x)$ is called remainder from dividing $A(x)$ by $B(x)$, and $Q(x)$ - incomplete private.

If the remainder of $R(x)$ is a zero polynomial, then $A(x)$ is said to be divisible by $B(x)$.

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 division of polynomials by a corner and multiplication by a column are analyzed in detail.

Content

Theorem

Let Pk (x), Qn (x) are polynomials in variable x of degrees k and n , respectively, with k ≥ n . Then the polynomial P k (x) can only be represented in the following way:
(1) P k (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 - known coefficients, s i , u i - unknown coefficients.

Let's introduce the notation:
.
Substitute in (1) :
;
(2) .
The first term on the right side is a polynomial of degree k. The sum of the second and third terms is a polynomial of degree at most k - 1 . Equate the coefficients at x k :
p k = s k-n q n .
Hence s k-n = p k / q n .

Let's transform the equation (2) :
.
Let's introduce the notation: .
Since s k-n = p k / q n , then the coefficient at x k is equal to zero. Therefore - this is a polynomial of degree at most 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 get the equation:
,
from which we determine the coefficients of the polynomial U n- 1(x).

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

Division of polynomials

Dividing both sides of the equation (1) on Q n (x), we get:
(4) .
By analogy with decimal numbers, S k-n (x) is called the integer part of the fraction or private, 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 an integer part and a proper fraction.

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

Therefore, polynomials can be applied to the rule of division by a corner (sometimes called division by a column), which is applied to the division of numbers. The only difference is that, when dividing polynomials, you do not need to convert numbers greater than nine to higher digits. Consider the process of dividing polynomials by a corner using specific examples.

An example of dividing polynomials by a corner

Here the numerator is a polynomial of the fourth degree. The denominator is a polynomial of the second degree. Because the 4 ≥ 2 , then the fraction is not correct. We select the integer part by dividing the polynomials with a corner (in a column):

Let us give a detailed description of the division process. The original polynomials are written in the left and right columns. Under the denominator polynomial, in the right column, we draw a horizontal line (corner). Below this line, at an angle, there will be an integer part of the fraction.

1.1 We find the first member of the integer part (under the corner). To do this, we divide the highest term of the numerator by the highest term of the denominator: .

1.2 Multiply 2x2 on x 2 - 3 x + 5:
. The result is written in the left column:

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

.

So, we got an intermediate result:
.

The fraction on the right side is incorrect 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 row of the left column.
2.1 Divide the senior member of the numerator by the senior member of the denominator: ;

2.2 We 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 denominator polynomial, 1 < 2 . Therefore, the fraction is correct.

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

Example 2

Select the integer 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:
.

Multiplication of polynomials by a column

You can also multiply polynomials by a column, similar to the multiplication of integers. Let's consider specific examples.

An example of multiplying polynomials by a column

Find the product of polynomials:
.

2.1
.

2.2
.

2.3
.
The result is written in a column, aligning the powers of x.

3
;
;
;
.

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

Example 2

Find the product of polynomials in a column:
.

When multiplying polynomials by a column, it is important to write the same powers of the variable x under each other. If some powers of x are omitted, then they should be written explicitly by multiplying by zero, or leave spaces.

In this example, some degrees are omitted. Therefore, we write them explicitly, multiplied by zero:
.
We multiply polynomials by a column.

1 We write the original polynomials under each other in a column and draw a line.

2.1 We multiply the lowest term of the second polynomial by the first polynomial:
.
The result is written in a column.

2.2 The next term of the second polynomial is equal to zero. Therefore, its product by the first polynomial is also equal to zero. The null line can be omitted.

2.3 We multiply the next term of the second polynomial by the first polynomial:
.
The result is written in a column, aligning the powers of x.

2.3 We multiply the next (highest) term of the second polynomial by the first polynomial:
.
The result is written in a column, aligning the powers of x.

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

General view of the monomial

f(x)=axn, where:

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

-x- variable

-n exponent that belongs to the set N

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

Examples: 3x2 and -5x2; ½ x 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)=3x2-5; h(x)=5x-1
A polynomial containing three terms is called a trinomial.

General form of a polynomial with one variable

where:

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

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

third degree polynomial with coefficients 5, -2, 7 and -1
5 - leading factor
-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 factor
-4 - free member
x- variable

Polynomial division

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 a division p(x) on the 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 descending order. 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 into leading member q(x), and the result is written below the dividing line (in the denominator).
We multiply the result by all terms q(x) and write the result with opposite signs under the terms p(x) with the corresponding degrees.
We add term by term the terms with the same degrees.
We assign the remaining terms to the result p(x).
We 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 under 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) x5 -3x4 +2x3 +7x2 -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

2x4 -2x3 +2x2

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

8) x5 -3x4 +2x3 +7x2 -3x+5

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

2x4 -2x3 +2x2

/ -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 first degree polynomial

This division can be done using the above algorithm, or even faster using Horner's method.
If a 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)- first degree polynomial ⇒ q(x)=mx+n
Then the polynomial in the quotient will have a 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 \u003d 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 \u003d 5
b 2 \u003d 3.5-2 \u003d 13
b 1 =3.13+4=43 ⇒ c(x)=5x 3 +13x 2 +43x+123; r=362
b 0 \u003d 3.43-6 \u003d 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 \u003d -2
p(x)=1+x(-4+x(1+x(0+x(3-2x))))

b 4 \u003d -2 b 1 =(-2).(-14)+1=29
b 3 =(-2).(-2)+3=7 b 0 =(-2).29-4=-62
b2=(-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))
b2=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 the remainder 1-9 .
If we divide by a polynomial of the first degree mx+n, then to find the quotient and the 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(x0).
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

Let it be required

(2x 3 - 7x 2 + x + 1) ÷ (2x - 1).

Here the product (2x 3 - 7x 2 + x + 1) and one factor (2x - 1) are given, - you need to find another factor. In this example, it is immediately clear (but this cannot be established in general) that the other, desired, factor, or quotient, is also a polynomial. This is clear because this product has 4 terms, and this multiplier is only 2. However, it is impossible to say in advance how many terms the desired multiplier has: there may be 2 terms, 3 terms, etc. Remembering that the highest term of the product always turns out from multiplying the highest term of one factor by the highest term of another (see multiplication of a polynomial by a polynomial) and that there cannot be terms like this, we are sure that 2x 3 (the highest term of this product) will come from multiplying 2x (the highest term of this factor ) by the unknown leading term of the sought multiplier. To find the last one, therefore, we have to divide 2x 3 by 2x - we get x 2 . This is the senior member of the private.

Recall then that when multiplying a polynomial by a polynomial, each term of one polynomial must be multiplied by each term of the other. Therefore, this product (2x 3 - 7x 2 + x + 1) is the product of the divisor (2x - 1) and all terms of the quotient. But we can now find the product of the divisor and the first (highest) member of the quotient, i.e. (2x - 1) ∙ x 2; we get 2x 3 - x 2 . Knowing the product of the divisor by all terms of the quotient (it = 2x 3 - 7x 2 + x + 1) and knowing the product of the divisor by the 1st term of the quotient (it = 2x 3 - x 2), by subtraction we can find the product of the divisor by all the others, except for the 1st, members of the private. Get

(2x 3 - 7x 2 + x + 1) - (2x 3 - x 2) = 2x 3 - 7x 2 + x + 1 - 2x 3 + x 2 = -6x 2 + x + 1.

The highest term (–6x 2) of this remaining product must be the product of the highest term of the divisor (2x) and the highest term of the rest (except the 1st term) of the quotient. From here we find the senior term of the remaining quotient. We need –6x 2 ÷ 2x, we get –3x. This is the second term of the desired quotient. We can again find the product of the divisor (2x - 1) and the second, just found, quotient term, i.e., -3x.

We get (2x - 1) ∙ (-3x) \u003d -6x 2 + 3x. From this entire product, we have already subtracted the product of the divisor by the 1st term of the quotient and got the remainder -6x 2 + x + 1, which is the product of the divisor by the rest, except for the 1st, terms of the quotient. Subtracting from it the product just found -6x 2 + 3x, we get the remainder, which is the product of the divisor by all the other, except for the 1st and 2nd, members of the quotient:

-6x 2 + x + 1 - (-6x 2 + 3x) = -6x 2 + x + 1 + 6x 2 - 3x = -2x + 1.

Dividing the senior term of this remaining product (–2x) by the senior term of the divisor (2x), we get the senior term of the rest of the quotient, or its third term, (–2x) ÷ 2x = –1, this is the 3rd term of the quotient.

Multiplying the divisor by it, we get

(2x – 1) ∙ (–1) = –2x + 1.

Subtracting this product of the divisor by the 3rd term of the quotient from the entire product remaining so far, i.e.

(–2x + 1) – (–2x + 1) = –2x + 1 + 2x – 1 = 0,

we will see that in our example the product is divided into the rest, except for the 1st, 2nd and 3rd, members of the quotient = 0, from which we conclude that the quotient has no more members, i.e.

(2x 3 - 7x 2 + x + 1) ÷ (2x - 1) = x 2 - 3x - 1.

From the previous we see: 1) it is convenient to arrange the terms of the dividend and divisor in descending powers, 2) it is necessary to establish some kind of order for performing calculations. Such a convenient order can be considered the one that is used in arithmetic when dividing multi-valued numbers. Following it, we arrange all the previous calculations as follows (more brief explanations are given on the side):

Those subtractions that are needed here are carried out by changing the signs of the terms of the subtrahend, and these variable signs are written on top.

Yes, it's written

This means: the subtrahend was 2x 3 - x 2, and after changing signs, we got -2x 3 + x 2.

Due to the accepted arrangement of calculations, due to the fact that the terms of the dividend and divisor are arranged in descending powers, and due to the fact that the degrees of the letter x in both polynomials go down each time by 1, it turned out that such terms are written under each other (for example: –7x 2 and +x 2) why it is easy to cast them. It can be noted that not all members of the dividend are needed at every moment of the calculation. For example, the term +1 is not needed at the moment where the 2nd term of the quotient was found, and this part of the calculation can be simplified.

More examples:

1. (2a 4 - 3ab 3 - b 4 - 3a 2 b 2) ÷ (b 2 + a 2 + ab).

Arrange the letters a in descending powers and the dividend and the divisor:

(Note that here, due to the absence of a term with a 3 in the dividend, in the first subtraction it turned out that not similar terms -a 2 b 2 and -2a 3 b are signed under each other. Of course, they cannot be reduced to one term and both are written below the line in seniority).

In both examples, one should be more attentive to similar terms: 1) not similar terms often turn out to be written under each other, and 2) sometimes (as, for example, in the last example, the terms -4a n and -a n at the first subtraction) similar terms come out written not one below the other.

It is possible to carry out the division of polynomials in a different order, namely: each time to look for the lowest term or the whole or the remaining quotient. It is convenient in this case to arrange these polynomials in ascending powers of some letter. For example:

This article will consider rational fractions, its allocation of integer parts. Fractions are right and wrong. When the numerator is less than the denominator in a fraction, it is a proper fraction, and vice versa.

Consider examples of proper fractions: 1 2, 9 29, 8 17, improper: 16 3, 21 20, 301 24.

We will calculate fractions that can be reduced, that is, 12 16 is 3 4, 21 14 is 3 2.

When selecting the integer part, the process of dividing the numerator by the denominator is performed. Then such a fraction can be represented as the sum of an integer and a fractional part, where the fractional part is considered the ratio of the remainder of the division and the denominator.

Example 1

Find the remainder when 27 is divided by 4.

Solution

It is necessary to make a division by a column, then we get that

So, 27 4 \u003d integer part + the rest of the n and m and miner \u003d 6 + 3 4

Answer: remainder 3 .

Example 2

Select whole parts 331 12 and 41 57 .

Solution

We divide the denominator by the numerator using a corner:

Therefore, we have that 331 12 \u003d 27 + 7 12.

The second fraction is correct, which means that the integer part is equal to zero.

Answer: integer parts 27 and 0 .

Consider the classification of polynomials, in other words, a fractional rational function. It is considered correct when the degree of the numerator is less than the degree of the denominator, otherwise it is considered incorrect.

Definition 1

Division of a polynomial by a polynomial occurs according to the principle of division by an angle, and the representation of the function as the sum of the integer and fractional parts.

To divide a polynomial into a linear binomial, Horner's scheme is used.

Example 3

Divide x 9 + 7 x 7 - 3 2 x 3 - 2 by the monomial 2 x 2.

Solution

Using the property of division, we write that

x 9 + 7 x 7 - 3 2 x 3 - 2 2 x 2 = x 9 2 x 2 + 7 x 7 2 x 2 - 3 2 x 3 2 x 2 + x 2 2 x 2 - 2 2 x 2 = = 1 2 x 7 + 7 2 x 5 - 3 4 x + 1 2 - 2 2 x - 2 .

Often this type of transformation is performed when taking integrals.

Example 4

Divide a polynomial by a polynomial: 2 x 3 + 3 by x 3 + x.

Solution

The division sign can be written as a fraction of the form 2 x 3 + 3 x 3 + x. Now you need to select the whole part. We do this by dividing by a column. We get that

So, we get that the integer part has the value - 2 x + 3, then the whole expression is written as 2 x 3 + 3 x 3 + x = 2 + - 2 x + 3 x 3 + x

Example 5

Divide and find the remainder after dividing 2 x 6 - x 5 + 12 x 3 - 72 x 2 + 3 by x 3 + 2 x 2 - 1 .

Solution

Let us fix a fraction of the form 2 x 6 - x 5 + 12 x 3 - 72 x 2 + 3 x 3 + 2 x 2 - 1 .

The degree of the numerator is greater than that of the denominator, which means that we have an improper fraction. Using division by a column, select the whole part. We get that

Let's do the division again and get:

From here we have that the remainder is - 65 x 2 + 10 x - 3, hence:

2 x 6 - x 5 + 12 x 3 - 72 x 2 + 3 x 3 + 2 x 2 - 1 = 2 x 3 - 5 x 2 + 10 x - 6 + - 65 x 2 + 10 x - 3 x 3 + 2 x 2 - 1

There are cases where it is necessary to additionally perform a fraction conversion in order to be able to reveal the remainder when dividing. It looks like this:

3 x 5 + 2 x 4 - 12 x 2 - 4 x 3 - 3 = 3 x 2 x 3 - 3 - 3 x 2 x 3 - 3 + 3 x 5 + 2 x 4 - 12 x 2 - 4 x 3 - 3 = = 3 x 2 x 3 - 3 + 2 x 4 - 3 x 2 - 4 x 3 - 3 = 3 x 2 + 2 x 4 - 3 x 2 - 4 x 3 - 3 = = 3 x 2 + 2 x x 3 - 3 - 2 x x 3 - 3 + 2 x 4 - 3 x 2 - 4 x 3 - 3 = = 3 x 2 + 2 x (x 3 - 3) - 3 x 2 + 6 x - 4 x 3 - 3 = 3 x 2 + 2 x + - 3 x 2 + 6 x - 4 x 3 - 3

This means that the remainder when dividing 3 x 5 + 2 x 4 - 12 x 2 - 4 by x 3 - 3 gives the value - 3 x 2 + 6 x - 4. To quickly find the result, abbreviated multiplication formulas are used.

Example 6

Divide 8 x 3 + 36 x 2 + 54 x + 27 by 2 x + 3 .

Solution

Let's write the division as a fraction. We get that 8 x 3 + 36 x 2 + 54 x + 27 2 x + 3 . Note that in the numerator, the expression can be added using the sum cube formula. We have that

8 x 3 + 36 x 2 + 54 x + 27 2 x + 3 = (2 x + 3) 3 2 x + 3 = (2 x + 3) 2 = 4 x 2 + 12 x + 9

The given polynomial is divisible without a remainder.

For the solution, a more convenient solution method is used, and the division of a polynomial by a polynomial is considered to be the most universal, therefore, it is often used when selecting an integer part. The final entry must contain the resulting polynomial from division.

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