Factorization scheme. Factoring Large Numbers

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.

1. Examples with the solution of a quadratic equation

Example 1.1

x 4 + x 3 - 6 x 2.

Decision

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

.

Answer

Example 1.2

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

Decision

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:
.

Answer

Example 1.3

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

Decision

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:
.

Answer

Examples of factoring polynomials using formulas

Examples with biquadratic polynomials

Example 2.1

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

Decision

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

;
.

Answer

Example 2.2

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

Decision

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 recursive polynomial

Factoring the recursive polynomial:
.

Decision

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:
, ;
;


;
.

Answer

Examples of Factoring Polynomials with Integer Roots

Example 3.1

Factoring a polynomial:
.

Decision

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
.

Answer

Example 3.2

Factoring a polynomial:
.

Decision

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 .
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:
.

Since the equation x 2 + 2 = 0 has no real roots, then the factorization of the polynomial has the form.

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Consider, using specific examples, how to factorize a polynomial.

We will expand polynomials in accordance with .

Factoring polynomials:

Check if there is a common factor. yes, it is equal to 7cd. Let's take it out of brackets:

The expression in brackets consists of two terms. There is no longer a common factor, the expression is not a formula for the sum of cubes, which means that the decomposition is completed.

Check if there is a common factor. No. The polynomial consists of three terms, so we check if there is a full square formula. Two terms are the squares of the expressions: 25x²=(5x)², 9y²=(3y)², the third term is equal to twice the product of these expressions: 2∙5x∙3y=30xy. So this polynomial is a perfect square. Since the double product is with a minus sign, then this is:

We check if it is possible to take the common factor out of brackets. There is a common factor, it is equal to a. Let's take it out of brackets:

There are two terms in brackets. We check if there is a formula for the difference of squares or the difference of cubes. a² is the square of a, 1=1². So, the expression in brackets can be written according to the difference of squares formula:

There is a common factor, it is equal to 5. We take it out of brackets:

in parentheses are three terms. Check if the expression is a perfect square. Two terms are squares: 16=4² and a² is the square of a, the third term is equal to twice the product of 4 and a: 2∙4∙a=8a. Therefore, it is a perfect square. Since all the terms are with a "+" sign, the expression in brackets is the full square of the sum:

The common factor -2x is taken out of brackets:

In parentheses is the sum of the two terms. We check if the given expression is the sum of cubes. 64=4³, x³-cube x. So, the binomial can be expanded according to the formula:

There is a common factor. But, since the polynomial consists of 4 members, we will first, and only then take the common factor out of brackets. We group the first term with the fourth, in the second - with the third:

From the first brackets we take out the common factor 4a, from the second - 8b:

There is no common multiplier yet. To get it, from the second brackets we will take out the brackets “-”, while each sign in the brackets will change to the opposite:

Now we take the common factor (1-3a) out of brackets:

In the second brackets there is a common factor 4 (this is the same factor that we did not take out of brackets at the beginning of the example):

Since the polynomial consists of four terms, we perform grouping. We group the first term with the second, the third with the fourth:

There is no common factor in the first brackets, but there is a formula for the difference of squares, in the second brackets the common factor is -5:

A common factor (4m-3n) has appeared. Let's take it out of brackets.

Expanding polynomials to get a product sometimes seems confusing. But it is not so difficult if you understand the process step by step. The article details how to factorize a square trinomial.

Many do not understand how to factorize a square trinomial, and why this is done. At first it may seem that this is a useless exercise. But in mathematics, nothing is done just like that. The transformation is necessary to simplify the expression and the convenience of calculation.

A polynomial having the form - ax² + bx + c, is called a square trinomial. The term "a" must be negative or positive. In practice, this expression is called a quadratic equation. Therefore, sometimes they say differently: how to expand a quadratic equation.

Interesting! A square polynomial is called because of its largest degree - a square. And a trinomial - because of the 3 component terms.

Some other kinds of polynomials:

• linear binomial (6x+8);
• cubic quadrilateral (x³+4x²-2x+9).

Factorization of a square trinomial

First, the expression is equal to zero, then you need to find the values ​​of the roots x1 and x2. There may be no roots, there may be one or two roots. The presence of roots is determined by the discriminant. Its formula must be known by heart: D=b²-4ac.

If the result of D is negative, there are no roots. If positive, there are two roots. If the result is zero, the root is one. The roots are also calculated by the formula.

If the calculation of the discriminant results in zero, you can apply any of the formulas. In practice, the formula is simply abbreviated: -b / 2a.

Formulas for different values ​​of the discriminant are different.

If D is positive:

If D is zero:

Online calculators

There is an online calculator on the Internet. It can be used to factorize. Some resources provide the opportunity to see the solution step by step. Such services help to better understand the topic, but you need to try to understand well.

Useful video: Factoring a square trinomial

Examples

We suggest looking at simple examples of how to factorize a quadratic equation.

Example 1

Here it is clearly shown that the result will be two x, because D is positive. They need to be substituted into the formula. If the roots are negative, the sign in the formula is reversed.

We know the formula for factoring a square trinomial: a(x-x1)(x-x2). We put the values ​​in brackets: (x+3)(x+2/3). There is no number before the term in the exponent. This means that there is a unit, it is lowered.

Example 2

This example clearly shows how to solve an equation that has one root.

Substitute the resulting value:

Example 3

Given: 5x²+3x+7

First, we calculate the discriminant, as in the previous cases.

D=9-4*5*7=9-140= -131.

The discriminant is negative, which means there are no roots.

After receiving the result, it is worth opening the brackets and checking the result. The original trinomial should appear.

Alternative solution

Some people have never been able to make friends with the discriminant. There is another way to factorize a square trinomial. For convenience, the method is shown in an example.

Given: x²+3x-10

We know that we should end up with 2 parentheses: (_)(_). When the expression looks like this: x² + bx + c, we put x at the beginning of each bracket: (x_) (x_). The remaining two numbers are the product that gives "c", i.e. -10 in this case. To find out what these numbers are, you can only use the selection method. Substituted numbers must match the remaining term.

For example, multiplying the following numbers gives -10:

• -1, 10;
• -10, 1;
• -5, 2;
• -2, 5.

1. (x-1)(x+10) = x2+10x-x-10 = x2+9x-10. No.
2. (x-10)(x+1) = x2+x-10x-10 = x2-9x-10. No.
3. (x-5)(x+2) = x2+2x-5x-10 = x2-3x-10. No.
4. (x-2)(x+5) = x2+5x-2x-10 = x2+3x-10. Fits.

So, the transformation of the expression x2+3x-10 looks like this: (x-2)(x+5).

Important! You should be careful not to confuse the signs.

Decomposition of a complex trinomial

If "a" is greater than one, difficulties begin. But everything is not as difficult as it seems.

In order to factorize, one must first see if it is possible to factor something out.

For example, given the expression: 3x²+9x-30. Here the number 3 is taken out of brackets:

3(x²+3x-10). The result is the already known trinomial. The answer looks like this: 3(x-2)(x+5)

How to decompose if the term that is squared is negative? In this case, the number -1 is taken out of the bracket. For example: -x²-10x-8. The expression will then look like this:

The scheme differs little from the previous one. There are only a few new things. Let's say the expression is given: 2x²+7x+3. The answer is also written in 2 brackets, which must be filled in (_) (_). X is written in the 2nd bracket, and what is left in the 1st. It looks like this: (2x_)(x_). Otherwise, the previous scheme is repeated.

The number 3 gives the numbers:

• -1, -3;
• -3, -1;
• 3, 1;
• 1, 3.

We solve equations by substituting the given numbers. The last option fits. So the transformation of the expression 2x²+7x+3 looks like this: (2x+1)(x+3).

Other cases

It is not always possible to transform an expression. In the second method, the solution of the equation is not required. But the possibility of converting terms into a product is checked only through the discriminant.

It is worth practicing solving quadratic equations so that there are no difficulties when using formulas.

Useful video: factorization of a trinomial

Conclusion

You can use it in any way. But it is better to work both to automatism. Also, those who are going to connect their lives with mathematics need to learn how to solve quadratic equations well and decompose polynomials into factors. All the following mathematical topics are built on this.

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 a common value is taken out 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).