§ 10. Factorization of polynomials by the method taking the common factor out of brackets
In the 6th grade, we decomposed composite numbers into prime factors, that is, we gave natural numbers as a product. For example, 12 = 2 2 ∙ 3; 105 = 3 ∙ 5 ∙ 7 others
Some polynomials can also be represented as a product. This means that these polynomials can be factorized. For example, 5a: - 5y - 5 (x - y); a 3 and 3a 2 = a 2 (a + 3) and the like.
Consider one of the ways to factorize polynomials - taking the common factor out of brackets. One of the examples of such a decomposition known to us is the distributive property of multiplication a(b + c) = ab + ac, if it is written in reverse order: ab + ac - a(b + c). This means that the polynomial ab + ac has been decomposed into two factors a and b + c.
When factoring polynomials with integer coefficients, the factor that is taken out of brackets is chosen so that the terms of the polynomial that remains in brackets do not have a common letter factor, and the modules of their coefficients do not have common divisors.
Let's look at a few examples.
Example 1. Factor the expression:
3) 15a 3 b - 10a 2 b 2.
Development
1) The common factor is 4, so
8m+4= 4 . 2m+ 4 ∙ 1 = 4(2m + 1).
2) The common factor is the variable a, therefore
at + 7ap = a(t + 7p).
3) In this case, the common numerical factor is the greatest common divisor of the numbers 10 and 15 - the number 5, and the common letter factor is the monomial a 2 b. So,
15a 3 b - 10a 2 b 2 \u003d 5a 2 b ∙ 3a - 5a 2 b ∙ b \u003d 5a 2 b (3a - 2b).
Example 2. Factorize:
1) 2m(b - c) + 3p(b - c);
2) x(y - t) + c(t - c).
Development
1) In this case, the common factor is the binomial b = c.
Therefore, 2m( b - With) + 3p( b - c) = (b - c)(2m + 3p).
2) The terms have factors in - t and t - in, which are opposite expressions. Therefore, in the second term, we take out the factor -1, we get: c (t - c) \u003d -c (y - t).
Therefore, x(y - t) + c(t - c) = x(y - t) - c(y - t) = (y - t) (x - c).
To check the correctness of factorization, multiply the resulting factors. The result must equal the given polynomial.
Factoring polynomials often simplifies the process of solving an equation.
Example 3. Find the roots of the equation 5x 2 - 7x \u003d 0.
Development Let us factorize the left side of the equation by taking the common factor out of brackets: x(5x - 7) = 0. Considering that the product is equal to zero if and only if at least one of the factors is equal to zero, we will have: x = 0 or 5x - 7 = 0, whence x = 0 or x = 1.4.
Answer: 0; 1.4.
What transformation is called the factorization of a polynomial? Using the polynomial ab + ac as an example, explain how factorization is performed by taking the common factor out of brackets.
- (Verbal) Find the common factor in the expression:
- (Verbal) Factor out:
- Take the common factor out of brackets:
- (Verbally) correctly factored:
1) 7a + 7 = 7a;
2) 5m - 5 = 5(m - 5);
3) 2a - 2 = 2(a - 1);
4) 7xy - 14x \u003d 7x - (y - 2);
5) 5mn + bn = 5m(n + 3);
6) 7ab + 8cb = 15b(a + c)?
- Write the sum as a product:
- Multiply:
- Multiply:
4) 7a + 21ay;
5) 9x 2 - 27x;
6) 3a - 9a 2;
8) 12ax - 4a 2;
9) -18xy + 24v 2;
10) a 2 b - ab 2;
11) rm - p 2 m;
12) -x 2 y 2 - xy.
- Take the common factor out of brackets:
4) 15xy + 5x;
6) 15m - 30m2;
7) 9xy + 6x 2;
9) -p 2 q - pq 2.
- Multiply:
5) 3b 2 - 9b 3;
7) 4y 2 + 12y 4 ;
8) 5m5 + 15m2;
9) -16a 4 - 20a.
- Multiply:
4) 18p 3 - 12p 2;
5) 14b 3 + 7b 4 ;
6) -25m 3 - 20m.
- Write the sum of 6x 2 in + 15x as a product and find its value if x = -0.5, y = 5.
- Write the expression 12a 2 b - 8a as a product and find its value if a = 2, 6 = .
- Take the common factor out of brackets:
1) a 4 + a 3 - a 2;
2) m 9 - m 2 + m 7;
3) b 6 + b 5 - b 9;
4) - in 7 - in 12 - in 3.
- Present as a product:
1) p 7 + p 3 - p 4;
2) a 10 - a 5 + a 8;
3) b 7 - b 5 - b 2;
4) -m 8 - m 2 - m 4.
- Calculate in a convenient way:
1) 132 ∙ 27 + 132 ∙ 73;
2) 119 ∙ 37 - 19 ∙ 37.
- Solve the equation:
1) x 2 - 2x = 0;
2) x 2 + 4x = 0.
- Find the roots of the equation:
1) x 2 + 3x = 0;
2) x 2 -7x \u003d 0.
1) 4a 3 + 2a 2 - 8a;
2) 9b 3 - 3b 2 - 27b 6 ;
3) 16m 2 - 24m 6 - 22m 3;
4) -5b 3 - 20b 2 - 25b 5.
- Take the common factor out of brackets:
1) 5s 8 - 5s 7 + 10s 4;
2) 9m 4 + 27m 3 - 81m;
3) 8r 7 - 4r 5 + 10r 3;
4) 21b - 28b 4 - 14b 3 .
- Take the common factor out of brackets:
1) 7m 4 - 21m 2 n 2 + 14m 3;
2) 12a 2 b - 18ab 2 + 30ab 3;
3) 8x 2 in 2 - 4x 3 in 5 + 12x 4 in 3;
4) 5p 4 q 2 - 10p 2 q 4 + 15pq 3.
- Factor the polynomial:
1) 12a - 6a 2 x 2 - 9a 3;
2) 12b 2 in - 18b 3 - 30b 4 in;
3) 16bx 2 - 8b 2 x 3 + 24b 3 x;
4) 60m 4 n 3 - 45m 2 n 4 + 30m 3 n 5 .
- Calculate in a convenient way:
1) 843 ∙ 743 - 743 2 ;
2) 1103 2 - 1103 ∙ 100 - 1103 ∙ 3.
- Find the value of the expression:
1) 4.23 a - a 2 if a = 5.23;
2) x 2 y + x 3, if x \u003d 2.51, c \u003d -2.51;
3) am 5 - m 6 if = -1, a = -5;
4) -xy - x 2, if x \u003d 2.7, c \u003d 7.3.
- Find the value of the expression:
1) 9.11 a + a 2 if a = -10.11;
2) 5x 2 + 5a 2 x, if a = ; x = .
- Factor the polynomial:
1) 2p(x - y) + q(x - y);
2) a(x + y) - (x + y);
3) (a - 7) - b (a - 7);
4) 5(a + 1) + (a + 1) 2;
5) (x + 2) 2 - x (x + 2);
6) -5m(m - 2) + 4(m - 2) 2 .
- Express the expression as a product:
1) a (x - y) + b (y - x);
2) g(b - 5) - n(5 - b);
3) 7x - (2b - 3) + 5y (3 - 2b);
4) (x - y) 2 - a (y - x);
5) 5(x - 3) 2 - (3 - x);
6) (a + 1)(2b - 3) - (a + 3)(3 - 2b).
- Multiply:
1) 3x(b - 2) + y(b - 2);
2) (m 2 - 3) - x (m 2 - 3);
3) a(b - 9) + c(9 - b);
4) 7(a + 2) + (a + 2) 2;
5) (s - m) 2 - 5 (m - s);
6) - (x + 2y) - 5 (x + 2y) 2.
- Find the roots of the equation:
1) 4x 2 - x \u003d 0;
2) 7x 2 + 28x = 0;
3) x 2 + x = 0;
4) x 2 - x \u003d 0.
- Solve the equation:
1) 12x 2 + x = 0;
2) 0.2 x 2 - 2x \u003d 0;
3) x 2 - x \u003d 0;
4) 1 - x 2 + - x \u003d 0.
- Solve the equation:
1) x(3x + 2) - 5(3x + 2) = 0;
2) 2x(x - 2) - 5(2 - x) = 0.
- Solve the equation:
1) x(4x + 5) - 7(4x + 5) = 0;
2) 7(x - 3) - 2x(3 - x) = 0.
1) 17 3 + 17 2 is a multiple of 18;
2) 9 14 - 81 6 multiple of 80.
- Prove that the value of the expression:
1) 39 9 - 39 8 is divisible by 38;
2) 49 5 - 7 8 is divisible by 48.
- Take the common factor out of brackets:
1) (5m - 10) 2 ;
2) (18a + 27b) 2 .
- Find the roots of the equation:
1) x (x - 3) \u003d 7x - 21;
2) 2x(x - 5) = 20 - 4x.
- Solve the equation:
1) x (x - 2) \u003d 4x - 8;
2) 3x (x - 4) \u003d 28 - 7x.
- Prove that the number
1) 10 4 + 5 3 is divisible by 9;
2) 4 15 - 4 14 + 4 13 is divisible by 13;
3) 27 3 - 3 7 + 9 3 is divisible by 25;
4) 21 3 + 14 a - 7 3 is divisible by 34.
Exercises to repeat
- Simplify the expression and find its value:
1) -3x 2 + 7x 3 - 4x 2 + 3x 2 if x = 0.1;
2) 8m + 5n - 7m + 15n if m = 7, n = -1.
- Instead of asterisks, write down such monomial coefficients so that equality turns into an identity:
1) 2m 2 - 4mn + n 2 + (*m 2 - *m - *n 2) = 3m 2 - 9mn - 5n 2;
2) 7x 2 - 10y 2 - xy - (* x 2 - * xy + * 2) \u003d -x 2 + 3y 2 + xy.
- The length of a rectangle is three times its width. If the length of a rectangle is reduced by 5 cm, then its area will decrease by 40 cm 2. Find the length and width of the rectangle.
Interesting tasks for lazy students
It is known that a< b < с. Могут ли одновременно выполняться неравенства |а| >|c| and |b|< |с|?
In the framework of the study of identical transformations, the topic of taking the common factor out of brackets is very important. In this article, we will explain what exactly this transformation is, derive the basic rule and analyze typical examples of problems.
Yandex.RTB R-A-339285-1
The concept of factoring out the brackets
To successfully apply this transformation, you need to know for which expressions it is used and what result you want to get as a result. Let's explain these points.
You can take the common factor out of brackets in expressions that are sums in which each term is a product, and in each product there is one factor that is common (the same) for everyone. This is what is called the common factor. That is what we will take out of brackets. So, if we have works 5 3 And 5 4 , then we can take the common factor 5 out of brackets.
What is this transformation? In the course of it, we represent the original expression as the product of a common factor and an expression in brackets containing the sum of all the original terms, except for the common factor.
Let's take the example above. We take out the common factor 5 in 5 3 And 5 4 and get 5 (3 + 4) . The final expression is the product of the common factor 5 and the expression in brackets, which is the sum of the original terms without 5 .
This transformation is based on the distributive property of multiplication, which we have already studied before. In literal form, it can be written as a (b + c) = a b + a c. By changing the right side from the left, we will see the scheme of taking the common factor out of brackets.
The rule for taking the common factor out of brackets
Using all of the above, we derive the basic rule for such a transformation:
Definition 1
To bracket the common factor, you need to write the original expression as a product of the common factor and brackets, which include the original sum without the common factor.
Example 1
Let's take a simple example of rendering. We have a numeric expression 3 7 + 3 2 − 3 5, which is the sum of three terms 3 · 7 , 3 · 2 and a common factor 3 . Taking as a basis the rule we have derived, we write the product as 3 (7 + 2 - 5). This is the result of our transformation. The solution entry looks like this: 3 7 + 3 2 − 3 5 = 3 (7 + 2 − 5).
We can take the factor out of brackets not only in numerical, but also in literal expressions. For example, in 3 x − 7 x + 2 you can take out the variable x and get 3 x − 7 x + 2 = x (3 − 7) + 2, in the expression (x 2 + y) x y − (x 2 + y) x 3- common multiplier (x 2 + y) and get in the end (x 2 + y) (x y − x 3).
It is not always possible to determine immediately which multiplier is common. Sometimes an expression needs to be preliminarily transformed by replacing numbers and expressions with products that are identically equal to them.
Example 2
So, for example, in the expression 6 x + 4 y you can take out the common factor 2 , not written explicitly. To find it, we need to transform the original expression, representing six as 2 3 and four as 2 2 . That is 6 x + 4 y = 2 3 x + 2 2 y = 2 (3 x + 2 y). Or in the expression x 3 + x 2 + 3 x can be bracketed by the common factor x , which is found after the replacement x 3 on x · x 2 . Such a transformation is possible due to the basic properties of the degree. As a result, we get the expression x (x 2 + x + 3).
Another case that should be dealt with separately is the bracketing of the minus. Then we take out not the sign itself, but minus one. For example, let's transform the expression in this way − 5 − 12 x + 4 x y. Let's rewrite the expression as (− 1) 5 + (− 1) 12 x − (− 1) 4 x y so that the total multiplier can be seen more clearly. Let's take it out of the brackets and get − (5 + 12 x − 4 x y) . This example shows that in brackets the same amount is obtained, but with opposite signs.
In the conclusions, we note that the transformation by taking the common factor out of brackets is very often used in practice, for example, to calculate the value of rational expressions. This method is also useful when you need to represent an expression as a product, for example, to decompose a polynomial into separate factors.
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Chichaeva Darina 8th grade
In the work, a student of the 8th grade painted the rule for factoring a polynomial by taking the common factor out of brackets with a detailed process for solving a set of examples on this topic. For each analyzed example, 2 examples are offered for an independent solution, to which there are answers. The work will help to study this topic for those students who, for some reason, did not learn it when passing the program material of the 7th grade and (or) when repeating the algebra course in the 8th grade after the summer holidays.
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Municipal budgetary educational institution
secondary school №32
"UNESCO Associated School "Eureka Development"
Volzhsky, Volgograd region
Work completed:
8B class student
Chichaeva Darina
Volzhsky
2014
Taking the common factor out of brackets
- - One of the ways to factorize a polynomial istaking the common factor out of brackets;
- - When taking the common factor out of brackets, thedistributive property;
- - If all members of the polynomial contain common factor, then this factor can be taken out of brackets.
When solving equations, in calculations, and in a number of other problems, it can be useful to replace a polynomial with the product of several polynomials (among which there may be monomials). The representation of a polynomial as a product of two or more polynomials is called the factorization of a polynomial.
Consider the polynomial 6a2b+15b2 . Each of its terms can be replaced by the product of two factors, one of which is equal to 3b: →6a 2 b = 3b*2a 2 , + 15b 2 = 3b*5b →from this we get: 6a 2 b + 15b 2 \u003d 3b * 2a 2 + 3b * 5b.
The resulting expression based on the distributive property of multiplication can be represented as a product of two factors. One of them is the common factor 3b , and the other is the sum 2а 2 and 5b→ 3b*2a 2 +3b*5b=3b(2a 2 +5b) →Thus, we expanded the polynomial: 6a2b+15b2 into factors, presenting it as a product of a monomial 3b and the polynomial 2a 2 +5b. This method of factoring a polynomial is called taking the common factor out of brackets.
Examples:
Multiply:
A) kx-px.
Multiplier x x take it out of brackets.
kx:x=k; px:x=p.
We get: kx-px=x*(k-p).
b) 4a-4b.
Multiplier 4 exists in term 1 and term 2. That's why 4 take it out of brackets.
4a:4=a; 4b:4=b.
We get: 4a-4b=4*(a-b).
c) -9m-27n.
9m and -27n are divided by -9 . Therefore, we take out the numerical factor-9.
9m: (-9)=m; -27n: (-9)=3n.
We have: -9m-27n=-9*(m+3n).
d) 5y 2 -15y.
5 and 15 are divisible by 5; y 2 and y are divisible by y.
Therefore, we take out the common factor 5u .
5y 2 : 5y=y; -15y: 5y=-3.
So: 5y 2 -15y=5y*(y-3).
Comment: From two degrees with the same base, we take out the degree with a lower exponent.
e) 16y 3 + 12y 2.
16 and 12 are divisible by 4; y 3 and y 2 are divisible by y 2 .
So the common factor 4y2.
16y 3 : 4y 2 =4y; 12y 2 : 4y 2 =3.
As a result, we will get: 16y 3 +12y 2 \u003d 4y 2 * (4y + 3).
f) Factor the polynomial 8b(7y+a)+n(7y+a).
In this expression, we see that there is the same factor(7y+a) , which can be bracketed. So, we get:8b(7y+a)+n(7y+a)=(8b+n)*(7y+a).
g) a(b-c)+d(c-b).
Expressions b-c and c-b are opposite. So to make them the same, before d change the "+" sign to "-":
a(b-c)+d(c-b)=a(b-c)-d(b-c).
a(b-c)+d(c-b)=a(b-c)-d(b-c)=(b-c)*(a-d).
Examples for an independent solution:
- mx+my;
- ah+ay;
- 5x+5y ;
- 12x+48y;
- 7ax+7bx;
- 14x+21y;
- –ma-a;
- 8mn-4m2;
- -12y 4 -16y;
- 15y 3 -30y 2 ;
- 5c(y-2c)+y 2 (y-2c);
- 8m(a-3)+n(a-3);
- x(y-5)-y(5-y);
- 3a(2x-7)+5b(7-2x);
Answers.
1) m(x+y); 2) a(x+y); 3) 5(x+y); 4) 12(x+4y); 5) 7x(a+b); 6) 7(2x+3y); 7) -а(m+1); 8) 4m(2n-m);
9) -4y(3y 3 +4); 10) 15y 2 (y-2); 11) (y-2c) (5c + y 2 ); 12) (a-3)(8m+n); 13) (y-5)(x+y); 14) (2x-7)(3a-5b).
Definition 1
First let's remember rules for multiplying a monomial by a monomial:
To multiply a monomial by a monomial, you must first multiply the coefficients of the monomials, then, using the rule of multiplication of powers with the same base, multiply the variables included in the monomials.
Example 1
Find the product of monomials $(2x)^3y^2z$ and $(\frac(3)(4)x)^2y^4$
Solution:
First, we calculate the product of the coefficients
$2\cdot\frac(3)(4) =\frac(2\cdot 3)(4)$ in this task we used the rule of multiplying a number by a fraction - to multiply an integer by a fraction, you need to multiply the number by the numerator of the fraction, and the denominator leave unchanged
Now let's use the main property of a fraction - the numerator and denominator of a fraction can be divided by the same number, different from $0$. Divide the numerator and denominator of this fraction by $2$, i.e., reduce the given fraction by $2$ $2\cdot\frac(3)(4)$ =$\frac(2\cdot 3)(4)=\ \frac(3 )(2)$
The resulting result turned out to be an improper fraction, that is, one in which the numerator is greater than the denominator.
Let us transform this fraction by means of extracting the integer part. Recall that in order to isolate the integer part, an incomplete quotient is necessary, which are obtained by dividing the numerator by the denominator, write down as an integer part, the remainder of the division into the numerator of the fractional part, the divisor into the denominator.
We have found the coefficient of the future product.
Now we will sequentially multiply the variables $x^3\cdot x^2=x^5$,
$y^2\cdot y^4 =y^6$. Here we used the rule for multiplying powers with the same base: $a^m\cdot a^n=a^(m+n)$
Then the result of multiplication of monomials will be:
$(2x)^3y^2z \cdot (\frac(3)(4)x)^2y^4=1\frac(1)(2)x^5y^6$.
Then, based on this rule, you can perform the following task:
Example 2
Represent the given polynomial as a product of a polynomial and a monomial $(4x)^3y+8x^2$
We represent each of the monomials that make up the polynomial as a product of two monomials in order to select a common monomial, which will be a factor in both the first and second monomials.
First, we start with the first monomial $(4x)^3y$. Let's factorize its coefficient into simple factors: $4=2\cdot 2$. Let's do the same with the coefficient of the second monomial $8=2\cdot 2 \cdot 2$. Note that two factors $2\cdot 2$ are included in both the first and second coefficients, so $2\cdot 2=4$--this number will be included in the general monomial as a coefficient
Now let's pay attention that in the first monomial $x^3$ , and in the second the same variable in the power of $2:x^2$. Hence, it is convenient to represent the variable $x^3$ as follows:
The variable $y$ is included in only one term of the polynomial, which means that it cannot be included in the general monomial.
Let's represent the first and second monomial entering the polynomial as a product:
$(4x)^3y=4x^2\cdot xy$
$8x^2=4x^2\cdot 2$
Note that the common monomial, which will be a factor in both the first and second monomials, is $4x^2$.
$(4x)^3y+8x^2=4x^2\cdot xy + 4x^2\cdot 2$
Now we apply the distributive law of multiplication, then the resulting expression can be represented as a product of two factors. One of the factors will be the common factor: $4x^2$ and the other will be the sum of the remaining factors: $xy + 2$. Means:
$(4x)^3y+8x^2 = 4x^2\cdot xy + 4x^2\cdot 2 = 4x^2(xy+2)$
This method is called factorization by taking out a common factor.
The common factor in this case was the monomial $4x^2$ .
Algorithm
Remark 1
Find the greatest common divisor of the coefficients of all monomials included in the polynomial - it will be the coefficient of the common monomial factor, which we will take out of brackets
The monomial consisting of the coefficient found in item 2, the variables found in item 3 will be a common factor. which can be bracketed as a common factor.
Example 3
Take out the common factor $3a^3-(15a)^2b+4(5ab)^2$
Solution:
We find the GCD of the coefficients for this, we decompose the coefficients into simple factors
$45=3\cdot 3\cdot 5$
And we find the product of those that enter into the expansion of each:
Identify the variables that are part of each monomial and choose the variable with the smallest exponent
$a^3=a^2\cdot a$
The variable $b$ only enters the second and third monomial, which means that it will not enter the common factor.
Let us compose a monomial consisting of the coefficient found in item 2, the variables found in item 3, we get: $3a$- this will be the common factor. Then:
$3a^3-(15a)^2b+4(5ab)^2=3a(a^2-5ab+15b^2)$
