Decomposition of a polynomial into simple factors over s. Polynomials
Factoring an equation is the process of finding those terms or expressions that, when multiplied, lead to the initial equation. Factoring is a useful skill for solving basic algebraic problems, and becomes almost essential when working with quadratic equations and other polynomials. Factoring is used to simplify algebraic equations to make them easier to solve. Factoring can help you eliminate certain possible answers faster than you would by solving an equation by hand.
Steps
Factoring numbers and basic algebraic expressions
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Factoring numbers. The concept of factoring is simple, but in practice, factoring can be challenging (if a complex equation is given). So first, let's look at the concept of factoring using numbers as an example, continue with simple equations, and then move on to complex equations. The factors of a given number are the numbers that, when multiplied, give the original number. For example, the factors of the number 12 are the numbers: 1, 12, 2, 6, 3, 4, since 1*12=12, 2*6=12, 3*4=12.
- Likewise, you can think of the factors of a number as its divisors, that is, the numbers that the number is divisible by.
- Find all the factors of the number 60. We often use the number 60 (for example, 60 minutes in an hour, 60 seconds in a minute, etc.) and this number has quite a large number of factors.
- 60 multipliers: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
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Remember: terms of an expression containing a coefficient (number) and a variable can also be factorized. To do this, find the coefficient factors of the variable. Knowing how to factor the terms of equations, you can easily simplify this equation.
- For example, the term 12x can be written as the product of 12 and x. You can also write 12x as 3(4x), 2(6x), etc., breaking down 12 into the factors that work best for you.
- You can deal 12x multiple times in a row. In other words, you shouldn't stop at 3(4x) or 2(6x); continue the expansion: 3(2(2x)) or 2(3(2x)) (obviously 3(4x)=3(2(2x)), etc.)
- For example, the term 12x can be written as the product of 12 and x. You can also write 12x as 3(4x), 2(6x), etc., breaking down 12 into the factors that work best for you.
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Apply the distributive property of multiplication to factor algebraic equations. Knowing how to factor numbers and expression terms (coefficients with variables), you can simplify simple algebraic equations by finding the common factor of a number and an expression term. Typically, to simplify an equation, you need to find the greatest common factor (GCD). This simplification is possible due to the distributive property of multiplication: for any numbers a, b, c, the equality a(b+c) = ab+ac is true.
- Example. Factor the equation 12x + 6. First, find the gcd of 12x and 6. 6 is the largest number that divides both 12x and 6, so you can factor this equation by: 6(2x+1).
- This process is also true for equations that have negative and fractional terms. For example, x/2+4 can be factored into 1/2(x+8); for example, -7x+(-21) can be factored into -7(x+3).
Factoring Quadratic Equations
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Make sure the equation is given in quadratic form (ax 2 + bx + c = 0). Quadratic equations have the form: ax 2 + bx + c = 0, where a, b, c are numerical coefficients other than 0. If you are given an equation with one variable (x) and in this equation there are one or more terms with a second-order variable , you can move all the terms of the equation to one side of the equation and set it equal to zero.
- For example, given the equation: 5x 2 + 7x - 9 = 4x 2 + x – 18. This can be converted into the equation x 2 + 6x + 9 = 0, which is a quadratic equation.
- Equations with variable x of large orders, for example, x 3, x 4, etc. are not quadratic equations. These are cubic equations, fourth-order equations, and so on (unless such equations can be simplified to quadratic equations with the variable x raised to the power of 2).
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Quadratic equations, where a = 1, are expanded into (x+d)(x+e), where d*e=c and d+e=b. If the quadratic equation given to you has the form: x 2 + bx + c = 0 (that is, the coefficient of x 2 is 1), then such an equation can (but is not guaranteed) be expanded into the above factors. To do this, you need to find two numbers that, when multiplied, give “c”, and when added, “b”. Once you find these two numbers (d and e), substitute them into the following expression: (x+d)(x+e), which, when opening the parentheses, leads to the original equation.
- For example, given a quadratic equation x 2 + 5x + 6 = 0. 3*2=6 and 3+2=5, so you can factor this equation into (x+3)(x+2).
- For negative terms, make the following minor changes to the factorization process:
- If a quadratic equation has the form x 2 -bx+c, then it expands into: (x-_)(x-_).
- If a quadratic equation has the form x 2 -bx-c, then it expands into: (x+_)(x-_).
- Note: Spaces can be replaced with fractions or decimals. For example, the equation x 2 + (21/2)x + 5 = 0 is expanded into (x+10)(x+1/2).
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Factorization by trial and error. Simple quadratic equations can be factored by simply substituting numbers into the possible solutions until you find the correct solution. If the equation has the form ax 2 +bx+c, where a>1, possible solutions are written in the form (dx +/- _)(ex +/- _), where d and e are non-zero numerical coefficients, which when multiplied give a. Either d or e (or both coefficients) can be equal to 1. If both coefficients are equal to 1, then use the method described above.
- For example, given the equation 3x 2 - 8x + 4. Here 3 has only two factors (3 and 1), so possible solutions are written as (3x +/- _)(x +/- _). In this case, substituting -2 for spaces, you will find the correct answer: -2*3x=-6x and -2*x=-2x; - 6x+(-2x)=-8x and -2*-2=4, that is, such an expansion when opening the brackets will lead to the terms of the original equation.
What should you do if, in the process of solving a problem from the Unified State Exam or on an entrance exam in mathematics, you received a polynomial that cannot be factorized using the standard methods that you learned at school? In this article, a math tutor will tell you about one effective method, the study of which is outside the scope of the school curriculum, but with the help of which factoring a polynomial is not difficult. Read this article to the end and watch the attached video tutorial. The knowledge you gain will help you in the exam.
Factoring a polynomial using the division method
In the event that you received a polynomial greater than the second degree and were able to guess the value of the variable at which this polynomial becomes equal to zero (for example, this value is equal to ), know! This polynomial can be divided by .
For example, it is easy to see that a fourth-degree polynomial vanishes at . This means that it can be divided without a remainder by , thereby obtaining a polynomial of third degree (less by one). That is, present it in the form:
Where A, B, C And D- some numbers. Let's expand the brackets:
Since the coefficients for the same degrees must be the same, we get:
So, we got:
Let's move on. It is enough to go through several small integers to see that the third-degree polynomial is again divisible by . This results in a polynomial of the second degree (less by one). Then move on to a new entry:
Where E, F And G- some numbers. We open the brackets again and arrive at the following expression:
Again, from the condition of equality of coefficients for the same degrees, we obtain:
Then we get:
That is, the original polynomial can be factorized as follows:
In principle, if desired, using the difference of squares formula, the result can also be represented in the following form:
Here is a simple and effective way to factor polynomials. Remember it, it may be useful to you in an exam or math competition. Check if you have learned how to use this method. Try to solve the following task yourself.
Factor the polynomial:
Write your answers in the comments.
Material prepared by Sergey Valerievich
The concepts of “polynomial” and “factorization of a polynomial” in algebra are encountered very often, because you need to know them in order to easily carry out calculations with large multi-digit numbers. This article will describe several decomposition methods. All of them are quite simple to use; you just need to choose the right one for each specific case.
The concept of a polynomial
A polynomial is a sum of monomials, that is, expressions containing only the operation of multiplication.
For example, 2 * x * y is a monomial, but 2 * x * y + 25 is a polynomial that consists of 2 monomials: 2 * x * y and 25. Such polynomials are called binomials.
Sometimes, for the convenience of solving examples with multivalued values, an expression needs to be transformed, for example, decomposed into a certain number of factors, that is, numbers or expressions between which the multiplication action is performed. There are a number of ways to factor a polynomial. It is worth considering them, starting with the most primitive, which is used in primary school.
Grouping (record in general form)
The formula for factoring a polynomial using the grouping method in general looks like this:
ac + bd + bc + ad = (ac + bc) + (ad + bd)
It is necessary to group the monomials so that each group has a common factor. In the first bracket this is the factor c, and in the second - d. This must be done in order to then move it out of the bracket, thereby simplifying the calculations.
Decomposition algorithm using a specific example
The simplest example of factoring a polynomial using the grouping method is given below:
10ac + 14bc - 25a - 35b = (10ac - 25a) + (14bc - 35b)
In the first bracket you need to take the terms with the factor a, which will be common, and in the second - with the factor b. Pay attention to the + and - signs in the finished expression. We put in front of the monomial the sign that was in the initial expression. That is, you need to work not with the expression 25a, but with the expression -25. The minus sign seems to be “glued” to the expression behind it and always taken into account when calculating.
In the next step, you need to take the multiplier, which is common, out of brackets. This is exactly what the grouping is for. To put outside the bracket means to write before the bracket (omitting the multiplication sign) all those factors that are exactly repeated in all the terms that are in the bracket. If there are not 2, but 3 or more terms in a bracket, the common factor must be contained in each of them, otherwise it cannot be taken out of the bracket.
In our case, there are only 2 terms in brackets. The overall multiplier is immediately visible. In the first bracket it is a, in the second it is b. Here you need to pay attention to the digital coefficients. In the first bracket, both coefficients (10 and 25) are multiples of 5. This means that not only a, but also 5a can be taken out of the bracket. Before the bracket, write 5a, and then divide each of the terms in brackets by the common factor that was taken out, and also write the quotient in brackets, not forgetting about the + and - signs. Do the same with the second bracket, take out 7b, as well as 14 and 35 multiple of 7.
10ac + 14bc - 25a - 35b = (10ac - 25a) + (14bc - 35b) = 5a(2c - 5) + 7b(2c - 5).
We got 2 terms: 5a(2c - 5) and 7b(2c - 5). Each of them contains a common factor (the entire expression in brackets is the same here, which means it is a common factor): 2c - 5. It also needs to be taken out of the bracket, that is, terms 5a and 7b remain in the second bracket:
5a(2c - 5) + 7b(2c - 5) = (2c - 5)*(5a + 7b).
So the full expression is:
10ac + 14bc - 25a - 35b = (10ac - 25a) + (14bc - 35b) = 5a(2c - 5) + 7b(2c - 5) = (2c - 5)*(5a + 7b).
Thus, the polynomial 10ac + 14bc - 25a - 35b is decomposed into 2 factors: (2c - 5) and (5a + 7b). The multiplication sign between them can be omitted when writing
Sometimes there are expressions of this type: 5a 2 + 50a 3, here you can put out of brackets not only a or 5a, but even 5a 2. You should always try to put the largest common factor out of the bracket. In our case, if we divide each term by a common factor, we get:
5a 2 / 5a 2 = 1; 50a 3 / 5a 2 = 10a(when calculating the quotient of several powers with equal bases, the base is preserved and the exponent is subtracted). Thus, the unit remains in the bracket (in no case do you forget to write one if you take one of the terms out of the bracket) and the quotient of division: 10a. It turns out that:
5a 2 + 50a 3 = 5a 2 (1 + 10a)
Square formulas
For ease of calculation, several formulas were derived. These are called abbreviated multiplication formulas and are used quite often. These formulas help factor polynomials containing degrees. This is another effective way to factorize. So here they are:
- a 2 + 2ab + b 2 = (a + b) 2 - a formula called the “square of the sum”, since as a result of decomposition into a square, the sum of numbers enclosed in brackets is taken, that is, the value of this sum is multiplied by itself 2 times, and therefore is a multiplier.
- a 2 + 2ab - b 2 = (a - b) 2 - the formula for the square of the difference, it is similar to the previous one. The result is the difference, enclosed in parentheses, contained in the square power.
- a 2 - b 2 = (a + b)(a - b)- this is a formula for the difference of squares, since initially the polynomial consists of 2 squares of numbers or expressions, between which subtraction is performed. Perhaps, of the three mentioned, it is used most often.
Examples for calculations using square formulas
The calculations for them are quite simple. For example:
- 25x 2 + 20xy + 4y 2 - use the formula “square of the sum”.
- 25x 2 is the square of 5x. 20xy is the double product of 2*(5x*2y), and 4y 2 is the square of 2y.
- Thus, 25x 2 + 20xy + 4y 2 = (5x + 2y) 2 = (5x + 2y)(5x + 2y). This polynomial is decomposed into 2 factors (the factors are the same, so it is written as an expression with a square power).
Actions using the squared difference formula are carried out similarly to these. The remaining formula is difference of squares. Examples of this formula are very easy to define and find among other expressions. For example:
- 25a 2 - 400 = (5a - 20)(5a + 20). Since 25a 2 = (5a) 2, and 400 = 20 2
- 36x 2 - 25y 2 = (6x - 5y) (6x + 5y). Since 36x 2 = (6x) 2, and 25y 2 = (5y 2)
- c 2 - 169b 2 = (c - 13b)(c + 13b). Since 169b 2 = (13b) 2
It is important that each of the terms is a square of some expression. Then this polynomial must be factorized using the difference of squares formula. For this, it is not necessary that the second degree be above the number. There are polynomials that contain large degrees, but still fit these formulas.
a 8 +10a 4 +25 = (a 4) 2 + 2*a 4 *5 + 5 2 = (a 4 +5) 2
In this example, a 8 can be represented as (a 4) 2, that is, the square of a certain expression. 25 is 5 2, and 10a is 4 - this is the double product of the terms 2 * a 4 * 5. That is, this expression, despite the presence of degrees with large exponents, can be decomposed into 2 factors in order to subsequently work with them.
Cube formulas
The same formulas exist for factoring polynomials containing cubes. They are a little more complicated than those with squares:
- a 3 + b 3 = (a + b)(a 2 - ab + b 2)- this formula is called the sum of cubes, since in its initial form the polynomial is the sum of two expressions or numbers enclosed in a cube.
- a 3 - b 3 = (a - b)(a 2 + ab + b 2) - a formula identical to the previous one is designated as the difference of cubes.
- a 3 + 3a 2 b + 3ab 2 + b 3 = (a + b) 3 - cube of a sum, as a result of calculations, the sum of numbers or expressions is enclosed in brackets and multiplied by itself 3 times, that is, located in a cube
- a 3 - 3a 2 b + 3ab 2 - b 3 = (a - b) 3 - the formula, compiled by analogy with the previous one, changing only some signs of mathematical operations (plus and minus), is called the “difference cube”.
The last two formulas are practically not used for the purpose of factoring a polynomial, since they are complex, and it is rare enough to find polynomials that fully correspond to exactly this structure so that they can be factored using these formulas. But you still need to know them, since they will be required when operating in the opposite direction - when opening parentheses.
Examples on cube formulas
Let's look at an example: 64a 3 − 8b 3 = (4a) 3 − (2b) 3 = (4a − 2b)((4a) 2 + 4a*2b + (2b) 2) = (4a−2b)(16a 2 + 8ab + 4b 2 ).
Quite simple numbers are taken here, so you can immediately see that 64a 3 is (4a) 3, and 8b 3 is (2b) 3. Thus, this polynomial is expanded according to the formula difference of cubes into 2 factors. Actions using the formula for the sum of cubes are carried out by analogy.
It is important to understand that not all polynomials can be expanded in at least one way. But there are expressions that contain greater powers than a square or a cube, but they can also be expanded into abbreviated multiplication forms. For example: x 12 + 125y 3 =(x 4) 3 +(5y) 3 =(x 4 +5y)*((x 4) 2 − x 4 *5y+(5y) 2)=(x 4 + 5y)( x 8 − 5x 4 y + 25y 2).
This example contains as many as 12 degrees. But even it can be factorized using the sum of cubes formula. To do this, you need to imagine x 12 as (x 4) 3, that is, as a cube of some expression. Now, instead of a, you need to substitute it in the formula. Well, the expression 125y 3 is a cube of 5y. Next, you need to compose the product using the formula and perform calculations.
At first, or in case of doubt, you can always check by inverse multiplication. You just need to open the parentheses in the resulting expression and perform actions with similar terms. This method applies to all of the reduction methods listed: both to working with a common factor and grouping, and to working with formulas of cubes and quadratic powers.
In general, this task requires a creative approach, since there is no universal method for solving it. But let's try to give a few tips.
In the overwhelming majority of cases, the factorization of a polynomial is based on a corollary of Bezout’s theorem, that is, the root is found or selected and the degree of the polynomial is reduced by one by dividing by . The root of the resulting polynomial is sought and the process is repeated until complete expansion.
If the root cannot be found, then specific expansion methods are used: from grouping to introducing additional mutually exclusive terms.
Further presentation is based on the skills of solving equations of higher degrees with integer coefficients.
Bracketing out the common factor.
Let's start with the simplest case, when the free term is equal to zero, that is, the polynomial has the form .
Obviously, the root of such a polynomial is , that is, we can represent the polynomial in the form .
This method is nothing more than putting the common factor out of brackets.
Example.
Factor a third degree polynomial.
Solution.
Obviously, what is the root of the polynomial, that is X can be taken out of brackets:
Let's find the roots of the quadratic trinomial
Thus,
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Factoring a polynomial with rational roots.
First, let's consider a method for expanding a polynomial with integer coefficients of the form , the coefficient of the highest degree is equal to one.
In this case, if a polynomial has integer roots, then they are divisors of the free term.
Example.
Solution.
Let's check if there are intact roots. To do this, write down the divisors of the number -18
: . That is, if a polynomial has integer roots, then they are among the written numbers. Let's check these numbers sequentially using Horner's scheme. Its convenience also lies in the fact that in the end we obtain the expansion coefficients of the polynomial:
That is, x=2 And x=-3 are the roots of the original polynomial and we can represent it as a product:
It remains to expand the quadratic trinomial.
The discriminant of this trinomial is negative, therefore it has no real roots.
Answer:
Comment:
Instead of Horner's scheme, one could use the selection of the root and subsequent division of the polynomial by a polynomial.
Now consider the expansion of a polynomial with integer coefficients of the form , and the coefficient of the highest degree is not equal to one.
In this case, the polynomial can have fractionally rational roots.
Example.
Factor the expression.
Solution.
By performing a variable change y=2x, let's move on to a polynomial with a coefficient equal to one at the highest degree. To do this, first multiply the expression by 4 .
If the resulting function has integer roots, then they are among the divisors of the free term. Let's write them down:
Let us sequentially calculate the values of the function g(y) at these points until zero is reached.
Online calculator.
Isolating the square of a binomial and factoring a square trinomial.
This math program distinguishes a square binomial from a square trinomial, i.e. does a transformation like: \(ax^2+bx+c \rightarrow a(x+p)^2+q \) and factorizes a quadratic trinomial: \(ax^2+bx+c \rightarrow a(x+n)(x+m) \)
Those. the problems boil down to finding the numbers \(p, q\) and \(n, m\)
The program not only gives the answer to the problem, but also displays the solution process.
This program can be useful for high school students in general education schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with detailed solutions.
In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.
If you are not familiar with the rules for entering a quadratic trinomial, we recommend that you familiarize yourself with them.
Rules for entering a quadratic polynomial
Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q\), etc.
Numbers can be entered as whole or fractional numbers.
Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.
Rules for entering decimal fractions.
In decimal fractions, the fractional part can be separated from the whole part by either a period or a comma.
For example, you can enter decimal fractions like this: 2.5x - 3.5x^2
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The whole part is separated from the fraction by the ampersand sign: &
Input: 3&1/3 - 5&6/5x +1/7x^2
Result: \(3\frac(1)(3) - 5\frac(6)(5) x + \frac(1)(7)x^2\)
When entering an expression you can use parentheses. In this case, when solving, the introduced expression is first simplified.
For example: 1/2(x-1)(x+1)-(5x-10&1/2)
Example of a detailed solution
Isolating the square of a binomial.$$ ax^2+bx+c \rightarrow a(x+p)^2+q $$ $$2x^2+2x-4 = $$ $$2x^2 +2 \cdot 2 \cdot\left( \frac(1)(2) \right)\cdot x+2 \cdot \left(\frac(1)(2) \right)^2-\frac(9)(2) = $$ $$2\left (x^2 + 2 \cdot\left(\frac(1)(2) \right)\cdot x + \left(\frac(1)(2) \right)^2 \right)-\frac(9 )(2) = $$ $$2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Answer:$$2x^2+2x-4 = 2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Factorization.$$ ax^2+bx+c \rightarrow a(x+n)(x+m) $$ $$2x^2+2x-4 = $$
$$ 2\left(x^2+x-2 \right) = $$
$$ 2 \left(x^2+2x-1x-1 \cdot 2 \right) = $$ $$ 2 \left(x \left(x +2 \right) -1 \left(x +2 \right ) \right) = $$ $$ 2 \left(x -1 \right) \left(x +2 \right) $$ Answer:$$2x^2+2x-4 = 2 \left(x -1 \right) \left(x +2 \right) $$
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A little theory.
Isolating the square of a binomial from a square trinomial
If the square trinomial ax 2 +bx+c is represented as a(x+p) 2 +q, where p and q are real numbers, then we say that from square trinomial, the square of the binomial is highlighted.
From the trinomial 2x 2 +12x+14 we extract the square of the binomial.
\(2x^2+12x+14 = 2(x^2+6x+7) \)
To do this, imagine 6x as a product of 2*3*x, and then add and subtract 3 2. We get:
$$ 2(x^2+2 \cdot 3 \cdot x + 3^2-3^2+7) = 2((x+3)^2-3^2+7) = $$ $$ = 2 ((x+3)^2-2) = 2(x+3)^2-4 $$
That. We extract the square binomial from the square trinomial, and showed that:
$$ 2x^2+12x+14 = 2(x+3)^2-4 $$
Factoring a quadratic trinomial
If the square trinomial ax 2 +bx+c is represented in the form a(x+n)(x+m), where n and m are real numbers, then the operation is said to have been performed factorization of a quadratic trinomial.
Let us show with an example how this transformation is done.
Let's factor the quadratic trinomial 2x 2 +4x-6.
Let us take the coefficient a out of brackets, i.e. 2:
\(2x^2+4x-6 = 2(x^2+2x-3) \)
Let's transform the expression in brackets.
To do this, imagine 2x as the difference 3x-1x, and -3 as -1*3. We get:
$$ = 2(x^2+3 \cdot x -1 \cdot x -1 \cdot 3) = 2(x(x+3)-1 \cdot (x+3)) = $$
$$ = 2(x-1)(x+3) $$
That. We factored the quadratic trinomial, and showed that:
$$ 2x^2+4x-6 = 2(x-1)(x+3) $$
Note that factoring a quadratic trinomial is possible only if the quadratic equation corresponding to this trinomial has roots.
Those. in our case, it is possible to factor the trinomial 2x 2 +4x-6 if the quadratic equation 2x 2 +4x-6 =0 has roots. In the process of factorization, we established that the equation 2x 2 + 4x-6 = 0 has two roots 1 and -3, because with these values, the equation 2(x-1)(x+3)=0 turns into a true equality.