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Factoring a polynomial in two unknowns. Methods for factoring a polynomial of degree higher than the second. Alternative solution

8 examples of factorization of polynomials are given. They include examples with solving quadratic and biquadratic equations, examples with recurrent polynomials, and examples with finding integer roots of third and fourth degree polynomials.

Content


See also: Methods for factoring polynomials
The roots of a quadratic equation
Solution of cubic equations

1. Examples with the solution of a quadratic equation

Example 1.1


x 4 + x 3 - 6 x 2.

Take out x 2 for brackets:
.
2 + x - 6 = 0:
.
Equation roots:
, .


.

Example 1.2

Factoring a third-degree polynomial:
x 3 + 6 x 2 + 9 x.

We take x out of brackets:
.
We solve the quadratic equation x 2 + 6 x + 9 = 0:
Its discriminant is .
Since the discriminant is equal to zero, the roots of the equation are multiples: ;
.

From here we obtain the decomposition of the polynomial into factors:
.

Example 1.3

Factoring a fifth-degree polynomial:
x 5 - 2 x 4 + 10 x 3.

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

The factorization of a polynomial has the form:
.

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

Examples of factoring polynomials using formulas

Examples with biquadratic polynomials

Example 2.1

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

Apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b).

;
.

Example 2.2

Factoring a polynomial that reduces to a biquadratic:
x 8 + x 4 + 1.

Apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b):

;

;
.

Example 2.3 with recursive polynomial

Factoring the recursive polynomial:
.

The recursive polynomial has an odd degree. Therefore it has a root x = - 1 . We divide the polynomial by x - (-1) = x + 1. As a result, we get:
.
We make a substitution:
, ;
;


;
.

Examples of Factoring Polynomials with Integer Roots

Example 3.1

Factoring a polynomial:
.

Suppose the equation

6
-6, -3, -2, -1, 1, 2, 3, 6 .
(-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;
6 3 - 6 6 2 + 11 6 - 6 = 60.

So, we have found three roots:
x 1 = 1 , x 2 = 2 , x 3 = 3 .
Since the original polynomial is of the third degree, it has no more than three roots. Since we have found three roots, they are simple. Then
.

Example 3.2

Factoring a polynomial:
.

Suppose the equation

has at least one integer root. Then it is the divisor of the number 2 (a member without x ). That is, the whole root can be one of the numbers:
-2, -1, 1, 2 .
Substitute these values ​​one by one:
(-2) 4 + 2 (-2) 3 + 3 (-2) 3 + 4 (-2) + 2 = 6 ;
(-1) 4 + 2 (-1) 3 + 3 (-1) 3 + 4 (-1) + 2 = 0 ;
1 4 + 2 1 3 + 3 1 3 + 4 1 + 2 = 12;
2 4 + 2 2 3 + 3 2 3 + 4 2 + 2 = 54.

So, we have found one root:
x 1 = -1 .
We divide the polynomial by x - x 1 = x - (-1) = x + 1:


Then,
.

Now we need to solve the equation of the third degree:
.
If we assume that this equation has an integer root, then it is a divisor of the number 2 (a member without x ). That is, the whole root can be one of the numbers:
1, 2, -1, -2 .
Substitute x = -1 :
.

So we have found another root x 2 = -1 . It would be possible, as in the previous case, to divide the polynomial by , but we will group the terms:
.

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Factoring an equation is the process of finding terms or expressions that, when multiplied, lead to the initial equation. Factoring is a useful skill for solving basic algebraic problems, and becomes a practical necessity 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 rule out certain possible answers faster than you can by manually solving the equation.

Steps

Factorization of numbers and basic algebraic expressions

  1. Factorization of numbers. The concept of factoring is simple, but factoring can be tricky in practice (given a complex equation). So let's start with 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.

    • Similarly, you can think of the factors of a number as its divisors, that is, the numbers that the given number is divisible by.
    • Find all 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 multipliers.
      • 60 multipliers: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.

  2. Remember: terms of an expression containing a coefficient (number) and a variable can also be factored. To do this, find the multipliers of the coefficient at the variable. Knowing how to factorize the terms of the 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. by factoring 12 into the factors that work best for you.
      • You can lay out 12x multiple times in a row. In other words, you shouldn't stop at 3(4x) or 2(6x); continue expansion: 3(2(2x)) or 2(3(2x)) (obviously, 3(4x)=3(2(2x)) etc.)

  3. Apply the distributive property of multiplication to factorize algebraic equations. Knowing how to factorize numbers and terms of an expression (coefficients with variables), you can simplify simple algebraic equations by finding the common factor of a number and a term of an expression. Usually, to simplify the equation, you need to find the greatest common divisor (gcd). Such a 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 into: 6(2x+1).
    • This process is also true for equations that have negative and fractional terms. For example, x/2+4 can be decomposed into 1/2(x+8); for example, -7x+(-21) can be decomposed into -7(x+3).

    Factorization of quadratic equations

    1. Make sure the equation is in quadratic form (ax 2 + bx + c = 0). Quadratic equations are: 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 this equation has 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 equate it to zero.

      • For example, given the equation: 5x 2 + 7x - 9 = 4x 2 + x - 18. It can be converted to the equation x 2 + 6x + 9 = 0, which is a quadratic equation.
      • Equations with a 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 (only if such equations cannot be simplified to quadratic equations with the variable x to the power of 2).

    2. Quadratic equations, where a \u003d 1, are decomposed into (x + d) (x + e), where d * e \u003d c and d + e \u003d b. If the quadratic equation given to you has the form: x 2 + bx + c \u003d 0 (that is, the coefficient at x 2 is equal to 1), then such an equation can (but not guaranteed) be decomposed 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 the brackets are opened, leads to the original equation.

      • For example, given the quadratic equation x 2 + 5x + 6 = 0. 3*2=6 and 3+2=5, so you can expand the equation into (x+3)(x+2).
      • For negative terms, make the following minor changes to the factorization process:
        • If the quadratic equation has the form x 2 -bx + c, then it decomposes into: (x-_) (x-_).
        • If the quadratic equation has the form x 2 -bx-c, then it decomposes into: (x + _) (x-_).
      • Note: spaces can be replaced with fractions or decimals. For example, the equation x 2 + (21/2)x + 5 = 0 is decomposed into (x + 10) (x + 1/2).

    3. Factorization by trial and error. Simple quadratic equations can be factored by simply substituting numbers into possible solutions until you find the correct solution. If the equation has the form ax 2 +bx+c, where a>1, the possible solutions are written as (dx +/- _)(ex +/- _), where d and e are numerical coefficients other than zero, 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 the 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.

    4. Full square. In some cases, quadratic equations can be factored quickly and easily using a special algebraic identity. Any quadratic equation of the form x 2 + 2xh + h 2 = (x + h) 2 . That is, if in your equation the factor b is equal to twice the square root of the factor c, then your equation can be decomposed into (x + (kV.root(c))) 2 .

      • For example, given the equation x 2 + 6x + 9. Here 3 2 =9 and 3*2=6. Therefore, this equation can be decomposed into (x+3)(x+3) or (x + 3) 2 .

    5. Use factorization to solve quadratic equations. By factoring the equation, you can set each factor to zero and calculate the value of x (by solving the equation, we mean finding the values ​​of x that make the equation early to zero).

      • Let's return to the equation x 2 + 5x + 6 \u003d 0. This equation is decomposed into factors (x + 3) (x + 2) \u003d 0. If one of the factors is 0, then the whole equation is 0. Therefore, we write: (x+3)=0 and (x+2)=0 and find x=-3 and x=-2 (respectively).

    6. Check the answer (some answers may be wrong). To do this, substitute the found x values ​​​​into the original equation. Sometimes when substituting the found values, the original equation is not equal to zero; this means that such values ​​of x are wrong.

      • For example, substitute x=-2 and x=-3 into x 2 + 5x + 6 = 0. First, substitute x=-2:
        • (-2) 2 + 5(-2) + 6 = 0
        • 4 + -10 + 6 = 0
        • 0 \u003d 0. That is, x \u003d -2 is the correct answer.
      • Now substitute x=-3:
        • (-3) 2 + 5(-3) + 6 = 0
        • 9 + -15 + 6 = 0
        • 0 \u003d 0. That is, x \u003d -3 is the correct answer.

    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 a number (real or complex) that turns the polynomial to zero. The roots of a polynomial can be both real roots and complex conjugate roots, then the polynomial can be represented in the following form:

    Consider methods for expanding polynomials of degree "n" into the product of factors of the first and second degrees.

    Method number 1.Method of indefinite 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 the given polynomial is decomposed is known in advance. When using the method of indeterminate coefficients, the following statements are true:

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

    P.2. Any third-degree polynomial decomposes into a product of linear and square factors.

    P.3. Any polynomial of the fourth degree decomposes into the product of two polynomials of the second degree.

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

    P.1. In accordance with the accepted statements, the identical equality is true 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 the coefficients for the corresponding powers of the cubic expression.

    This system of equations can be solved by the method of selection of coefficients (if a simple academic problem) or methods for solving nonlinear systems of equations can be used. Solving this system of equations, we obtain that the uncertain coefficients are defined as follows:

    Thus, the original expression is decomposed into factors 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 formulas are formulas relating the coefficients of algebraic equations of degree n and its roots. These formulas were implicitly presented in the works of the French mathematician Francois Vieta (1540 - 1603). Due to the fact that Viet considered only positive real roots, therefore, he 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, which 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 compose a polynomial from given roots.

    Example 2.1. Consider how the roots of a polynomial are related to its coefficients using the cubic equation as an example

    In accordance with the Vieta formulas, the relationship between the roots of a polynomial and its coefficients is as follows:

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

    Method number 3. Factorization of a quadratic equation with rational roots

    It follows from Vieta's last formula that the roots of a polynomial are divisors of its free term and the leading coefficient. In this regard, if the condition of the problem contains 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 whose degree is 1 less than the degree of the initial polynomial is determined by dividing the polynomial of degree n by a binomial, for example, using Horner's scheme or most in a simple way- "column".

    Example 3.1. It is necessary to factorize the polynomial

    P.1. Due to the fact that the coefficient at the highest term is equal to one, then the rational roots of this polynomial are divisors of the free term of the expression, i.e. can be whole numbers . Substituting each of the presented numbers into the original expression, we find that the root of the presented polynomial is .

    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.

    The found root is written in the first cell of the second line (in this example, 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 will result from dividing the polynomial by the binomial. The unknown coefficients are defined as follows:

    The value from the corresponding cell of the first row is transferred to the second cell of the second row (in this example, 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 this example, 2 ∙ 1 -5 = -3).

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

    Thus, the original polynomial is factorized:

    Method number 4.Using Shorthand Multiplication Formulas

    Abbreviated multiplication formulas are used to simplify calculations, as well as the decomposition of polynomials into factors. Abbreviated multiplication formulas make it possible to simplify the solution of individual problems.

    Formulas Used for Factoring

    In order to factorize, it is necessary to simplify the expressions. This is necessary in order to be able to further reduce. The decomposition of a polynomial makes sense when its degree is not lower than the second. A polynomial with the first degree is called linear.

    The article will reveal all the concepts of decomposition, theoretical basis and methods for factoring a polynomial.

    Theory

    Theorem 1

    When any polynomial with degree n having the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 , are represented as a product with a constant factor with the highest degree a n and n linear factors (x - x i) , i = 1 , 2 , … , n , then P n (x) = a n (x - x n) (x - x n - 1) . . . · (x - x 1) , where x i , i = 1 , 2 , … , n - these are the roots of the polynomial.

    The theorem is intended for roots of complex type x i , i = 1 , 2 , … , n and for complex coefficients a k , k = 0 , 1 , 2 , … , n . This is the basis of any decomposition.

    When coefficients of the form a k , k = 0 , 1 , 2 , … , n are real numbers, then complex roots will occur in conjugate pairs. For example, the roots x 1 and x 2 related to a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 are considered complex conjugate, then the other roots are real, hence we get that the polynomial takes the form P n (x) = a n (x - x n) (x - x n - 1) · . . . (x - x 3) x 2 + p x + q, where x 2 + p x + q = (x - x 1) (x - x 2) .

    Comment

    The roots of a polynomial can be repeated. Consider the proof of the theorem of algebra, the consequences of Bezout's theorem.

    Fundamental theorem of algebra

    Theorem 2

    Any polynomial with degree n has at least one root.

    Bezout's theorem

    After dividing a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 on (x - s) , then we get the remainder, which is equal to the polynomial at the point s , then we get

    P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = (x - s) Q n - 1 (x) + P n (s) , where Q n - 1 (x) is a polynomial with degree n - 1 .

    Corollary from Bezout's theorem

    When the root of the polynomial P n (x) is considered to be s , then P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = (x - s) Q n - 1 (x) . This corollary is sufficient when used to describe the solution.

    Factorization of a square trinomial

    A square trinomial of the form a x 2 + b x + c can be factored into linear factors. then we get that a x 2 + b x + c \u003d a (x - x 1) (x - x 2) , where x 1 and x 2 are roots (complex or real).

    This shows that the decomposition itself reduces to solving the quadratic equation later.

    Example 1

    Factorize a square trinomial.

    Decision

    It is necessary to find the roots of the equation 4 x 2 - 5 x + 1 = 0. To do this, you need to find the value of the discriminant according to the formula, then we get D \u003d (- 5) 2 - 4 4 1 \u003d 9. Hence we have that

    x 1 = 5 - 9 2 4 = 1 4 x 2 = 5 + 9 2 4 = 1

    From here we get that 4 x 2 - 5 x + 1 = 4 x - 1 4 x - 1.

    To perform the check, you need to open the brackets. Then we get an expression of the form:

    4 x - 1 4 x - 1 = 4 x 2 - x - 1 4 x + 1 4 = 4 x 2 - 5 x + 1

    After verification, we arrive at the original expression. That is, we can conclude that the expansion is correct.

    Example 2

    Factorize a square trinomial of the form 3 x 2 - 7 x - 11 .

    Decision

    We get that it is necessary to calculate the resulting quadratic equation of the form 3 x 2 - 7 x - 11 = 0.

    To find the roots, you need to determine the value of the discriminant. We get that

    3 x 2 - 7 x - 11 = 0 D = (- 7) 2 - 4 3 (- 11) = 181 x 1 = 7 + D 2 3 = 7 + 181 6 x 2 = 7 - D 2 3 = 7 - 1816

    From here we get that 3 x 2 - 7 x - 11 = 3 x - 7 + 181 6 x - 7 - 181 6 .

    Example 3

    Factorize the polynomial 2 x 2 + 1.

    Decision

    Now you need to solve the quadratic equation 2 x 2 + 1 = 0 and find its roots. We get that

    2 x 2 + 1 = 0 x 2 = - 1 2 x 1 = - 1 2 = 1 2 i x 2 = - 1 2 = - 1 2 i

    These roots are called complex conjugate, which means that the decomposition itself can be represented as 2 x 2 + 1 = 2 x - 1 2 · i x + 1 2 · i.

    Example 4

    Expand the square trinomial x 2 + 1 3 x + 1 .

    Decision

    First you need to solve a quadratic equation of the form x 2 + 1 3 x + 1 = 0 and find its roots.

    x 2 + 1 3 x + 1 = 0 D = 1 3 2 - 4 1 1 = - 35 9 x 1 = - 1 3 + D 2 1 = - 1 3 + 35 3 i 2 = - 1 + 35 i 6 = - 1 6 + 35 6 i x 2 = - 1 3 - D 2 1 = - 1 3 - 35 3 i 2 = - 1 - 35 i 6 = - 1 6 - 35 6 i

    Having obtained the roots, we write

    x 2 + 1 3 x + 1 = x - - 1 6 + 35 6 i x - - 1 6 - 35 6 i = = x + 1 6 - 35 6 i x + 1 6 + 35 6 i

    Comment

    If the value of the discriminant is negative, then the polynomials will remain second-order polynomials. Hence it follows that we will not decompose them into linear factors.

    Methods for factoring a polynomial of degree higher than the second

    The decomposition assumes a universal method. Most of all cases are based on a corollary of Bezout's theorem. To do this, you need to select the value of the root x 1 and lower its degree by dividing by a polynomial by 1 by dividing by (x - x 1) . The resulting polynomial needs to find the root x 2 , and the search process is cyclical until we get a complete expansion.

    If the root is not found, then other methods of factorization are used: grouping, additional terms. This topic assumes the solution of equations with higher powers and integer coefficients.

    Taking the common factor out of brackets

    Consider the case when the free term is equal to zero, then the form of the polynomial becomes P n (x) = a n x n + a n - 1 x n - 1 + . . . + a 1 x .

    It can be seen that the root of such a polynomial will be equal to x 1 \u003d 0, then you can represent the polynomial in the form of an expression P n (x) \u003d a n x n + a n - 1 x n - 1 +. . . + a 1 x = = x (a n x n - 1 + a n - 1 x n - 2 + . . . + a 1)

    This method is considered to be taking the common factor out of brackets.

    Example 5

    Factorize the third degree polynomial 4 x 3 + 8 x 2 - x.

    Decision

    We see that x 1 \u003d 0 is the root of the given polynomial, then we can bracket x out of the entire expression. We get:

    4 x 3 + 8 x 2 - x = x (4 x 2 + 8 x - 1)

    Let's move on to finding the roots of the square trinomial 4 x 2 + 8 x - 1. Let's find the discriminant and the roots:

    D = 8 2 - 4 4 (- 1) = 80 x 1 = - 8 + D 2 4 = - 1 + 5 2 x 2 = - 8 - D 2 4 = - 1 - 5 2

    Then it follows that

    4 x 3 + 8 x 2 - x = x 4 x 2 + 8 x - 1 = = 4 x x - - 1 + 5 2 x - - 1 - 5 2 = = 4 x x + 1 - 5 2 x + 1 + 5 2

    To begin with, let's take for consideration a decomposition method containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 , where the coefficient of the highest power is 1 .

    When the polynomial has integer roots, then they are considered divisors of the free term.

    Example 6

    Expand the expression f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18.

    Decision

    Consider whether there are integer roots. It is necessary to write out the divisors of the number - 18. We get that ± 1 , ± 2 , ± 3 , ± 6 , ± 9 , ± 18 . It follows that this polynomial has integer roots. You can check according to the Horner scheme. It is very convenient and allows you to quickly obtain the expansion coefficients of a polynomial:

    It follows that x \u003d 2 and x \u003d - 3 are the roots of the original polynomial, which can be represented as a product of the form:

    f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18 = (x - 2) (x 3 + 5 x 2 + 9 x + 9) = = (x - 2) (x + 3) (x 2 + 2 x + 3)

    We turn to the decomposition of a square trinomial of the form x 2 + 2 x + 3 .

    Since the discriminant is negative, it means that there are no real roots.

    Answer: f (x) \u003d x 4 + 3 x 3 - x 2 - 9 x - 18 \u003d (x - 2) (x + 3) (x 2 + 2 x + 3)

    Comment

    It is allowed to use root selection and division of a polynomial by a polynomial instead of Horner's scheme. Let us turn to the consideration of the decomposition of a polynomial containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 , the highest of which does not equal one.

    This case takes place for fractional rational fractions.

    Example 7

    Factorize f (x) = 2 x 3 + 19 x 2 + 41 x + 15 .

    Decision

    It is necessary to change the variable y = 2 x , one should pass to a polynomial with coefficients equal to 1 at the highest degree. You need to start by multiplying the expression by 4 . We get that

    4 f (x) = 2 3 x 3 + 19 2 2 x 2 + 82 2 x + 60 = = y 3 + 19 y 2 + 82 y + 60 = g (y)

    When the resulting function of the form g (y) \u003d y 3 + 19 y 2 + 82 y + 60 has integer roots, then their finding is among the divisors of the free term. The entry will look like:

    ± 1 , ± 2 , ± 3 , ± 4 , ± 5 , ± 6 , ± 10 , ± 12 , ± 15 , ± 20 , ± 30 , ± 60

    Let's proceed to the calculation of the function g (y) at these points in order to get zero as a result. We get that

    g (1) = 1 3 + 19 1 2 + 82 1 + 60 = 162 g (- 1) = (- 1) 3 + 19 (- 1) 2 + 82 (- 1) + 60 = - 4 g (2) = 2 3 + 19 2 2 + 82 2 + 60 = 308 g (- 2) = (- 2) 3 + 19 (- 2) 2 + 82 (- 2) + 60 = - 36 g (3) = 3 3 + 19 3 2 + 82 3 + 60 = 504 g (- 3) = (- 3) 3 + 19 (- 3) 2 + 82 (- 3) + 60 = - 42 g (4) = 4 3 + 19 4 2 + 82 4 + 60 = 756 g (- 4) = (- 4) 3 + 19 (- 4) 2 + 82 (- 4) + 60 = - 28 g (5) = 5 3 + 19 5 2 + 82 5 + 60 = 1070 g (- 5) = (- 5) 3 + 19 (- 5) 2 + 82 (- 5) + 60

    We get that y \u003d - 5 is the root of the equation of the form y 3 + 19 y 2 + 82 y + 60, which means that x \u003d y 2 \u003d - 5 2 is the root of the original function.

    Example 8

    It is necessary to divide by a column 2 x 3 + 19 x 2 + 41 x + 15 by x + 5 2.

    Decision

    We write and get:

    2 x 3 + 19 x 2 + 41 x + 15 = x + 5 2 (2 x 2 + 14 x + 6) = = 2 x + 5 2 (x 2 + 7 x + 3)

    Checking the divisors will take a lot of time, so it is more profitable to take the factorization of the resulting square trinomial of the form x 2 + 7 x + 3. By equating to zero, we find the discriminant.

    x 2 + 7 x + 3 = 0 D = 7 2 - 4 1 3 = 37 x 1 = - 7 + 37 2 x 2 = - 7 - 37 2 ⇒ x 2 + 7 x + 3 = x + 7 2 - 37 2 x + 7 2 + 37 2

    Hence it follows that

    2 x 3 + 19 x 2 + 41 x + 15 = 2 x + 5 2 x 2 + 7 x + 3 = = 2 x + 5 2 x + 7 2 - 37 2 x + 7 2 + 37 2

    Artificial tricks when factoring a polynomial

    Rational roots are not inherent in all polynomials. To do this, you need to use special methods to find factors. But not all polynomials can be decomposed or represented as a product.

    Grouping method

    There are cases when you can group the terms of a polynomial to find a common factor and take it out of brackets.

    Example 9

    Factorize the polynomial x 4 + 4 x 3 - x 2 - 8 x - 2.

    Decision

    Because the coefficients are integers, then the roots can presumably also be integers. To check, we take the values ​​1 , - 1 , 2 and - 2 in order to calculate the value of the polynomial at these points. We get that

    1 4 + 4 1 3 - 1 2 - 8 1 - 2 = - 6 ≠ 0 (- 1) 4 + 4 (- 1) 3 - (- 1) 2 - 8 (- 1) - 2 = 2 ≠ 0 2 4 + 4 2 3 - 2 2 - 8 2 - 2 = 26 ≠ 0 (- 2) 4 + 4 (- 2) 3 - (- 2) 2 - 8 (- 2) - 2 = - 6 ≠ 0

    This shows that there are no roots, it is necessary to use a different method of decomposition and solution.

    Grouping is required:

    x 4 + 4 x 3 - x 2 - 8 x - 2 = x 4 + 4 x 3 - 2 x 2 + x 2 - 8 x - 2 = = (x 4 - 2 x 2) + (4 x 3 - 8 x) + x 2 - 2 = = x 2 (x 2 - 2) + 4 x (x 2 - 2) + x 2 - 2 = = (x 2 - 2) (x 2 + 4 x + 1)

    After grouping the original polynomial, it is necessary to represent it as a product of two square trinomials. To do this, we need to factorize. we get that

    x 2 - 2 = 0 x 2 = 2 x 1 = 2 x 2 = - 2 ⇒ x 2 - 2 = x - 2 x + 2 x 2 + 4 x + 1 = 0 D = 4 2 - 4 1 1 = 12 x 1 = - 4 - D 2 1 = - 2 - 3 x 2 = - 4 - D 2 1 = - 2 - 3 ⇒ x 2 + 4 x + 1 = x + 2 - 3 x + 2 + 3

    x 4 + 4 x 3 - x 2 - 8 x - 2 = x 2 - 2 x 2 + 4 x + 1 = = x - 2 x + 2 x + 2 - 3 x + 2 + 3

    Comment

    The simplicity of grouping does not mean that it is easy enough to choose terms. There is no definite way to solve it, therefore it is necessary to use special theorems and rules.

    Example 10

    Factorize the polynomial x 4 + 3 x 3 - x 2 - 4 x + 2.

    Decision

    The given polynomial has no integer roots. The terms should be grouped. We get that

    x 4 + 3 x 3 - x 2 - 4 x + 2 = = (x 4 + x 3) + (2 x 3 + 2 x 2) + (- 2 x 2 - 2 x) - x 2 - 2 x + 2 = = x 2 (x 2 + x) + 2 x (x 2 + x) - 2 (x 2 + x) - (x 2 + 2 x - 2) = = (x 2 + x) (x 2 + 2 x - 2) - (x 2 + 2 x - 2) = (x 2 + x - 1) (x 2 + 2 x - 2)

    After factoring, we get that

    x 4 + 3 x 3 - x 2 - 4 x + 2 = x 2 + x - 1 x 2 + 2 x - 2 = = x + 1 + 3 x + 1 - 3 x + 1 2 + 5 2 x + 1 2 - 5 2

    Using abbreviated multiplication and Newton's binomial formulas to factorize a polynomial

    Appearance often does not always make it clear which way to use during decomposition. After the transformations have been made, you can build a line consisting of Pascal's triangle, otherwise they are called Newton's binomial.

    Example 11

    Factorize the polynomial x 4 + 4 x 3 + 6 x 2 + 4 x - 2.

    Decision

    It is necessary to convert the expression to the form

    x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3

    The sequence of coefficients of the sum in brackets is indicated by the expression x + 1 4 .

    So we have x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 .

    After applying the difference of squares, we get

    x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3

    Consider the expression that is in the second parenthesis. It is clear that there are no horses there, so the formula for the difference of squares should be applied again. We get an expression like

    x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3 = = x + 1 - 3 4 x + 1 + 3 4 x 2 + 2 x + 1 + 3

    Example 12

    Factorize x 3 + 6 x 2 + 12 x + 6 .

    Decision

    Let's change the expression. We get that

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

    It is necessary to apply the formula for abbreviated multiplication of the difference of cubes. We get:

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

    A method for replacing a variable when factoring a polynomial

    When changing a variable, the degree is reduced and the polynomial is factorized.

    Example 13

    Factorize a polynomial of the form x 6 + 5 x 3 + 6 .

    Decision

    By the condition, it is clear that it is necessary to make a replacement y = x 3 . We get:

    x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6

    The roots of the resulting quadratic equation are y = - 2 and y = - 3, then

    x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3

    It is necessary to apply the formula for the abbreviated multiplication of the sum of cubes. We get expressions of the form:

    x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3 = = x + 2 3 x 2 - 2 3 x + 4 3 x + 3 3 x 2 - 3 3 x + 9 3

    That is, we have obtained the desired expansion.

    The cases discussed above will help in considering and factoring a polynomial in various ways.

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    A polynomial is an expression consisting of the sum of monomials. The latter are the product of a constant (number) and the root (or roots) of the expression to the power k. In this case, one speaks of a polynomial of degree k. The decomposition of a polynomial involves the transformation of the expression, in which the terms are replaced by factors. Let us consider the main ways of carrying out this kind of transformation.

    Method for expanding a polynomial by extracting a common factor

    This method is based on the laws of the distribution law. So, mn + mk = m * (n + k).

    • Example: expand 7y 2 + 2uy and 2m 3 – 12m 2 + 4lm.

    7y 2 + 2uy = y * (7y + 2u),

    2m 3 - 12m 2 + 4lm = 2m(m 2 - 6m + 2l).

    However, the factor that is necessarily present in each polynomial may not always be found, so this method is not universal.

    Polynomial expansion method based on abbreviated multiplication formulas

    Abbreviated multiplication formulas are valid for a polynomial of any degree. In general, the transformation expression looks like this:

    u k – l k = (u – l)(u k-1 + u k-2 * l + u k-3 *l 2 + … u * l k-2 + l k-1), where k is a representative of natural numbers .

    Most often in practice, formulas for polynomials of the second and third orders are used:

    u 2 - l 2 \u003d (u - l) (u + l),

    u 3 - l 3 \u003d (u - l) (u 2 + ul + l 2),

    u 3 + l 3 = (u + l)(u 2 - ul + l 2).

    • Example: expand 25p 2 - 144b 2 and 64m 3 - 8l 3 .

    25p 2 - 144b 2 \u003d (5p - 12b) (5p + 12b),

    64m 3 - 8l 3 = (4m) 3 - (2l) 3 = (4m - 2l)((4m) 2 + 4m * 2l + (2l) 2) = (4m - 2l)(16m 2 + 8ml + 4l 2 ).


    Polynomial decomposition method - grouping terms of an expression

    This method in some way echoes the technique of deriving a common factor, but has some differences. In particular, before isolating the common factor, one should group the monomials. Grouping is based on the rules of associative and commutative laws.

    All monomials presented in the expression are divided into groups, in each of which general meaning such that the second factor will be the same in all groups. In general, such a decomposition method can be represented as an expression:

    pl + ks + kl + ps = (pl + ps) + (ks + kl) ⇒ pl + ks + kl + ps = p(l + s) + k(l + s),

    pl + ks + kl + ps = (p + k)(l + s).

    • Example: expand 14mn + 16ln - 49m - 56l.

    14mn + 16ln - 49m - 56l = (14mn - 49m) + (16ln - 56l) = 7m * (2n - 7) + 8l * (2n - 7) = (7m + 8l)(2n - 7).


    Polynomial Decomposition Method - Full Square Formation

    This method is one of the most efficient in the course of polynomial decomposition. At the initial stage, it is necessary to determine the monomials that can be “folded” into the square of the difference or sum. For this, one of the following relations is used:

    (p - b) 2 \u003d p 2 - 2pb + b 2,

    • Example: expand the expression u 4 + 4u 2 – 1.

    Among its monomials, we single out the terms that form a complete square: u 4 + 4u 2 - 1 = u 4 + 2 * 2u 2 + 4 - 4 - 1 =

    \u003d (u 4 + 2 * 2u 2 + 4) - 4 - 1 \u003d (u 4 + 2 * 2u 2 + 4) - 5.

    Complete the transformation using the rules of abbreviated multiplication: (u 2 + 2) 2 - 5 = (u 2 + 2 - √5) (u 2 + 2 + √5).

    That. u 4 + 4u 2 - 1 = (u 2 + 2 - √5)(u 2 + 2 + √5).