Online calculator.
Isolating the square of a binomial and factoring a square trinomial.

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. 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.

8 examples of factoring polynomials are given. They include examples of solving quadratic and biquadratic equations, examples of reciprocal polynomials, and examples of finding integer roots of third- and fourth-degree polynomials.

1. Examples with solving a quadratic equation

Example 1.1

x 4 + x 3 - 6 x 2.

Solution

We take out x 2 outside of brackets:
.
2 + x - 6 = 0:
.
Roots of the equation:
, .

.

Answer

Example 1.2

Factor the third degree polynomial:
x 3 + 6 x 2 + 9 x.

Solution

Let's take x out of brackets:
.
Solving the quadratic equation x 2 + 6 x + 9 = 0:
Its discriminant: .
Since the discriminant is zero, the roots of the equation are multiples: ;
.

From this we obtain the factorization of the polynomial:
.

Answer

Example 1.3

Factor the fifth degree polynomial:
x 5 - 2 x 4 + 10 x 3.

Solution

We take out x 3 outside of brackets:
.
Solving the quadratic equation x 2 - 2 x + 10 = 0.
Its discriminant: .
Since the discriminant is less than zero, the roots of the equation are complex: ;
, .

The factorization of the polynomial has the form:
.

If we are interested in factorization with real coefficients, then:
.

Answer

Examples of factoring polynomials using formulas

Examples with biquadratic polynomials

Example 2.1

Factor the biquadratic polynomial:
x 4 + x 2 - 20.

Solution

Let's apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b).

;
.

Answer

Example 2.2

Factor the polynomial that reduces to a biquadratic one:
x 8 + x 4 + 1.

Solution

Let's apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b):

;

;
.

Answer

Example 2.3 with recurrent polynomial

Factor the reciprocal polynomial:
.

Solution

A reciprocal polynomial has odd degree. Therefore it has root x = - 1 . Divide the polynomial by x -(-1) = x + 1
.
.
, ;
;


;
.

Answer

As a result we get:

Let's make a substitution:

Examples of factoring polynomials with integer roots
.

Solution

Example 3.1

6
-6, -3, -2, -1, 1, 2, 3, 6 .
Factor the polynomial:;
Let's assume that the equation;
(-6) 3 - 6·(-6) 2 + 11·(-6) - 6 = -504;
(-3) 3 - 6·(-3) 2 + 11·(-3) - 6 = -120;
(-2) 3 - 6·(-2) 2 + 11·(-2) - 6 = -60;
(-1) 3 - 6·(-1) 2 + 11·(-1) - 6 = -24;
1 3 - 6 1 2 + 11 1 - 6 = 0;
2 3 - 6 2 2 + 11 2 - 6 = 0.

3 3 - 6 3 2 + 11 3 - 6 = 0
x 1 = 1 6 3 - 6 6 2 + 11 6 - 6 = 60 2 = 2 6 3 - 6 6 2 + 11 6 - 6 = 60 3 = 3 .
So, we found three roots:
.

Answer

, x

Examples of factoring polynomials with integer roots
.

Solution

Example 3.1

Since the original polynomial is of the third degree, it has no more than three roots. Since we found three roots, they are simple. Then 2 Example 3.2
-2, -1, 1, 2 .
has at least one whole root. Then it is a divisor of the number
(member without x). That is, the whole root can be one of the numbers: 6 ;
We substitute these values ​​one by one: 0 ;
(-2) 4 + 2·(-2) 3 + 3·(-2) 3 + 4·(-2) + 2 =;
(-1) 4 + 2·(-1) 3 + 3·(-1) 3 + 4·(-1) + 2 = .
1 4 + 2 1 3 + 3 1 3 + 4 1 + 2 = 12 2 Example 3.2
1, 2, -1, -2 .
2 4 + 2 2 3 + 3 2 3 + 4 2 + 2 = 54 -1 :
.

If we assume that this equation has an integer root, then it is a divisor of the number 2 = -1 Let's substitute x =
.

So, we have found another root x 2 + 2 = 0 .

It would be possible, as in the previous case, to divide the polynomial by , but we will group the terms:

Since the equation xhas no real roots, then the factorization of the polynomial has the form.

In this lesson, we will recall all the previously studied methods of factoring a polynomial and consider examples of their application, in addition, we will study a new method - the method of isolating a complete square and learn how to use it in solving various problems.Subject:

Factoring polynomials

Lesson:

Factoring polynomials. Method for selecting a complete square. Combination of methods

Let us recall the basic methods of factoring a polynomial that were studied earlier:

;

The method of putting a common factor out of brackets, that is, a factor that is present in all terms of the polynomial. Let's look at an example:

Grouping method. It is not always possible to extract a common factor in a polynomial. In this case, you need to divide its members into groups in such a way that in each group you can take out a common factor and try to break it down so that after taking out the factors in the groups, a common factor appears in the entire expression, and you can continue the decomposition. Let's look at an example:

Let's group the first term with the fourth, the second with the fifth, and the third with the sixth:

Let's take out the common factors in the groups:

The expression now has a common factor. Let's take it out:

Application of abbreviated multiplication formulas. Let's look at an example:

;

Let's write the expression in detail:

Obviously, we have before us the formula for the squared difference, since it is the sum of the squares of two expressions and their double product is subtracted from it. Let's use the formula:

Today we will learn another method - the method of selecting a complete square. It is based on the formulas of the square of the sum and the square of the difference. Let's remind them:

Formula for the square of the sum (difference);

The peculiarity of these formulas is that they contain the squares of two expressions and their double product. Let's look at an example:

Let's write down the expression:

So, the first expression is , and the second is .

In order to create a formula for the square of a sum or difference, the double product of expressions is not enough. It needs to be added and subtracted:

Let's complete the square of the sum:

Let's transform the resulting expression:

Let us apply the formula for the difference of squares, recall that the difference of the squares of two expressions is the product of and the sum of their difference:

So, this method consists, first of all, in identifying the expressions a and b that are squared, that is, determining which expressions are squared in this example. After this, you need to check for the presence of a double product and if it is not there, then add and subtract it, this will not change the meaning of the example, but the polynomial can be factorized using the formulas for the square of the sum or difference and difference of squares, if possible.

Let's move on to solving examples.

Example 1 - factorize:

Let's find expressions that are squared:

Let us write down what their double product should be:

Let's add and subtract double the product:

Let's complete the square of the sum and give similar ones:

Let's write it using the difference of squares formula:

Example 2 - solve the equation:

;

On the left side of the equation is a trinomial. You need to factor it into factors. We use the squared difference formula:

We have the square of the first expression and the double product, the square of the second expression is missing, let’s add and subtract it:

Let's fold a complete square and give similar terms:

Let's apply the difference of squares formula:

So we have the equation

We know that a product is equal to zero only if at least one of the factors is equal to zero. Let's create the following equations based on this:

Let's solve the first equation:

Let's solve the second equation:

Answer: or

;

We proceed similarly to the previous example - select the square of the difference.

Any algebraic polynomial of degree n can be represented as a product of n-linear factors of the form and a constant number, which is the coefficients of the polynomial at the highest degree x, i.e.

Where - are the roots of the polynomial.

The root of a polynomial is the number (real or complex) that makes the polynomial vanish. The roots of a polynomial can be either real roots or complex conjugate roots, then the polynomial can be represented in the following form:

Let's consider methods for decomposing polynomials of degree “n” into the product of factors of the first and second degrees.

Method number 1.Method of undetermined coefficients.

The coefficients of such a transformed expression are determined by the method of indefinite coefficients. The essence of the method is that the type of factors into which a given polynomial is decomposed is known in advance. When using the method of uncertain coefficients, the following statements are true:

P.1. Two polynomials are identically equal if their coefficients are equal for the same powers of x.

P.2. Any polynomial of the third degree is decomposed into the product of linear and quadratic factors.

P.3. Any fourth-degree polynomial can be decomposed into the product of two second-degree polynomials.

Example 1.1. It is necessary to factorize the cubic expression:

P.1. In accordance with the accepted statements, the identical equality holds for the cubic expression:

P.2. The right side of the expression can be represented as terms as follows:

P.3. We compose a system of equations from the condition of equality of coefficients at the corresponding powers of the cubic expression.

This system of equations can be solved by selecting coefficients (if it is a simple academic problem) or methods for solving nonlinear systems of equations can be used. Solving this system of equations, we find that the uncertain coefficients are determined as follows:

Thus, the original expression is factorized in the following form:

This method can be used both in analytical calculations and in computer programming to automate the process of finding the root of an equation.

Method number 2.Vieta formulas

Vieta's formulas are formulas connecting the coefficients of algebraic equations of degree n and its roots. These formulas were implicitly presented in the works of the French mathematician François Vieta (1540 - 1603). Due to the fact that Vieth considered only positive real roots, he therefore did not have the opportunity to write these formulas in a general explicit form.

For any algebraic polynomial of degree n that has n-real roots,

The following relations are valid that connect the roots of a polynomial with its coefficients:

Vieta's formulas are convenient to use to check the correctness of finding the roots of a polynomial, as well as to construct a polynomial from given roots.

Example 2.1. Let us consider how the roots of a polynomial are related to its coefficients using the example of a cubic equation

In accordance with Vieta’s formulas, the relationship between the roots of a polynomial and its coefficients has the following form:

Similar relations can be made for any polynomial of degree n.

Method No. 3. Factoring a quadratic equation with rational roots

From Vieta's last formula it follows that the roots of a polynomial are divisors of its free term and leading coefficient. In this regard, if the problem statement specifies a polynomial of degree n with integer coefficients

then this polynomial has a rational root (irreducible fraction), where p is the divisor of the free term, and q is the divisor of the leading coefficient. In this case, a polynomial of degree n can be represented as (Bezout’s theorem):

A polynomial, the degree of which is 1 less than the degree of the initial polynomial, is determined by dividing a polynomial of degree n binomial, for example, using Horner's scheme or in the simplest way - "column".

Example 3.1. It is necessary to factor the polynomial

P.1. Due to the fact that the coefficient of the highest term is equal to one, the rational roots of this polynomial are divisors of the free term of the expression, i.e. can be integers . We substitute each of the presented numbers into the original expression and find that the root of the presented polynomial is equal to .

Let's divide the original polynomial by a binomial:

Let's use Horner's scheme

The coefficients of the original polynomial are set in the top line, while the first cell of the top line remains empty.

In the first cell of the second line, the found root is written (in the example under consideration, the number “2” is written), and the following values ​​in the cells are calculated in a certain way and they are the coefficients of the polynomial, which is obtained by dividing the polynomial by the binomial. The unknown coefficients are determined as follows:

The value from the corresponding cell of the first row is transferred to the second cell of the second row (in the example under consideration, the number “1” is written).

The third cell of the second row contains the value of the product of the first cell and the second cell of the second row plus the value from the third cell of the first row (in the example under consideration 2 ∙1 -5 = -3).

The fourth cell of the second row contains the value of the product of the first cell and the third cell of the second row plus the value from the fourth cell of the first row (in the example under consideration, 2 ∙ (-3) +7 = 1).

Thus, the original polynomial is factorized:

Method number 4.Using abbreviated multiplication formulas

Abbreviated multiplication formulas are used to simplify calculations, as well as factoring polynomials. Abbreviated multiplication formulas allow you to simplify the solution of individual problems.

Formulas used to factorize

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 signs + and - 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:

1. 25x 2 + 20xy + 4y 2 - use the formula “square of the sum”.
2. 25x 2 is the square of 5x. 20xy is the double product of 2*(5x*2y), and 4y 2 is the square of 2y.
3. 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 much as the 12th degree. 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.