## THE DISTRIBUTIVE LAW

If we want to multiply a sum by another number, either we can multiply each term of the sum by the number before we địa chỉ cửa hàng or we can first địa chỉ cửa hàng the terms và then multiply. For example, In either case the result is the same.

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

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

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

Our first example involves the hàng hóa of a monomial & binomial.

Example 1 Write 2x(x - 3) without parentheses.

Solution

We think of 2x(x - 3) as 2x & then apply the distributive law lớn obtain The above method works equally as well with the sản phẩm of a monomial & trinomial.

Example 2 Write - y(y2 + 3y - 4) without parentheses.

Solution

Applying the distributive property yields When simplifying expressions involving parentheses, we first remove the parentheses and then combine lượt thích terms.

Example 3 Simplify a(3 - a) - 2(a + a2).

We begin by removing parentheses lớn obtain Now, combining like terms yields a - 3a2.

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

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

Solution Notice that in Example 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 used the procedures that we introduced in Section 2.5 lớn simplify expressions.

## FACTORING MONOMIALS FROM POLYNOMIALS

From the symmetric property of equality, we know that if

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

Thus, if there is a monomial factor common to lớn all terms in a polynomial, we can write the polynomial as the sản phẩm of the common factor & another polynomial. For instance, 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, & the number x is said lớn be factored "from" or "out of" the polynomial x2 + 3x.

To factor a monomial from a polynomial:Write a set of parentheses preceded by the monomial common khổng lồ each term in the polynomial.Divide the monomial factor into each term in the polynomial và write the quotient in the parentheses.Generally, 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 verifyingthat the sản phẩm is the original polynomial. Using Example 1, we get If the common monomial is hard lớn find, we can write each term in prime factored size and note the common factors.

Example 2 Factor 4x3 - 6x2 + 2x.

Solution We can write We now see that 2x is a common monomial factor to all three terms. Then we factor 2x out of the polynomial, & write 2x()

Now, we divide each term in the polynomial by 2x and write the quotients inside the parentheses to lớn get

2x(2x2 - 3x + 1)

We can kiểm tra our answer in Example 2 by multiplying the factors khổng lồ obtain In this book, we will restrict the common factors to monomials consisting of numerical coefficients that are integers và to integral powers of the variables. The choice of sign for the monomial factor is a matter of convenience. Thus,

-3x2 - 6x

can be factored either as

-3x(x + 2) or as 3x(-x - 2)

The first size is usually more convenient.

Example 3Factor out the common monomial, including -1.

a. - 3x2 - 3 xyb. -x3 - x2 + x Solution Sometimes it is convenient to write formulas in factored form.

Example 4 a. A = p + PRT = P(1 + RT) b. S = 4kR2 - 4kr2 = 4k(R2 - r2)

## 4.3BINOMIAL PRODUCTS I

We can use the distributive law khổng lồ multiply two binomials. Although there is little need to multiply binomials in arithmetic as shown in the example below, the distributive law also applies khổng lồ expressions containing variables. We will now apply the above procedure for an expression containing variables.

Example 1

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

Solution First, apply the distributive property lớn get Now, combine like terms khổng lồ obtain x2 + x - 6

With practice, you will be able lớn mentally địa chỉ cửa hàng the second & third products. Theabove process is sometimes called the FOIL method. F, O, I, và L stand for: 1.The hàng hóa of the First terms.2.The product of the Outer terms.3.The hàng hóa of the Inner terms.4.The sản phẩm of the Last terms.

The FOIL method can also be used to lớn square binomials.

Example 2

Write (x + 3)2 without parentheses.Solution

First, rewrite (x + 3)2 as (x + 3)(x + 3). Next, apply the FOIL method to get Combining lượt thích terms yieldsx2 + 6x + 9

When we have a monomial factor and two binomial factors, it is easiest lớn first multiply the binomials.

Example 3

Write 3x(x - 2)(x + 3) without parentheses.Solution First, multiply the binomials to obtain3x(x2 + 3x - 2x - 6) = 3x(x2 + x - 6)

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

Common Errors

Notice in Example 2 Similarly, In general, ## 4.4FACTORING TRINOMIALS I

In Section 4.3, we saw how lớn find the hàng hóa of two binomials. Now we will reverse this process. That is, given the hàng hóa of two binomials, we will find the binomial factors. The process involved is another example of factoring. As before,we will only consider factors in which the terms have integral numerical coefficients. Such factors vị not always exist, but we will study the cases where they do.

Consider the following product. Notice that the first term in the trinomial, x2, is sản phẩm (1); the last term in thetrinomial, 12, is product and the middle term in the trinomial, 7x, is the sum of products (2) & (3).In general, We use this equation (from right to left) to factor any trinomial of the khung x2 + Bx + C. We find two numbers whose hàng hóa is C and whose sum is B.

Example 1 Factor x2 + 7x + 12.Solution We look for two integers whose hàng hóa is 12 và whose sum is 7. Consider the following pairs of factors whose product is 12. We see that the only pair of factors whose hàng hóa is 12 và whose sum is 7 is 3 & 4. Thus,

x2 + 7x + 12 = (x + 3)(x + 4)

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

x2 + 7x + 12would be of the form

( + )( + )

When the first và third terms of a trinomial are positive but the middle term is negative, we need only 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 of the form

(-)(-)

Example 2 Factor x2 - 5x + 6.

Solution Because the third term is positive và 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 & whose sum is -5 is -3 & -2. Thus,

x2 - 5x + 6 = (x - 3)(x - 2)

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

x2 - x - 12

would be of the size

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

Example 3

Factor x2 - x - 12.

Solution We must find two integers whose product is -12 & whose sum is -1. We danh mục the possibilities. We see that the only pair of factors whose sản phẩm is -12 & whose sum is -1 is -4 and 3. Thus,

x2 - x - 12 = (x - 4)(x + 3)

It is easier khổng lồ factor a trinomial completely if any monimial factor common to each term of the trinomial is factored first. For example, we can factor

12x2 + 36x + 24

as A monomial can then be factored from these binomial factors. However, first factoring the common factor 12 from the original expression yields

12(x2 + 3x + 2)

Factoring again, we have

12(* + 2)(x + 1)

which is said khổng lồ be in completely factored form. In such cases, it is not necessary khổng lồ factor the numerical factor itself, that is, we bởi vì not write 12 as 2 * 2 * 3.

Example 4

Factor 3x2 + 12x + 12 completely.

SolutionFirst we factor out the 3 from the trinomial to lớn get

3(x2 + 4x + 4)

Now, we factor the trinomial và obtain

3(x + 2)(x + 2)

The techniques we have developed are also valid for a trinomial such as x2 + 5xy + 6y2.

Example 5Factor x2 + 5xy + 6y2.

Solution We find two positive factors whose product is 6y2 and whose sum is 5y (the coefficient of x). The two factors are 3y and 2y. Thus,

x2 + 5xy + 6y2 = (x + 3y)(x + 2y)

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

Example 6

Factor 8 + 2x - x2.

Solution We first rewrite the trinomial in descending powers of x khổng lồ get

-x2 + 2x + 8

Now, we can factor out the -1 to lớn obtain

-(x2 - 2x - 8)

Finally, we factor the trinomial khổng lồ yield

-(x- 4)(x + 2)

Sometimes, trinomials are not factorable.

Example 7

Factor x2 + 5x + 12.

Solution We look for two integers whose product is 12 & whose sum is 5. From the table in Example 1 on page 149, we see that there is no pair of factors whose product is 12 & whose sum is 5. In this case, the trinomial is not factorable.

Skill at factoring is usually the result of extensive practice. If possible, vị the factoring process mentally, writing your answer directly. You can kiểm tra the results of a factorization by multiplying the binomial factors and verifying that the hàng hóa is equal to lớn the given trinomial.

## 4.5BINOMIAL PRODUCTS II

In this section, we use the procedure developed in Section 4.3 khổng lồ 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 - 2y)(3x + y)

Solution

We first apply the FOIL method và then combine lượt thích terms. As before, if we have a squared binomial, we first rewrite it as a product, then apply the FOIL method.

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 may have seen in Section 4.3, the product of two bionimals may have no first-degree term in the answer.

Example 3

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

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

When a monomial factor và two binomial factors are being multiplied, it iseasiest to multiply the binomials first.

Example 4

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

Solution We first multiply the binomials to lớn get3x(2x2 + 4x - x - 2) = 3x(2x2 + 3x - 2)Now multiplying by the monomial yields3x(2x2) + 3x(3x) + 3x(-2) = 6x3 + 9x2 - 6x

## 4.6FACTORING TRINOMIALS II

In Section 4.4 we factored trinomials of the khung x2 + Bx + C where the second-degree term had a coefficient of 1. Now we want to extend our factoring techniquesto trinomials of the form Ax2 + Bx + C, where the second-degree term has acoefficient other than 1 or -1.

First, we consider a demo to determine if a trinomial is factorable. A trinomial ofthe khung Ax2 + Bx + C is factorable if we can find two integers whose sản phẩm isA * C & whose sum is B.

Example 1

Determine if 4x2 + 8x + 3 is factorable.

Solution We kiểm tra to see if there are two integers whose product is (4)(3) = 12 and whosesum is 8 (the coefficient of x). Consider the following possibilities. Since the factors 6 and 2 have a sum of 8, the value of B in the trinomialAx2 + Bx + C, the trinomial is factorable.

Example 2

The trinomial 4x2 - 5x + 3 is not factorable, since the above table shows thatthere is no pair of factors whose sản phẩm is 12 và whose sum is -5. The demo tosee if the trinomial is factorable can usually be done mentally.

Once we have determined that a trinomial of the size Ax2 + Bx + C is fac-torable, we proceed to find a pair of factors whose hàng hóa is A, a pair of factorswhose product is C, & an arrangement that yields the proper middle term. Weillustrate by examples.

Example 3

Factor 4x2 + 8x + 3.

Solution Above, we determined that this polynomial is factorable. We now proceed. 1. We consider all pairs of factors whose hàng hóa is 4. Since 4 is positive, only positive integers need lớn be considered. The possibilities are 4, 1 & 2, 2.2. We consider all pairs of factors whose product is 3. Since the middle term is positive, consider positive pairs of factors only. The possibilities are 3, 1. We write all possible arrangements of the factors as shown. 3. We select the arrangement in which the sum of products (2) và (3) yields a middle term of 8x.

Now, we consider the factorization of a trinomial in which the constant term is negative.

Example 4

Factor 6x2 + x - 2.

Solution First, we demo to see if 6x2 + x - 2 is factorable. We look for two integers that havea product of 6(-2) = -12 and a sum of 1 (the coefficient of x). The integers 4 and-3 have a sản phẩm of -12 và a sum of 1, so the trinomial is factorable. We nowproceed. We consider all pairs of factors whose product is 6. Since 6 is positive, only positive integers need lớn be considered. Then possibilities are 6, 1 và 2, 3.We consider all pairs of factors whose hàng hóa is -2. The possibilities are 2, -1 and -2, 1. We write all possible arrange ments of the factors as shown.We select the arrangement in which the sum of products (2) & (3) yields a middle term of x.

With practice, you will be able lớn mentally kiểm tra the combinations và will notneed to lớn write out all the possibilities. Paying attention lớn the signs in the trinomialis particularly helpful for mentally eliminating possible combinations.

It is easiest lớn factor a trinomial written in descending powers of the variable.

Example 5

Factor.

a. 3 + 4x2 + 8x b. X - 2 + 6x2

Solution Rewrite each trinomial in descending powers of x và then follow the solutions ofExamples 3 & 4.

a. 4x2 + 8x + 3 b. 6x2 + x - 2

As we said in Section 4.4, if a polynomial contains a common monomial factorin each of its terms, we should factor this monomial from the polynomial beforelooking for other factors.

Example 6

Factor 242 - 44x - 40.

Solution We first factor 4 from each term to lớn get

4(6x2 - 11x - 10)

We then factor the trinomial, khổng lồ obtain

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

ALTERNATIVE METHOD OF FACTORING TRINOMIALS

If the above "trial & error" method of factoring does not yield quick results, analternative method, which we will now demonstrate using the earlier example4x2 + 8x + 3, may be helpful.

We know that the trinomial is factorable because we found two numbers whoseproduct is 12 and whose sum is 8. Those numbers are 2 & 6. We now proceedand use these numbers lớn rewrite 8x as 2x + 6x. We now factor the first two terms, 4*2 + 2x & the last two terms, 6x + 3.A common factor, 2x + 1, is in each term, so we can factor again.This is the same result that we obtained before.

## 4.7FACTORING THE DIFFERENCE OF TWO SQUARES

Some polynomials occur so frequently that it is helpful khổng lồ recognize these specialforms, which in tum enables us to directly write their factored form. Observe that In this section we are interested in viewing this relationship from right khổng lồ left, fromthe polynomial a2 - b2 to lớn its factored size (a + b)(a - b).

The difference of two squares, a2 - b2, equals the product of the sum a + b & the difference a - b.

Example 1

a. X2 - 9 = x2 - 32 = (x + 3)(x - 3) b. X2 - 16 = x2 - 42 = (x + 4)(x - 4)

Since

(3x)(3x) = 9x2

we can view a binomial such as 9x2 - 4 as (3x)2 - 22 & use the above methodto factor.

Example 2

a.9x2 - 4 = (3x)2 - 22= (3x + 2)(3x - 2)b.4y2 - 25x2 = (2y)2 - (5x)2= (2y + 5x)(2y - 5x)

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

Example 3

a.x3 - x5 = x3(l - x2) = x3(1 + x)(l - x)b.a2x2y - 16y = y(a2x2 - 16) = y<(ax)2 - 42>= y(ax - 4 )(ax + 4)

## 4.8EQUATIONS INVOLVING PARENTHESES

Often we must solve equations in which the variable occurs within parentheses. Wecan solve these equations in the usual manner after we have simplified them byapplying the distributive law khổng lồ remove the parentheses.

Example 1

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

Solution We first apply the distributive law lớn get

20 - 4y + 6y - 3 = 3

Now combining lượt thích terms and solving for y yields

2y + 17 = 3

2y = -14

y=-l

The same method can be applied khổng lồ equations involving binomial products.

Example 2

Solve (x + 5)(x + 3) - x = x2 + 1.

Solution First, we apply the FOIL method lớn remove parentheses và obtain

x2 + 8x + 15 - x = x2 + 1

Now, combining like terms & solving for x yields

x2 + 7x + 15 = x2 + 1

7x = -14

x = -2

## 4.9WORD PROBLEMS INVOLVING NUMBERS

Parentheses are useful in representing products in which the variable is containedin one or more terms in any factor.

Example 1

One integer is three more than another. If x represents the smaller integer, representin terms of x

a. The larger integer.b. Five times the smaller integer.c. Five times the larger integer.

Solution a. X + 3b. 5x c. 5(x + 3)

Let us say we know the sum of two numbers is 10. If we represent one number byx, then the second number must be 10 - x as suggested by the following table. In general, if we know the sum of two numbers is 5 and x represents one number,the other number must be S - x.

Example 2

The sum of two integers is 13. If x represents the smaller integer, represent in termsof X

a. The larger integer.b. Five times the smaller integer.c. Five times the larger integer.

Solution a. 13 - x b. 5x c. 5(13 - x)

The next example concerns the notion of consecutive integers that was consid-ered in Section 3.8.

Example 3

The difference of the squares of two consecutive odd integers is 24. If x representsthe smaller integer, represent in terms of x

a. The larger integerb. The square of the smaller integer c. The square of the larger integer.

Solution

a. X + 2b. X2 c. (x + 2)2

Sometimes, the mathematical models (equations) for word problems involveparentheses. We can use the approach outlined on page 115 lớn obtain the equation.Then, we proceed to lớn solve the equation by first writing equivalently the equationwithout parentheses.

Example 4

One integer is five more than a second integer. Three times the smaller integer plustwice the larger equals 45. Find the integers.

Solution

Steps 1-2 First, we write what we want khổng lồ find (the integers) as word phrases. Then, we represent the integers in terms of a variable.The smaller integer: x The larger integer: x + 5

Step 3 A sketch is not applicable.

Step 4 Now, we write an equation that represents the condition in the problemand get

3x + 2(x + 5) = 45

Step 5 Applying the distributive law to lớn remove parentheses yields Step 6 The integers are 7 and 7 + 5 or 12.

## 4.10 APPLICATIONS

In this section, we will examine several applications of word problems that lead toequations that involve parentheses. Once again, we will follow the six steps out-lined on page 115 when we solve the problems.

COIN PROBLEMS

The basic idea of problems involving coins (or bills) is that the value of a numberof coins of the same denomination is equal to lớn the sản phẩm of the value of a singlecoin & the total number of coins. A table like the one shown in the next example is helpful in solving coin problems.

Example 1

A collection of coins consisting of dimes & quarters has a value of \$5.80. Thereare 16 more dimes than quarters. How many dimes và quarters are in the col-lection?

Solution

Steps 1-2 We first write what we want lớn find as word phrases. Then, werepresent each phrase in terms of a variable.The number of quarters: x The number of dimes: x + 16

Step 3 Next, we make a table showing the number of coins and their value. Step 4 Now we can write an equation. Step 5 Solving the equation yields Step 6 There are 12 quarters and 12 + 16 or 28 dimes in the collection.

INTEREST PROBLEMS

The basic idea of solving interest problems is that the amount of interest i earnedin one year at simple interest equals the hàng hóa of the rate of interest r & theamount of money p. Invested (i = r * p). For example, \$1000 invested for one yearat 9% yields i = (0.09)(1000) = \$90.

A table like the one shown in the next example is helpful in solving interestproblems.

Example 2

Two investments produce an annual interest of \$320. \$1000 more is invested at11% than at 10%. How much is invested at each rate?

Solution

Steps 1-2 We first write what we want khổng lồ find as word phrases. Then, werepresent each phrase in terms of a variable. Amount invested at 10%: x Amount invested at 11%: x + 100

Step 3 Next, we make a table showing the amount of money invested, therates of interest, & the amounts of interest. Step 4 Now, we can write an equation relating the interest from each in-vestment và the total interest received. Step 5 to lớn solve for x, first multiply each member by 100 to lớn obtain Step 6 \$1000 is invested at 10%; \$1000 + \$1000, or \$2000, is invested at11%.

MIXTURE PROBLEMS

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

A table lượt thích the ones shown in the following examples is helpful in solvingmixture problems.

Example 3

How much candy worth 80c a kilogram (kg) must a grocer blend with 60 kg ofcandy worth \$1 a kilogram lớn make a mixture worth 900 a kilogram?

Solution

Steps 1-2 We first write what we want lớn find as a word phrase. Then, werepresent the phrase in terms of a variable.Kilograms of 80c candy: x

Step 3 Next, we make a table showing the types of candy, the amount of each,and the total values of each. Step 4 We can now write an equation. Step 5 Solving the equation yields Step 6 The grocer should use 60 kg of the 800 candy.

Another type of mixture problem is one that involves the mixture of the two liquids.

Example 4

How many quarts of a 20% solution of acid should be added to 10 quarts of a 30%solution of acid khổng lồ obtain a 25% solution?

Solution

Steps 1-2 We first write what we want lớn find as a word phrase. Then, werepresent the phrase in terms of a variable.

Number of quarts of 20% solution to be added: x

Step 3 Next, we make a table or drawing showing the percent of each solu-tion, the amount of each solution, and the amount of pure acid in eachsolution.  Step 4 We can now write an equation relating the amounts of pure acid beforeand after combining the solutions. Step 5 to solve for x, first multiply each member by 100 to obtain

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

Step 6 add 10 quarts of 20% solution lớn produce the desired solution.

## CHAPTER SUMMARY

Algebraic expressions containing parentheses can be written without parentheses byapplying the distributive law in the forma(b + c) = ab + ac

A polynomial that contains a monomial factor common to lớn all terms in thepolynomial can be written as the sản phẩm of the common factor và anotherpolynomial by applying the distributive law in the formab + ac = a(b + c)

The distributive law can be used to lớn multiply binomials; the FOIL method suggeststhe four products involved. Given a trinomial of the form x2 + Bx + C, if there are two numbers, a & b,whose hàng hóa is C và whose sum is B, then x2 + Bx + C = (x + a)(x + b) otherwise, the trinomial is not factorable.

A trinomial of the khung Ax2 + Bx + C is factorable if there are two numbers whoseproduct is A * C và whose sum is B.

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The difference of squaresa2 - b2 = (a + b)(a - b)

Equations involving parentheses can be solved in the usual way after the equationhas been rewritten equivalently without parentheses.