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If we want to multiply a sum by another number, we can multiply each term of the sum by the number before adding them, or we can first add the terms and then multiply them. For example,

In both cases the result is the same.

This property, which we first introduced in Section 1.8, is called the distributive law. in symbols,

a(b + c) = ab + ac o (b + c)a = ba + ca

By applying the distributive law to algebraic expressions containing parentheses, we can obtain equivalent expressions without parentheses.

Our first example concerns the product of a monomial and a binomial.

example 1Write 2x(x - 3) without brackets.

solution

We think of 2x(x - 3) as 2x[x + (-3)] and then apply the distributive property to get

The above method works just as well with the product of a monomial and a trinomial.

example 2Writing - and (and²+ 3a - 4) without brackets.

solution

Applying the law of distribution, we get

When simplifying expressions that contain parentheses, we first remove the parentheses and then combine similar terms.

Example 3Vereinfachung a(3 - a) - 2(a + a)²).

We start by removing the brackets to obtain

Now the combination of equal terms gives a - 3a².

We can use the distributive property to rewrite expressions where the coefficient of an expression in parentheses is +1 or -1.

example 4Write each expression without parentheses.
uma. +(3a - 2b)
b. -(2a - 3b)

solution

Please note that inexample 4b, the sign of each term is changed when the expression is written without parentheses. This is the same result that we would have obtained if we had used the expression simplification techniques presented in Section 2.5.

FACTORIZATION OF MONOMIALS FROM POLYNOMIALS

By the symmetry property of equality, we know that if

a(b + c) = ab + ac, entonces ab + ac = a(b + c)

So if there is a common monomial factor for all terms of a polynomial, we can write the polynomial as the product of the common factor and another polynomial. For example, since each term in x²+ 3x contains x as a factor, we can write the expression as the product x(x + 3). Rewriting a polynomial in this way is called factoring, and the number x is said to have been factored "into" or "out of" the polynomial x.²+ 3x.

To factor a monomial from a polynomial:

1. Write a series of parentheses preceded by the monomial common to each term of the polynomial.
2. Divide the factor of the monomial by each term of the polynomial and write the quotient in parentheses.

Usually we can find the common monomial factor by inspection.

example 1
a. 4x + 4y = 4(x + y)
b. 3xy -6y - 3y(x - 2)

We can check that we factored correctly by multiplying the factors and checking that the product is the original polynomial. wear and tearexample 1, We have

If the common monomial is hard to find, we can write each term in prime factor form and find the common factors.

example 2Factor 4x³- 6x²+2x.

solution
we can write

Now we see that 2x is a monomial factor common to all three terms. Then we factor 2x of the polynomial and write
2x()

Now we divide each term of the polynomial by 2x

and write the quotients in brackets to get

2x (2x²- 3x + 1)

We can check in our responseexample 2Multiplication of the factors to be obtained

In this book, we restrict common factors to monomials, which consist of integer numerical coefficients and integer powers of the variables. The choice of the sign of the monomial factor is a matter of convenience. Hence,

-3x²- 6x

can be factored as

-3x(x + 2) oder 3x(-x - 2)

The first way is generally more convenient.

Example 3Factor the common monomial by including -1.

a. - 3x²- 3xy
b. -X³- X²+x
solution

Sometimes it is convenient to write formulas in factored form.

4.3 BINOMAL PRODUCTS I

We can use the distributive property to multiply two binomials. Although it is rarely necessary to multiply binomials in arithmetic, as the following example shows, the distributive property also applies to expressions involving variables.

We now apply the above procedure to an expression containing variables.

example 1

Write (x - 2)(x + 3) without parentheses.

solution
First apply the distribution law to get

Now combine similar terms to get
X²+ x - 6

With practice, you'll be able to add the second and third products in your head. The above process is sometimes referred to as the FOIL process. F, O, I and L stand for:

1. 1. The product of the first terms.
2. 2. The product of external terms.
3. 3. The product of internal terms.
4. 4.The product of the latest conditions.

The FOIL method can also be used for quadratic binomials.

example 2

Write (x + 3)²no brackets.
solution

First rewrite (x + 3).²as (x + 3)(x + 3). Then use the FOIL method to get

Combining the same terms yields
X²+ 6x + 9

If we have a monomial factor and two binomial factors, it's easier to multiply the binomial factors first.

Example 3

Write 3x(x - 2)(x + 3) without parentheses.
solution
First multiply the binomials to get
3x(x²+ 3x - 2x - 6) = 3x(x²+ x - 6)

Now apply the distributive property to get
3x(x²+ x - 6) = 3x³+ 3x²- 18x

Common mistakes

hint aexample 2

Similar,

Usually,

4.4 Factorization of trinomials I

In Section 4.3 we saw how to find the product of two binomials. Now we will reverse this process. That is, given the product of two binomials, we find the binomial factors. The process involved is another example of factoring. As before, we will only consider factors where the terms have integer numerical coefficients. These factors are not always present, but we will examine the cases where they are present.

Consider the following product.

Note that the first term in the trinomial, x², is the product (1); the last term of the trinomial, 12, is the product, and the middle term of the trinomial, 7x, is the sum of the products (2) and (3). Usually,

We use this equation (from right to left) to factor any trinomial of the form x²+ Bx + C. We find two numbers whose product is C and whose sum is B.

We see that the only pair of factors whose product is 12 and whose sum is 7 are 3 and 4.

X²+ 7x + 12 = (x + 3)(x + 4)

Note that if all terms of a trinomial are positive, we only need to consider pairs of positive factors since we are looking for a pair of factors whose product and sum are positive. That is, the factored notion of

X²+ 7x + 12 seria de forma

( + ) ( + )

If the first and third terms of a trinomial are positive but the middle term is negative, we only need to consider pairs of negative factors because we are looking for a pair of factors whose product is positive but whose sum is negative. That is, the factored form of

X²- 5x + 6

would be the way

(-)(-)

example 2X Factor²- 5x + 6.

solution
Since the third term is positive and the middle term is negative, we find two negative integers whose product is 6 and whose sum is -5. We list the possibilities.

We see that the only pair of factors whose product is 6 and whose sum is -5 is -3 and -2. Hence,

X²- 5x + 6 = (x - 3)(x - 2)

When the first term of a trinomial is positive and the third is negative, the factored signs are opposites. That is, the factored form of

X²- x - 12

would be the way

(+)(-) or (-)(+)

Example 3

X Factor²- x - 12.

Solution We need to find two integers whose product is -12 and whose sum is -1. We list the possibilities.

We see that the only pair of factors whose product is -12 and whose sum is -1 is -4 and 3.

X²- x - 12 = (x - 4)(x + 3)

It is easier to fully factor a trinomial if you first factor all the factors common to each term of the trinomial. For example, we can factor

12x²+ 36x + 24

As

A monomial can then be factored from these binomial factors. However, if one first factors out the common factor 12 from the original expression, one obtains

12 (x²+ 3x + 2)

We have factoring again

12(* + 2)(x + 1)

which is said to be in fully factored form. In these cases, the number factor itself does not need to be factored; H. we don't write 12 as 2 * 2 * 3.

example 4

Factor 3x²+ 12x + 12 total.

solution
First we factor the 3 of the trinomial to get

3 (x²+ 4x + 4)

Now we factor the trinomial and get

3(x + 2)(x + 2)

The techniques we have developed also apply to a trinomial such as x²+ 5xy + 6y².

Example 5X Factor²+ 5xy + 6y².

solution
We find two positive factors whose product is 6y²y whose sum is 5y (the coefficient of x). The two factors are 3y and 2y. Hence,

X²+ 5xy + 6y²= (x + 3y)(x + 2y)

When factoring, it is best to write the trinomial in decreasing powers of x. If the coefficient of x²-term is negative, factor a negative before proceeding.

Example 6

Factor 8 + 2x - x².

solution
We first rewrite the trinomial in decreasing powers of x to get it

-X²+ 2x + 8

Now we can factor the -1 to get

-(X²- 2x - 8)

Finally, we factor the trinomial to get

-(x-4)(x + 2)

Sometimes trinomials are not factorable.

Example 7

X Factor²+ 5x + 12.

solution
We are looking for two integers whose product is 12 and whose sum is 5. From the table inexample 1On page 149 we see that there is no pair of factors whose product is 12 and whose sum is 5. In this case the trinomial cannot be factored.

The ability to factor is usually the result of extensive practice. If possible, go through the factoring process in your head by typing your answer directly. You can check the results of a factorization by multiplying the binomial factors and checking that the product equals the given trinomial.

4.5 BINOMAL PRODUCTS II

In this section we use the procedure developed in Section 4.3 to multiply binomial factors whose first degree terms have numerical coefficients other than 1 or -1.

example 1

Write as a polynomial.

a. (2x - 3)(x + 1)
b. (3x - 2a)(3x + a)

solution

We first apply the FOIL method and then combine the same terms.

As before, when we have a squared binomial, we first rewrite it as a product and then FOIL.

example 2

a. (3x + 2)²
= (3x + 2)(3x + 2)
= 9x²+ 6x + 6x + 4
= 9x²+ 12x + 4

b. (2x - y)²
= (2x - y)(2x - y)
= 4x²- 2xy - 2xy + y²
- 4x²- 4xy + y²

As you saw in Section 4.3, the product of two bioanimals may not contain a first-degree term in the answer.

Example 3

nach. (2x - 3)(2x + 3)
= 4x²+ 6x - 6x - 9
= 4x²-9

b. (3x - y)(3x + y)
- 9x²+ 3xy - 3xy - y²= 9x²-y²

When multiplying a monomial factor and two binomial factors, it is easier to multiply the binomial factors first.

example 4

Write 3x(2x - l)(x + 2) as a polynomial.

solution
First we multiply the binomials to get 3x(2x²+ 4x - x - 2) = 3x(2x²+ 3x - 2)
Now multiplying by the monomial gives 3x(2x²) + 3x(3x) + 3x(-2) = 6x³+9x²- 6x

4.6 FACTORING OF TRINOMES II

In Section 4.4 we factored trinomials of the form x²+ Bx + C, where the quadratic term had a coefficient of 1. We now want to extend our factorization techniques to trinomials of the form Ax²+ Bx + C, where the quadratic term has a coefficient other than 1 or -1.

First, let's consider a test to see if a trinomial is factorable. A trinomial of the form Ax²+ Bx + C is factorable if we can find two integers whose product is A * C and whose sum is B.

example 1

determine 4 times²+ 8x + 3 is factorable.

solution
We check whether there are two integers whose product (4)(3) = 12 and whose sum is 8 (the coefficient of x). Consider the following possibilities.

Since the factors 6 and 2 add up to 8, the value of B in the trinomial is Ax²+ Bx + C the trinomial is factorable.

example 2

Das 4x Trinom²- 5x + 3 is not factorable because the table above shows that there is no pair of factors whose product is 12 and whose sum is -5. The test of whether the trinomial is factorable can usually be done mentally.

Once we have established that a trinomial of the form Ax²+ Bx + C is factorable, we proceed to find a pair of factors whose product is A, a pair of factors whose product is C, and an arrangement that produces the corresponding mean term. We illustrate with examples.

Example 3

Factor 4x²+ 8x + 3.

solution
Above we found that this polynomial is factorizable. Now it goes on.

1. We consider all pairs of factors whose product is 4. Since 4 is positive, only positive integers should be considered. The possibilities are 4, 1 and 2, 2.
2. We consider all pairs of factors whose product is 3. Since the middle term is positive, we only consider pairs of positive factors. The possibilities are 3, 1. We write all possible arrangements of the factors as shown.

3. We choose the arrangement where the sum of the products (2) and (3) gives an average of 8x.

We now consider factoring a trinomial where the constant term is negative.

example 4

Factor 6x²+ x - 2.

solution
First we test whether 6x²+ x - 2 is factorable. We are looking for two integers that have a product of 6(-2) = -12 and a sum of 1 (the coefficient of x). The integers 4 and -3 have a product of -12 and a sum of 1, so the trinomial is factorable. Now it goes on.

1. We consider all pairs of factors whose product is 6. Since 6 is positive, only positive integers should be considered. So the possibilities are 6, 1 and 2, 3.
2. We consider all pairs of factors whose product is -2. The possibilities are 2, -1 and -2, 1. We write all possible arrangements of the factors as shown.
3. We choose the arrangement where the sum of the products (2) and (3) gives a mean term of x.

With practice you will be able to check the combinations in your head and you will not have to write down all the possibilities. Paying attention to the trinomial signs is particularly helpful in mentally eliminating possible combinations.

It's easier to factor a trinomial written in decreasing powers of the variable.

Example 5

Factor.

a. 3 + 4x²+8x
b. x - 2 + 6x²

solution
Rewrite each trinomial in decreasing powers of x and follow the solutions in Examples 3 and 4.

a. 4x²+ 8x + 3
b. 6x²+ x - 2

As we said in Section 4.4, if a polynomial contains a common monomial factor in each of its terms, we must factor that monomial out of the polynomial before looking for other factors.

Example 6

Factor 24²- 44x - 40.

solution
First we factor 4 from each term to get

4 (6x²- 11x - 10)

We then factor the trinomial to get

4(3x + 2)(2x - 5)

ALTERNATIVE METHOD OF FACTORIZING THE TRINOME

If the previous "trial and error" factoring method does not provide quick results, there is an alternative method which we will now demonstrate using the 4x example above²+8x +3, could be useful.

We know that the trinomial is factorable because we found two numbers whose product is 12 and whose sum is 8. Those numbers are 2 and 6. We'll go ahead and use those numbers to rewrite 8x as 2x + 6x.

4.7 FACTORIZATION OF THE DIFFERENCES OF TWO SQUARE PARTS

Some polynomials are so common that it makes sense to recognize these special forms, which in turn allows us to write their factored form directly. Realize

In this section we are interested in considering this relationship from right to left, from polynomial to²-b²into its factored form (a + b)(a - b).

The difference of two squares, one²-b², is equal to the product of the sum a + b and the difference a - b.

example 1

uma. x2 - 9 = x2 - 32
= (x + 3)(x - 3)
b. x2 - 16 = x2 - 42
= (x + 4)(x - 4)

Since

(3x)(3x) = 9x²

we can see a binomial as 9x²- 4 Likes (3x)²- 2²and use the factoring method above.

example 2

a.9x²- 4 = (3x)²- 2²
= (3x + 2)(3x - 2)
b.4a²- 25x²= (2 years)²- (x5)²= (2a + 5x)(2a - 5x)

As before, we always factor a common monomial first whenever possible.

Example 3

Machado³- X⁵= x³(l-x²)
= x³(1 + x)(l - x)
licensed in letters²X²y - 16y = y(a²X²- sixteen)
= y[(ax)²- 4²]
= y(ax - 4)(ax + 4)

4.8 EQUATIONS WITH BRACKETS

We often need to solve equations where the variable is in parentheses. We can solve these equations in the usual way, after simplifying them, using the distributive property to remove the brackets.

example 1

Solve 4(5 - y) + 3(2y - 1) = 3.

solution
We first apply the distributive property to get

20 - 4 years + 6 years - 3 = 3

Now combine similar terms and solve for and gives

2 years + 17 = 3

2 years = -14

y=-l

The same procedure can be applied to equations involving binomial products.

example 2

Solution (x + 5)(x + 3) - x = x²+ 1.

solution
First we apply the FOIL method to remove and preserve the brackets

X²+ 8x + 15 - x = x²+1

Now combining equal terms and solving for x gives

X²+ 7x + 15 = x²+1

7x = -14

x = -2

4.9 WORD TASKS WITH NUMBERS

Parentheses are useful to represent products where the variable is included in one or more terms in any factor.

example 1

An integer is three more than another. If x represents the smallest integer, express it in terms of x

a. The largest integer.
B. five times the smallest integer.
C. Five times the largest integer.

solution
a. x+3
b. 5x
C. 5(x + 3)

Suppose we know that the sum of two numbers is 10. If we represent a number by x, then the second number must be 10 - x according to the table below.

In general, if we know that the sum of two numbers is 5 and x represents a number, the other number must be S - x.

example 2

The sum of two integers is 13. If x is the smallest integer, express it in terms of X

a. The largest integer.
B. five times the smallest integer.
C. Five times the largest integer.

solution
a. 13-x
b. 5x
C. 5(13 - x)

The following example refers to the notion of consecutive integers considered in Section 3.8.

Example 3

The difference of the squares of two consecutive odd integers is 24. If x is the smallest integer, express it in terms of x

a. the largest integer
B. The square of the smallest whole.
C. The square of the greatest whole.

solution

a. x + 2
b. X²
C. (x + 2)²

Mathematical models (equations) for word problems sometimes contain parentheses. We can use the method described on page 115 to get the equation. We then solve the equation by first writing the equation equivalently without parentheses.

example 4

An integer is five more than an integer second. Three times the smallest integer plus twice the largest equals 45. Find the integers.

solution

Steps 1-2
First we write what we want to find (the integers) as word phrases. Next we represent the integers in terms of a variable.
The smallest integer: x
The largest whole number: x + 5

level 3
A scheme is not applicable.

step 4
Now we write an equation representing the condition of the problem and get

3x + 2(x + 5) = 45

step 5
Applying the distributive property to remove the brackets, we get

step 6
The whole numbers are 7 and 7 + 5 or 12.

4.10 APPLICATIONS

In this section we consider several uses of word problems that lead to equations with parentheses. Again, we will follow the six troubleshooting steps outlined on page 115.

PROBLEMS WITH COINS

The basic idea with coin (or banknote) problems is that the value of multiple coins of the same denomination is equal to the product of the value of a single coin and the total number of coins.

A table like the one shown in the example below is useful for solving currency problems.

example 1

A collection of dimes and quarters is worth $5.80. It's 16 cents more than pennies. How many dimes and quarters are in the collection?

solution

Steps 1-2 First we write what we want to find as word phrases. Then each set was represented as a function of a variable.
Or number of quarters: x
The number of coins: x + 16

Step 3 Next we create a table showing the number of coins and their value.

Step 4 Now we can write an equation.

Step 5 Solving the equation we get

Step 6 There are 12 quarters and 12 + 16 or 28 coins in the collection.

PROBLEMS OF INTEREST

The basic idea for solving interest rate problems is that the amount of interest i earned in one year is equal to the product of the interest rate r and the invested amount p (i = r * p). For example, $1,000 invested for one year with a 9% return i = (0.09)(1,000) = $90.

A table like the one shown in the example below is useful for solving problems of interest.

example 2

Two investments yield annual interest of $320. 11% will invest $1,000 more than 10%. How much is invested in each installment?

solution

Steps 1-2 First we write what we want to find as word phrases. Then each set was represented as a function of a variable.
Investment amount at 10%: x
Investment amount at 11%: x + 100

Step 3 Next, let's create a table showing the amount of money invested, the interest rates, and the interest amounts.

Step 4 We can now set up an equation that relates the interest on each investment to the total interest received.

Step 5 To find x, first multiply each member by 100 to get

Step 6 $1,000 invested at 10%; $1,000 + $1,000 or $2,000 is invested at 11%.

MIXED PROBLEMS

The basic idea for solving mixing problems is that the amount (or value) of the substances mixed must equal the amount (or value) of the final mixture.

A table like the one shown in the examples below is helpful in solving mixing problems.

Example 3

How many 80 cents per kilogram (kg) candy does a grocer need to mix with 60 kg of $1 per kilogram candy to make a $900 per kilogram mix?

solution

Steps 1-2 First we write what we want to find as a phrase. Then we represent the sentence in terms of a variable.
Kilogram 80c bullets: x

Step 3 Next, let's create a table showing the types of candy, the amount of each, and the total values ​​of each candy.

Step 4 Now we can write an equation.

Step 5 Solving the equation we get

Step 6 The grocery store needs to use 60 kg of the 800 chocolates.

Another type of mixing problem is mixing two liquids.

example 4

How many liters of 20% acid solution must be added to 10 liters of 30% acid solution to make a 25% solution?

solution

Steps 1-2 First we write what we want to find as a phrase. Then we represent the sentence in terms of a variable.

Number of liters of 20% solution to add: x

Step 3 Next, let's create a table or drawing that shows the percentage of each solution, the amount of each solution, and the amount of pure acid in each solution.

Step 4 We can now set up an equation relating the amounts of pure acid before and after combining the solutions.

Step 5 To find x, first multiply each member by 100 to get

20x + 30(10) = 25(x + 10)
20x + 300 = 25x + 250
50 = 5x
10 = x

Step 6 Add 10 liters of a 20% solution to make the desired solution.

CHAPTER SUMMARY

1. Algebraic expressions that contain parentheses can be written without parentheses by applying the distributive property in the form
   a(b + c) = ab + ac
   

   A polynomial containing a monomial factor common to all terms of the polynomial can be written as the product of the common factor and another polynomial by applying the distributive law in the form
   ab + ac = a(b + c)

2. The distributive law can be used to multiply binomials; The FOIL method proposes all four products involved.

   

3. Given is a trinomial of the form x²+ Bx + C if there are two numbers, a and b, whose product is C and whose sum is B, then
   X²+ Bx + C = (x + a)(x + b)
   Otherwise the trinomial is not factorable.

4. A trinomial of the form Ax²+ Bx + C is factorable if there are two numbers whose product is A * C and whose sum is B.

5. the difference of the squares
   a²-b²= (a + b)(a - b)

   (Video) Solving Polynomial Equations By Factoring and Using Synthetic Division

6. Equations with parentheses can be solved in the usual way, after rewriting the equation equivalently without parentheses.