Bracketing examples. Taking the common factor out of brackets
Chichaeva Darina 8th grade
In the work, an 8th grade student described the rule for factoring a polynomial by factoring common multiplier behind the brackets with a detailed course of solving many examples on this topic. For each example discussed, 2 examples are offered for independent decision, to which there are answers. The work will help study this topic for those students who, for some reason, did not master it when passing the 7th grade program material and (or) when repeating the algebra course in 8th grade after the summer holidays.
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Municipal budgetary educational institution
secondary school No. 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 way to factor a polynomial isputting the common factor out of brackets;
- - When taking the general multiplier out of brackets, it is applieddistributive property;
- - If all terms of a polynomial contain common factor then this factor can be taken out of brackets.
When solving equations, in calculations and a number of other problems, it can be useful to replace a polynomial with the product of several polynomials (which may include monomials). Representing a polynomial as a product of two or more polynomials is called factoring the polynomial.
Consider the polynomial 6a 2 b+15b 2 . 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 =3b*2a 2 +3b*5b.
The resulting expression based on the distribution property of multiplication can be represented as a product of two factors. One of them is the common multiplier 3b , and the other is the sum 2a 2 and 5b→ 3b*2a 2 +3b*5b=3b(2a 2 +5b) →Thus, we expanded the polynomial: 6a 2 b+15b 2 into factors, representing it as a product of a monomial 3b and the polynomial 2a 2 +5b. This method factoring a polynomial is called taking the common factor out of brackets.
Examples:
Factor it out:
A) kx-px.
Multiplier x x we put it out of brackets.
kx:x=k; px:x=p.
We get: kx-px=x*(k-p).
b) 4a-4b.
Multiplier 4 exists in both the 1st term and the 2nd term. That's why 4 we put it out of brackets.
4a:4=a; 4b:4=b.
We get: 4a-4b=4*(a-b).
c) -9m-27n.
9m and -27n are divisible by -9 . Therefore, we take the numerical factor out of brackets-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 divided by y.
Therefore, we take the common factor out of brackets 5у.
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 the smaller exponent.
e) 16у 3 +12у 2.
16 and 12 are divisible by 4; y 3 and y 2 are divided by y 2.
So the common factor 4y 2 .
16y 3 : 4y 2 =4y; 12y 2 : 4y 2 =3.
As a result we get: 16y 3 +12y 2 =4y 2 *(4y+3).
f) Factor the polynomial 8b(7y+a)+n(7y+a).
IN this expression we see the same factor is present(7y+a) , which can be taken out of brackets. 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. Therefore, 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 independent solutions:
- 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) 7х(a+b); 6) 7(2x+3y); 7) -а(m+1); 8) 4m(2n-m);
9) -4y(3y 3 +4); 10) 15у 2 (у-2); 11) (y-2c)(5c+y 2); 12) (a-3)(8m+n); 13) (y-5)(x+y); 14) (2x-7)(3a-5b).
Among the various expressions that are considered in algebra are important place occupy sums of monomials. Here are examples of such expressions:
\(5a^4 - 2a^3 + 0.3a^2 - 4.6a + 8\)
\(xy^3 - 5x^2y + 9x^3 - 7y^2 + 6x + 5y - 2\)
The sum of monomials is called a polynomial. The terms in a polynomial are called terms of the polynomial. Monomials are also classified as polynomials, considering a monomial to be a polynomial consisting of one member.
For example, a polynomial
\(8b^5 - 2b \cdot 7b^4 + 3b^2 - 8b + 0.25b \cdot (-12)b + 16 \)
can be simplified.
Let us represent all terms in the form of monomials of the standard form:
\(8b^5 - 2b \cdot 7b^4 + 3b^2 - 8b + 0.25b \cdot (-12)b + 16 = \)
\(= 8b^5 - 14b^5 + 3b^2 -8b -3b^2 + 16\)
Let us present similar terms in the resulting polynomial:
\(8b^5 -14b^5 +3b^2 -8b -3b^2 + 16 = -6b^5 -8b + 16 \)
The result is a polynomial, all terms of which are monomials of the standard form, and among them there are no similar ones. Such polynomials are called polynomials of standard form.
Behind degree of polynomial of a standard form take the highest of the powers of its members. Thus, the binomial \(12a^2b - 7b\) has the third degree, and the trinomial \(2b^2 -7b + 6\) has the second.
Typically, the terms of standard form polynomials containing one variable are arranged in descending order of exponents. For example:
\(5x - 18x^3 + 1 + x^5 = x^5 - 18x^3 + 5x + 1\)
The sum of several polynomials can be transformed (simplified) into a polynomial of standard form.
Sometimes the terms of a polynomial need to be divided into groups, enclosing each group in parentheses. Since enclosing parentheses is the inverse transformation of opening parentheses, it is easy to formulate rules for opening brackets:
If a “+” sign is placed before the brackets, then the terms enclosed in brackets are written with the same signs.
If a “-” sign is placed before the brackets, then the terms enclosed in the brackets are written with opposite signs.
Transformation (simplification) of the product of a monomial and a polynomial
Using the distributive property of multiplication, you can transform (simplify) the product of a monomial and a polynomial into a polynomial. For example:
\(9a^2b(7a^2 - 5ab - 4b^2) = \)
\(= 9a^2b \cdot 7a^2 + 9a^2b \cdot (-5ab) + 9a^2b \cdot (-4b^2) = \)
\(= 63a^4b - 45a^3b^2 - 36a^2b^3 \)
The product of a monomial and a polynomial is identically equal to the sum of the products of this monomial and each of the terms of the polynomial.
This result is usually formulated as a rule.
To multiply a monomial by a polynomial, you must multiply that monomial by each of the terms of the polynomial.
We have already used this rule several times to multiply by a sum.
Product of polynomials. Transformation (simplification) of the product of two polynomials
In general, the product of two polynomials is identically equal to the sum of the product of each term of one polynomial and each term of the other.
Usually the following rule is used.
To multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other and add the resulting products.
Abbreviated multiplication formulas. Sum squares, differences and difference of squares
You have to deal with some expressions in algebraic transformations more often than others. Perhaps the most common expressions are \((a + b)^2, \; (a - b)^2 \) and \(a^2 - b^2 \), i.e. the square of the sum, the square of the difference and difference of squares. You noticed that the names of these expressions seem to be incomplete, for example, \((a + b)^2 \) is, of course, not just the square of the sum, but the square of the sum of a and b. However, the square of the sum of a and b does not occur very often; as a rule, instead of the letters a and b, it contains various, sometimes quite complex, expressions.
The expressions \((a + b)^2, \; (a - b)^2 \) can be easily converted (simplified) into polynomials of the standard form; in fact, you have already encountered this task when multiplying polynomials:
\((a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = \)
\(= a^2 + 2ab + b^2 \)
It is useful to remember the resulting identities and apply them without intermediate calculations. Brief verbal formulations help this.
\((a + b)^2 = a^2 + b^2 + 2ab \) - the square of the sum is equal to the sum of the squares and the double product.
\((a - b)^2 = a^2 + b^2 - 2ab \) - the square of the difference is equal to the sum of squares without the doubled product.
\(a^2 - b^2 = (a - b)(a + b) \) - the difference of squares is equal to the product of the difference and the sum.
These three identities allow one to replace its left-hand parts with right-hand ones in transformations and vice versa - right-hand parts with left-hand ones. The most difficult thing is to see the corresponding expressions and understand how the variables a and b are replaced in them. Let's look at several examples of using abbreviated multiplication formulas.
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 multiplying powers with the same base, multiply the variables included in the monomials.
Example 1
Find the product of the monomials $(2x)^3y^2z$ and $(\frac(3)(4)x)^2y^4$
Solution:
First, let's calculate the product of the coefficients
$2\cdot\frac(3)(4) =\frac(2\cdot 3)(4)$ in this task we used the rule for multiplying a number by a fraction - to multiply a whole number by a fraction, you need to multiply the number by the numerator of the fraction, and the denominator put without changes
Now let's use the basic property of a fraction - the numerator and denominator of a fraction can be divided by the same number, different from $0$. Let's divide the numerator and denominator of this fraction by $2$, that is, reduce this 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's transform this fraction by isolating the whole part. Let us remember that to isolate an integer part, it is necessary to write down the remainder of the division into the numerator of the fractional part, the divisor into the denominator.
We 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 multiplying monomials will be:
$(2x)^3y^2z \cdot (\frac(3)(4)x)^2y^4=1\frac(1)(2)x^5y^6$.
Then based on of this rule you can perform the following task:
Example 2
Represent a given polynomial as the product of a polynomial and a monomial $(4x)^3y+8x^2$
Let us represent each of the monomials included in the polynomial as the product of two monomials in order to isolate a common monomial, which will be a factor in both the first and second monomials.
First, let's start with the first monomial $(4x)^3y$. Let's factorize its coefficient into simple factors: $4=2\cdot 2$. We will 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, which means $2\cdot 2=4$ - this number will be included in the general monomial as a coefficient
Now let us note that in the first monomial there is $x^3$, and in the second there is the same variable to the power of $2:x^2$. This means that it is convenient to represent the variable $x^3$ like this:
The variable $y$ is included in only one term of the polynomial, which means it cannot be included in the general monomial.
Let's imagine the first and second monomial included in 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 monomial, 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 multipliers will be the total multiplier: $4x^2$ and the other will be the sum of the remaining multipliers: $xy + 2$. Means:
$(4x)^3y+8х^2 = 4x^2\cdot xy + 4x^2\cdot 2 = 4x^2(xy+2)$
This method is called factorization by taking out a common factor.
Common factor in in this case the monomial $4x^2$ was used.
Algorithm
Note 1
Find the greatest common divisor of the coefficients of all monomials included in the polynomial - it will be the coefficient of the common factor-monomial, which we will put out of brackets
A monomial consisting of the coefficient found in paragraph 2 and the variables found in paragraph 3 will be a common factor. which can be taken out of brackets as a common factor.
Example 3
Take out the common factor $3a^3-(15a)^2b+4(5ab)^2$
Solution:
Let's find the gcd of the coefficients; for this we will decompose the coefficients into simple factors
$45=3\cdot 3\cdot 5$
And we find the product of those that are included in the expansion of each:
Identify the variables that make up each monomial and select the variable with the smallest exponent
$a^3=a^2\cdot a$
The variable $b$ is included only in the second and third monomial, which means it will not be included in the common factor.
Let's compose a monomial consisting of the coefficient found in step 2, the variables found in step 3, we get: $3a$ - this will be the common factor. Then:
$3a^3-(15a)^2b+4(5ab)^2=3a(a^2-5ab+15b^2)$
In this article we will focus on taking the common factor out of brackets. First, let's figure out what this expression transformation consists of. Next, we will present the rule for placing the common factor out of brackets and consider in detail examples of its application.
Page navigation.
For example, the terms in the expression 6 x + 4 y have a common factor 2, which is not written down explicitly. It can be seen only after representing the number 6 as a product of 2·3, and 4 as a product of 2·2. So, 6 x+4 y=2 3 x+2 2 y=2 (3 x+2 y). Another example: in the expression x 3 +x 2 +3 x the terms have a common factor x, which becomes clearly visible after replacing x 3 with x x 2 (in this case we used) and x 2 with x x. After taking it out of brackets, we get x·(x 2 +x+3) .
Let’s separately say about putting the minus out of brackets. In fact, putting the minus out of the brackets means putting the minus one out of the brackets. For example, let’s take out the minus in the expression −5−12·x+4·x·y. The original expression can be rewritten as (−1) 5+(−1) 12 x−(−1) 4 x y, from where the common factor −1 is clearly visible, which we take out of the brackets. As a result, we arrive at the expression (−1)·(5+12·x−4·x·y) in which the coefficient −1 is replaced simply by a minus before the brackets, as a result we have −(5+12·x−4·x· y) . From here it is clearly seen that when the minus is taken out of brackets, the original sum remains in brackets, in which the signs of all its terms have been changed to the opposite.
In conclusion of this article, we note that bracketing the common factor is used very widely. For example, it can be used to more efficiently calculate the values of numeric expressions. Also, putting a common factor out of brackets allows you to represent expressions in the form of a product; in particular, one of the methods for factoring a polynomial is based on bracketing out.
Bibliography.
- Mathematics. 6th grade: educational. for general education institutions / [N. Ya. Vilenkin and others]. - 22nd ed., rev. - M.: Mnemosyne, 2008. - 288 p.: ill. ISBN 978-5-346-00897-2.
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 multiplying powers with the same base, multiply the variables included in the monomials.
Example 1
Find the product of the monomials $(2x)^3y^2z$ and $(\frac(3)(4)x)^2y^4$
Solution:
First, let's calculate the product of the coefficients
$2\cdot\frac(3)(4) =\frac(2\cdot 3)(4)$ in this task we used the rule for multiplying a number by a fraction - to multiply a whole number by a fraction, you need to multiply the number by the numerator of the fraction, and the denominator put without changes
Now let's use the basic property of a fraction - the numerator and denominator of a fraction can be divided by the same number, different from $0$. Let's divide the numerator and denominator of this fraction by $2$, that is, reduce this 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's transform this fraction by isolating the whole part. Let us remember that to isolate an integer part, it is necessary to write down the remainder of the division into the numerator of the fractional part, the divisor into the denominator.
We 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 multiplying 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 a given polynomial as the product of a polynomial and a monomial $(4x)^3y+8x^2$
Let us represent each of the monomials included in the polynomial as the product of two monomials in order to isolate a common monomial, which will be a factor in both the first and second monomials.
First, let's start with the first monomial $(4x)^3y$. Let's factorize its coefficient into simple factors: $4=2\cdot 2$. We will 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, which means $2\cdot 2=4$ - this number will be included in the general monomial as a coefficient
Now let us note that in the first monomial there is $x^3$, and in the second there is the same variable to the power of $2:x^2$. This means that it is convenient to represent the variable $x^3$ like this:
The variable $y$ is included in only one term of the polynomial, which means it cannot be included in the general monomial.
Let's imagine the first and second monomial included in 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 monomial, 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 multipliers will be the total multiplier: $4x^2$ and the other will be the sum of the remaining multipliers: $xy + 2$. Means:
$(4x)^3y+8х^2 = 4x^2\cdot xy + 4x^2\cdot 2 = 4x^2(xy+2)$
This method is called factorization by taking out a common factor.
The common factor in this case was the monomial $4x^2$.
Algorithm
Note 1
Find the greatest common divisor of the coefficients of all monomials included in the polynomial - it will be the coefficient of the common factor-monomial, which we will put out of brackets
A monomial consisting of the coefficient found in paragraph 2 and the variables found in paragraph 3 will be a common factor. which can be taken out of brackets as a common factor.
Example 3
Take out the common factor $3a^3-(15a)^2b+4(5ab)^2$
Solution:
Let's find the gcd of the coefficients; for this we will decompose the coefficients into simple factors
$45=3\cdot 3\cdot 5$
And we find the product of those that are included in the expansion of each:
Identify the variables that make up each monomial and select the variable with the smallest exponent
$a^3=a^2\cdot a$
The variable $b$ is included only in the second and third monomial, which means it will not be included in the common factor.
Let's compose a monomial consisting of the coefficient found in step 2, the variables found in step 3, we get: $3a$ - this will be the common factor. Then:
$3a^3-(15a)^2b+4(5ab)^2=3a(a^2-5ab+15b^2)$