Factoring literal expressions. How to Factor an Algebraic Equation

When solving equations and inequalities, it is often necessary to factor a polynomial whose degree is three or higher. In this article we will look at the easiest way to do this.

As usual, let's turn to theory for help.

Bezout's theorem states that the remainder when dividing a polynomial by a binomial is .

But what is important for us is not the theorem itself, but corollary from it:

If the number is the root of a polynomial, then the polynomial is divisible by the binomial without a remainder.

We are faced with the task of somehow finding at least one root of the polynomial, then dividing the polynomial by , where is the root of the polynomial. As a result, we obtain a polynomial whose degree is one less than the degree of the original one. And then, if necessary, you can repeat the process.

This task breaks down into two: how to find the root of a polynomial, and how to divide a polynomial by a binomial.

Let's take a closer look at these points.

1. How to find the root of a polynomial.

First, we check whether the numbers 1 and -1 are roots of the polynomial.

The following facts will help us here:

If the sum of all the coefficients of a polynomial is zero, then the number is the root of the polynomial.

For example, in a polynomial the sum of the coefficients is zero: . It's easy to check what the root of a polynomial is.

If the sum of the coefficients of a polynomial at even powers is equal to the sum of the coefficients at odd powers, then the number is the root of the polynomial. The free term is considered a coefficient for an even degree, since , a is an even number.

For example, in a polynomial the sum of coefficients for even powers is: , and the sum of coefficients for odd powers is: . It's easy to check what the root of a polynomial is.

If neither 1 nor -1 are roots of the polynomial, then we move on.

For a reduced polynomial of degree (that is, a polynomial in which the leading coefficient - the coefficient at - is equal to unity), the Vieta formula is valid:

Where are the roots of the polynomial.

There are also Vieta formulas concerning the remaining coefficients of the polynomial, but we are interested in this one.

From this Vieta formula it follows that if the roots of a polynomial are integers, then they are divisors of its free term, which is also an integer.

Based on this, we need to factor the free term of the polynomial into factors, and sequentially, from smallest to largest, check which of the factors is the root of the polynomial.

Consider, for example, the polynomial

Divisors of the free term: ; ; ;

The sum of all coefficients of a polynomial is equal to , therefore, the number 1 is not the root of the polynomial.

Sum of coefficients for even powers:

Sum of coefficients for odd powers:

Therefore, the number -1 is also not a root of the polynomial.

Let's check whether the number 2 is the root of the polynomial: therefore, the number 2 is the root of the polynomial. This means, according to Bezout’s theorem, the polynomial is divisible by a binomial without a remainder.

2. How to divide a polynomial into a binomial.

A polynomial can be divided into a binomial by a column.

Divide the polynomial by a binomial using a column:

There is another way to divide a polynomial by a binomial - Horner's scheme.

Watch this video to understand how to divide a polynomial by a binomial with a column, and using Horner's scheme.

I note that if, when dividing by a column, some degree of the unknown is missing in the original polynomial, we write 0 in its place - the same way as when compiling a table for Horner’s scheme.

So, if we need to divide a polynomial by a binomial and as a result of the division we get a polynomial, then we can find the coefficients of the polynomial using Horner’s scheme:

We can also use Horner scheme in order to check whether a given number is the root of a polynomial: if the number is the root of a polynomial, then the remainder when dividing the polynomial by is equal to zero, that is, in the last column of the second row of Horner’s diagram we get 0.

Using Horner's scheme, we "kill two birds with one stone": we simultaneously check whether the number is the root of a polynomial and divide this polynomial by a binomial.

Example. Solve the equation:

1. Let's write down the divisors of the free term and look for the roots of the polynomial among the divisors of the free term.

Divisors of 24:

2. Let's check whether the number 1 is the root of the polynomial.

The sum of the coefficients of a polynomial, therefore, the number 1 is the root of the polynomial.

3. Divide the original polynomial into a binomial using Horner's scheme.

A) Let’s write down the coefficients of the original polynomial in the first row of the table.

Since the containing term is missing, in the column of the table in which the coefficient should be written we write 0. On the left we write the found root: the number 1.

B) Fill in the first row of the table.

In the last column, as expected, we got zero; we divided the original polynomial by a binomial without a remainder. The coefficients of the polynomial resulting from division are shown in blue in the second row of the table:

It's easy to check that the numbers 1 and -1 are not roots of the polynomial

B) Let's continue the table. Let's check whether the number 2 is the root of the polynomial:

So the degree of the polynomial, which is obtained as a result of division by one, is less than the degree of the original polynomial, therefore, the number of coefficients and the number of columns are one less.

In the last column we got -40 - a number that is not equal to zero, therefore, the polynomial is divisible by a binomial with a remainder, and the number 2 is not the root of the polynomial.

C) Let's check whether the number -2 is the root of the polynomial. Since the previous attempt failed, to avoid confusion with the coefficients, I will erase the line corresponding to this attempt:

Great! We got zero as a remainder, therefore, the polynomial was divided into a binomial without a remainder, therefore, the number -2 is the root of the polynomial. The coefficients of the polynomial that is obtained by dividing a polynomial by a binomial are shown in green in the table.

As a result of division we get a quadratic trinomial , whose roots can easily be found using Vieta’s theorem:

So, the roots of the original equation are:

{}

Answer: ( }

Expanding polynomials to obtain a product can sometimes seem confusing. But it's not that difficult if you understand the process step by step. The article describes in detail how to factor a quadratic trinomial.

Many people do not understand how to factor a square trinomial, and why this is done. At first it may seem like a futile exercise. But in mathematics nothing is done for nothing. The transformation is necessary to simplify the expression and ease of calculation.

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

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

Some other types of polynomials:

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

Factoring a quadratic 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. You need to know its formula by heart: D=b²-4ac.

If the result 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 using the formula.

If, when calculating the discriminant, the result is zero, you can use any of the formulas. In practice, the formula is simply shortened: -b / 2a.

Formulas for different meanings discriminants differ.

If D is positive:

If D is zero:

Online calculators

On the Internet there is online calculator. It can be used to perform factorization. Some resources provide the opportunity to view the solution step by step. Such services help to better understand the topic, but you need to try to understand it well.

Useful video: Factoring a quadratic trinomial

Examples

We invite you to view simple examples, how to factor a quadratic equation.

Example 1

This clearly shows that the result is two x's because D is positive. They need to be substituted into the formula. If the roots turn out to be negative, the sign in the formula changes to the opposite.

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

Example 2

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

We substitute the resulting value:

Example 3

Given: 5x²+3x+7

First, let's calculate the discriminant, as in previous cases.

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

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

After receiving the result, you should open the brackets and check the result. The original trinomial should appear.

Alternative solution

Some people were never able to make friends with the discriminator. There is another way to factorize a quadratic trinomial. For convenience, the method is shown with an example.

Given: x²+3x-10

We know that we should get 2 brackets: (_)(_). When the expression looks like this: x²+bx+c, at the beginning of each bracket we put x: (x_)(x_). The remaining two numbers are the product that gives “c”, i.e. in this case -10. The only way to find out what numbers these are is by selection. The substituted numbers must correspond to 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.

This means that 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.

Expansion of a complex trinomial

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

To factorize, you first need to see if anything can be factored 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 well-known trinomial. The answer looks like this: 3(x-2)(x+5)

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

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

The number 3 is given by the numbers:

• -1, -3;
• -3, -1;
• 3, 1;
• 1, 3.

We solve equations by substituting these numbers. The last option is suitable. This means that the transformation of the expression 2x²+7x+3 looks like this: (2x+1)(x+3).

Other cases

It is not always possible to convert an expression. With the second method, solving the equation is not required. But the possibility of transforming terms into a product is checked only through the discriminant.

It's worth practicing to decide quadratic equations so that there are no difficulties when using formulas.

Useful video: factoring a trinomial

Conclusion

You can use it in any way. But it’s better to practice both until they become automatic. Also, learning how to solve quadratic equations well and factor polynomials is necessary for those who are planning to connect their lives with mathematics. All the following mathematical topics are built on this.

Factoring polynomials is identity transformation, as a result of which the polynomial is transformed into the product of several factors - polynomials or monomials.

There are several ways to factor polynomials.

Method 1. Taking the common factor out of brackets.

This transformation is based on the distributive law of multiplication: ac + bc = c(a + b). The essence of the transformation is to isolate the common factor in the two components under consideration and “take” it out of brackets.

Let us factor the polynomial 28x 3 – 35x 4.

Solution.

1. Find a common divisor for the elements 28x3 and 35x4. For 28 and 35 it will be 7; for x 3 and x 4 – x 3. In other words, our common factor is 7x 3.

2. We represent each of the elements as a product of factors, one of which
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x.

3. We take the common factor out of brackets
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x = 7x 3 (4 – 5x).

Method 2. Using abbreviated multiplication formulas. The “mastery” of using this method is to notice one of the abbreviated multiplication formulas in the expression.

Let us factor the polynomial x 6 – 1.

Solution.

1. We can apply the difference of squares formula to this expression. To do this, imagine x 6 as (x 3) 2, and 1 as 1 2, i.e. 1. The expression will take the form:
(x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1).

2. We can apply the formula for the sum and difference of cubes to the resulting expression:
(x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).

So,
x 6 – 1 = (x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).

Method 3. Grouping. The grouping method is to combine the components of a polynomial in such a way that it is easy to perform operations on them (addition, subtraction, subtraction of a common factor).

Let's factor the polynomial x 3 – 3x 2 + 5x – 15.

Solution.

1. Let's group the components in this way: 1st with 2nd, and 3rd with 4th
(x 3 – 3x 2) + (5x – 15).

2. In the resulting expression, we take the common factors out of brackets: x 2 in the first case and 5 in the second.
(x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3).

3. We take the common factor x – 3 out of brackets and get:
x 2 (x – 3) + 5(x – 3) = (x – 3)(x 2 + 5).

So,
x 3 – 3x 2 + 5x – 15 = (x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3) = (x – 3) ∙ (x 2 + 5 ).

Let's secure the material.

Factor the polynomial a 2 – 7ab + 12b 2 .

Solution.

1. Let us represent the monomial 7ab as the sum 3ab + 4ab. The expression will take the form:
a 2 – (3ab + 4ab) + 12b 2.

Let's open the brackets and get:
a 2 – 3ab – 4ab + 12b 2.

2. Let's group the components of the polynomial in this way: 1st with 2nd and 3rd with 4th. We get:
(a 2 – 3ab) – (4ab – 12b 2).

3. Let’s take the common factors out of brackets:
(a 2 – 3ab) – (4ab – 12b 2) = a(a – 3b) – 4b(a – 3b).

4. Let’s take the common factor (a – 3b) out of brackets:
a(a – 3b) – 4b(a – 3b) = (a – 3 b) ∙ (a – 4b).

So,
a 2 – 7ab + 12b 2 =
= a 2 – (3ab + 4ab) + 12b 2 =
= a 2 – 3ab – 4ab + 12b 2 =
= (a 2 – 3ab) – (4ab – 12b 2) =
= a(a – 3b) – 4b(a – 3b) =
= (a – 3 b) ∙ (a – 4b).

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Factoring a large number is not an easy task. Most people have trouble figuring out four or five digit numbers. To make the process easier, write the number above the two columns.

• Let's factorize the number 6552.

Divide the given number by the smallest prime divisor (other than 1) that divides the given number without leaving a remainder. Write this divisor in the left column, and write the result of the division in the right column. As noted above, even numbers are easy to factor because their smallest prime factor will always be 2 (odd numbers have different smallest prime factors).

• In our example, 6552 is an even number, so 2 is its smallest prime factor. 6552 ÷ 2 = 3276. Write 2 in the left column and 3276 in the right column.

Next, divide the number in the right column by the smallest prime factor (other than 1) that divides the number without a remainder. Write this divisor in the left column, and in the right column write the result of the division (continue this process until there are no 1 left in the right column).

• In our example: 3276 ÷ 2 = 1638. Write 2 in the left column, and 1638 in the right column. Next: 1638 ÷ 2 = 819. Write 2 in the left column, and 819 in the right column.

You got an odd number; For such numbers, finding the smallest prime divisor is more difficult. If you get an odd number, try dividing it by the smallest prime odd numbers: 3, 5, 7, 11.

• In our example, you received an odd number 819. Divide it by 3: 819 ÷ 3 = 273. Write 3 in the left column and 273 in the right column.
• When selecting dividers, try everything prime numbers up to square root from the greatest divisor you found. If no divisor divides the number by a whole, then you most likely have a prime number and can stop calculating.

Continue the process of dividing numbers by prime factors until you are left with a 1 in the right column (if you get a prime number in the right column, divide it by itself to get a 1).

• Let's continue the calculations in our example:
  • Divide by 3: 273 ÷ 3 = 91. There is no remainder. Write down 3 in the left column and 91 in the right column.
  • Divide by 3. 91 is divisible by 3 with a remainder, so divide by 5. 91 is divisible by 5 with a remainder, so divide by 7: 91 ÷ 7 = 13. No remainder. Write down 7 in the left column and 13 in the right column.
  • Divide by 7. 13 is divisible by 7 with a remainder, so divide by 11. 13 is divisible by 11 with a remainder, so divide by 13: 13 ÷ 13 = 1. There is no remainder. Write 13 in the left column and 1 in the right column. Your calculations are complete.

The left column shows the prime factors of the original number. In other words, when you multiply all the numbers in the left column, you will get the number written above the columns. If the same factor appears more than once in the list of factors, use exponents to indicate it. In our example, 2 appears 4 times in the list of multipliers; write these factors as 2 4 rather than 2*2*2*2.

• In our example, 6552 = 2 3 × 3 2 × 7 × 13. You factored 6552 into prime factors (the order of the factors in this notation does not matter).

Factoring an equation is the process of finding those terms or expressions that, when multiplied, lead to initial equation. Factoring is a useful skill for solving basic algebra 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

1. Factoring numbers. The concept of factoring is simple, but in practice, factoring can be challenging (if a complex equation is given). Therefore, first, let's look at the concept of factorization using numbers as an example, and continue with simple equations, and then move on to complex equations. Multipliers given number- These are 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 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 factorized. To do this, find the coefficient factors for 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.)

3. 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 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

   1. 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).

   2. 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 decimal numbers. For example, the equation x 2 + (21/2)x + 5 = 0 is expanded into (x+10)(x+1/2).

   3. Factorization by trial and error. Simple quadratic equations can be factored by simply plugging the numbers into the possible solutions until you find the right decision. 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.