This article will consider rational fractions, its selection 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.
Decision
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 .
Decision
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.
Decision
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.
Decision
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 .
Decision
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 .
Decision
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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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; 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)=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
Today we will learn how to divide polynomials into each other, and we will perform the division by a corner by analogy with ordinary numbers. This is a very useful technique, which, unfortunately, is not taught in most schools. So listen carefully to this video tutorial. 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 digits that the resulting numerical value is greater than the one by which we divide. If we cut off one bit, we get five. Obviously, seventeen in five does not fit, so this is not enough. We take two digits - we will 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 got 51. We subtract and we got “eight”. Now we demolish the next digit - five. Divide 85 by seventeen. We take five. Multiply seventeen by five and get 85. Subtract and we get zero.
Solving real examples
Task #1
Now let's follow the same steps, but not with numbers, but with polynomials. For example, let's take this:
\[\frac(((x)^(2))+8x+15)(x+5)=x+3\]
Pay attention, if when dividing numbers by each other, we meant 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 degree.
So, the first step: compare the first elements. Question: what should $x$ be multiplied by to get $((x)^(2))$? Obviously, for one more $x$. Multiply $x+5$ by the number $x$ just found. We have $((x)^(2))+5$ which is subtracted from the dividend. There are $3x$ left. Now we demolish 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 whole operation of dividing by a corner has been reduced to comparing the highest coefficients for the dividend and the divisor. It's even easier than when you divide numbers. There is no need to allocate a certain number of digits - we just compare the highest elements at each step. That's the whole algorithm.
Task #2
Let's try again:
\[\frac(((x)^(2))+x-2)(x-1)=x+2\]
First step: look at the higher coefficients. How much should $x$ be multiplied to write $((x)^(2))$? We multiply term by term. Note that when subtracting, we get exactly $2x$, because
We demolish -2 and again compare the first coefficient obtained with the highest element of the divisor. In total, we got 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 third-degree polynomial acts as a dividend. Let's compare the first elements. To get $((x)^(3))$, you need to multiply $x$ by $((x)^(2))$. After subtracting, we demolish $9x$. We multiply the divisor by $-x$ and subtract. As a result, our expression is completely divided. We write down the answer.
Task #3
Let's move on to the last task:
\[\frac(((x)^(3))+3((x)^(2))+50)(x+5)=((x)^(2))-2x+10\]
Compare $((x)^(3))$ and $x$. Obviously, you need to multiply by $((x)^(2))$. As a result, we see that we got 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 each 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 wrong.
There are no more tricks and tricks in such a division.
The material of today's lesson is nowhere and never found in a "pure" form. It is rarely taught in schools. However, the ability to divide polynomials into each other will greatly help you in solving equations of higher degrees, as well as all sorts of problems of "increased difficulty". Without this technique, you will have to factorize polynomials, select coefficients - and the result is by no means guaranteed. However, polynomials can also be divided by a corner - just like regular numbers! Unfortunately, this technique is not taught in schools. Many teachers believe that dividing polynomials by a corner is something insanely complicated, from the field of higher mathematics. I hasten to assure you: it is not. Moreover, dividing polynomials is even easier than ordinary numbers! Watch the lesson and see for yourself. :) In general, be sure to take this technique into service. The ability to divide polynomials into 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 students and university students. And that's all for me. See you!
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 $A(x)$ as
$A(x) = xQ"(x)B(x) + Q""(x)B(x) + R""(x) + a_(0) =$
$= (xQ"(x)+xQ""(x))B(x) + R""(x) + a_(0).$
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 $B(x)$.
Let's start with some definitions. 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 a free member of the polynomial $P_n(x)$. For example, for $4x^(14)+87x^2+4x-11$ the intercept is $(-11)$. Now let's 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 $G_m(x)$ the divisor, the polynomial $Q_p(x)$ the quotient of $P_n(x)$ divided by $G_m(x)$, and the polynomial $ R_k(x)$ - remainder after dividing $P_n(x)$ by $G_m(x)$. For example, for 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 this 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 after dividing $P_6(x)$ by $G_4(x)$. I 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 a remainder, since the equality holds:
$$ 21x^6+6x^5+105x^2+30x=(3x^4+15)\cdot(7x^2+2x) $$
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. We will analyze the implementation of this method with examples.
Before moving on to the examples, I will introduce one more term. He is not generally accepted, and we will use it solely for the convenience of presenting the material. Until the end of this page, we will call the leading 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 is $4x^(14)$.
Example #1
Divide $10x^5+3x^4-12x^3+25x^2-2x+5$ by $5x^2-x+2$ using "column" 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 the remainder. The problem will be solved step by step. We will use the same notation as for dividing numbers:
First step
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 $2x^3$ to get:
$$ 2x^3\cdot (5x^2-x+2)=10x^5-2x^4+4x^3 $$
Let's write down the result:
Now subtract the polynomial $10x^5-2x^4+4x^3$ from the polynomial $10x^5+3x^4-12x^3+25x^2-2x+5$:
$$ 10x^5+3x^4-12x^3+25x^2-2x+5-(10x^5-2x^4+4x^3)=5x^4-16x^3+25x^2-2x+ 5 $$
This is where the first step ends. The result that 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), the process division 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 the same way as in the first step, we divide the leading element of the first polynomial (i.e. $5x^4$) by the leading 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. Add to the quotient $x^2$
Multiply the polynomial $5x^2-x+2$ by $x^2$ to get:
$$ x^2\cdot (5x^2-x+2)=5x^4-x^3+2x^2 $$
Let's write down the result:
Now subtract the polynomial $5x^4-x^3+2x^2$ from the polynomial $5x^4-16x^3+25x^2-2x+5$:
$$ 5x^4-16x^3+25x^2-2x+5-(5x^4-x^3+2x^2)=-15x^3+23x^2-2x+5 $$
We add this polynomial already under the line:
This is where the second step ends. 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 $-15x^3+23x^2-2x+5$ (ie 3) is greater than the degree of the polynomial $5x^2-x+2$ (ie 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 the same way as in the previous steps, we divide the leading element of the first polynomial (i.e. $-15x^3$) by the leading 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 to the quotient $-3x$
Multiply the polynomial $5x^2-x+2$ by $(-3x)$ to get:
$$ -3x\cdot (5x^2-x+2)=-15x^3+3x^2-6x $$
Let's write down the result:
Now subtract the polynomial $-15x^3+3x^2-6x$ from the polynomial $-15x^3+23x^2-2x+5$:
$$ -15x^3+23x^2-2x+5-(-15x^3+3x^2-6x)=20x^2+4x+5 $$
We add this polynomial already under the line:
This is where the third step ends. 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 $20x^2+4x+5$ (ie 2) is equal to the degree of the polynomial $5x^2-x+2$ (ie 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 the same way as in the previous steps, we divide the leading element of the first polynomial (i.e. $20x^2$) by the leading 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 to the quotient $4$
Multiply the polynomial $5x^2-x+2$ by $4$ to get:
$$ 4\cdot (5x^2-x+2)=20x^2-4x+8 $$
Let's write down the result:
Now we subtract the polynomial $20x^2-4x+8$ from the polynomial $20x^2+4x+5$.