


Enter an expression and click the Factor button. PRODUCTS AND FACTORSTHE DISTRIBUTIVE LAWIf we want to multiply a sum by another number, either we can multiply each term of the sum by the number before we add or we can first add the terms and then multiply. For example, In either case 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 or (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 involves the product of a monomial and binomial. Example 1 Write 2x(x  3) without parentheses. Solution We think of 2x(x  3) as 2x[x + (3)] and then apply the distributive law to obtain The above method works equally as well with the product of a monomial and trinomial. Example 2 Write  y(y^{2} + 3y  4) without parentheses. Solution Applying the distributive property yields When simplifying expressions involving parentheses, we first remove the parentheses and then combine like terms. Example 3 Simplify a(3  a)  2(a + a^{2}). We begin by removing parentheses to obtain Now, combining like terms yields a  3a^{2}. 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. 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 to simplify expressions. FACTORING MONOMIALS FROM POLYNOMIALSFrom 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 all terms in a polynomial, we can write the polynomial as the product of the common factor and another polynomial. For instance, since each term in x^{2} + 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 be factored "from" or "out of' the polynomial x^{2} + 3x. To factor a monomial from a polynomial:
Generally, we can find the common monomial factor by inspection. Example 1 We can check that we factored correctly by multiplying the factors and verifying that the product is the original polynomial. Using Example 1, we get If the common monomial is hard to find, we can write each term in prime factored form and note the common factors. Example 2 Factor 4x^{3}  6x^{2} + 2x. Solution We now see that 2x is a common monomial factor to all three terms. Then we factor 2x out of the polynomial, and write Now, we divide each term in the polynomial by 2x and write the quotients inside the parentheses to get 2x(2x^{2}  3x + 1) We can check our answer in Example 2 by multiplying the factors to obtain In this book, we will restrict the common factors to monomials consisting of numerical coefficients that are integers and to integral powers of the variables. The choice of sign for the monomial factor is a matter of convenience. Thus, 3x^{2}  6x can be factored either as 3x(x + 2) or as 3x(x  2) The first form is usually more convenient. Example 3 Factor out the common monomial, including 1. a.  3x^{2}  3 xy
Sometimes it is convenient to write formulas in factored form. Example 4 4.3 BINOMIAL PRODUCTS IWe can use the distributive law to multiply two binomials. Although there is little need to multiply binomials in arithmetic as shown in the example below, the distributive law also applies to 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
Now, combine like terms to obtain With practice, you will be able to mentally add the second and third products. Theabove process is sometimes called the FOIL method. F, O, I, and L stand for:
The FOIL method can also be used to square binomials. Example 2 Write (x + 3)^{2} without parentheses. First, rewrite (x + 3)^{2} as (x + 3)(x + 3). Next, apply the FOIL method to get Combining like terms yields When we have a monomial factor and two binomial factors, it is easiest to first multiply the binomials. Example 3 Write 3x(x  2)(x + 3) without parentheses. Now, apply the distributive law to get Common Errors Notice in Example 2 Similarly, In general, 4.4 FACTORING TRINOMIALS IIn 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 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 do not always exist, but we will study the cases where they do. Consider the following product. Notice that the first term in the trinomial, x^{2}, is product (1); the last term in the trinomial, 12, is product and the middle term in the trinomial, 7x, is the sum of products (2) and (3). In general, We use this equation (from right to left) to factor any trinomial of the form x^{2} + Bx + C. We find two numbers whose product is C and whose sum is B. Example 1 Factor x^{2} + 7x + 12. We see that the only pair of factors whose product is 12 and whose sum is 7 is 3 and 4. Thus, x^{2} + 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 x^{2} + 7x + 12 would be of the form ( + )( + ) When the first and 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 x^{2}  5x + 6 would be of the form ()() Example 2 Factor x^{2}  5x + 6. Solution We see that the only pair of factors whose product is 6 and whose sum is 5 is 3 and 2. Thus, x^{2}  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 x^{2}  x  12 would be of the form ( + )(  ) or (  )( + ) Example 3 Factor x^{2}  x  12. Solution We must 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. Thus, x^{2}  x  12 = (x  4)(x + 3) It is easier to factor a trinomial completely if any monimial factor common to each term of the trinomial is factored first. For example, we can factor 12x^{2} + 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(x^{2} + 3x + 2) Factoring again, we have 12(* + 2)(x + 1) which is said to be in completely factored form. In such cases, it is not necessary to factor the numerical factor itself, that is, we do not write 12 as 2 * 2 * 3. Example 4 Factor 3x^{2} + 12x + 12 completely. Solution 3(x^{2} + 4x + 4) Now, we factor the trinomial and obtain 3(x + 2)(x + 2) The techniques we have developed are also valid for a trinomial such as x^{2} + 5xy + 6y^{2}. Example 5Factor x^{2} + 5xy + 6y^{2}. Solution x^{2} + 5xy + 6y^{2} = (x + 3y)(x + 2y) When factoring, it is best to write the trinomial in descending powers of x. If the coefficient of the x^{2}term is negative, factor out a negative before proceeding. Example 6 Factor 8 + 2x  x^{2}. Solution x^{2} + 2x + 8 Now, we can factor out the 1 to obtain (x^{2}  2x  8) Finally, we factor the trinomial to yield (x 4)(x + 2) Sometimes, trinomials are not factorable. Example 7 Factor x^{2} + 5x + 12. Solution Skill at factoring is usually the result of extensive practice. If possible, do the factoring process mentally, writing your answer directly. You can check the results of a factorization by multiplying the binomial factors and verifying that the product is equal to the given trinomial. 4.5 BINOMIAL PRODUCTS IIIn this section, we use the procedure developed in Section 4.3 to multiply binomial factors whose firstdegree terms have numerical coefficients other than 1 or  1. Example 1 Write as a polynomial.
a. (2x  3)(x + 1) Solution We first apply the FOIL method and then combine like 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} b. (2x  y)^{2} As you may have seen in Section 4.3, the product of two bionimals may have no firstdegree term in the answer. Example 3 a. (2x  3)(2x + 3) b. (3x  y)(3x + y) When a monomial factor and two binomial factors are being multiplied, it is easiest to multiply the binomials first. Example 4 Write 3x(2x  l)(x + 2) as a polynomial. Solution 4.6 FACTORING TRINOMIALS IIIn Section 4.4 we factored trinomials of the form x^{2} + Bx + C where the second degree term had a coefficient of 1. Now we want to extend our factoring techniques to trinomials of the form Ax^{2} + Bx + C, where the seconddegree term has a coefficient other than 1 or 1. First, we consider a test to determine if a trinomial is factorable. A trinomial of the form Ax^{2} + Bx + C is factorable if we can find two integers whose product is A * C and whose sum is B. Example 1 Determine if 4x^{2} + 8x + 3 is factorable. Solution Since the factors 6 and 2 have a sum of 8, the value of B in the trinomial Ax^{2} + Bx + C, the trinomial is factorable. Example 2 The trinomial 4x^{2}  5x + 3 is not factorable, since the above table shows that there is no pair of factors whose product is 12 and whose sum is 5. The test to see if the trinomial is factorable can usually be done mentally. Once we have determined that a trinomial of the form Ax^{2} + Bx + C is fac torable, we proceed to find a pair of factors whose product is A, a pair of factors whose product is C, and an arrangement that yields the proper middle term. We illustrate by examples. Example 3 Factor 4x^{2} + 8x + 3. Solution
Now, we consider the factorization of a trinomial in which the constant term is negative. Example 4 Factor 6x^{2} + x  2. Solution
With practice, you will be able to mentally check the combinations and will not need to write out all the possibilities. Paying attention to the signs in the trinomial is particularly helpful for mentally eliminating possible combinations. It is easiest to factor a trinomial written in descending powers of the variable. Example 5 Factor. a. 3 + 4x^{2} + 8x Solution a. 4x^{2} + 8x + 3 As we said in Section 4.4, if a polynomial contains a common monomial factor in each of its terms, we should factor this monomial from the polynomial before looking for other factors. Example 6 Factor 24^{2}  44x  40. Solution 4(6x^{2}  11x  10) We then factor the trinomial, to obtain 4(3x + 2)(2x  5) ALTERNATIVE METHOD OF FACTORING TRINOMIALS If the above "trial and error" method of factoring does not yield quick results, an alternative method, which we will now demonstrate using the earlier example 4x^{2} + 8x + 3, may be helpful. 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 now proceed and use these numbers to rewrite 8x as 2x + 6x.
4.7 FACTORING THE DIFFERENCE OF TWO SQUARESSome polynomials occur so frequently that it is helpful to recognize these special forms, 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 to left, from the polynomial a^{2}  b^{2} to its factored form (a + b)(a  b). The difference of two squares, a^{2}  b^{2}, equals the product of the sum a + b and the difference a  b. Example 1 a. x2  9 = x2  32 Since (3x)(3x) = 9x^{2} we can view a binomial such as 9x^{2}  4 as (3x)^{2}  2^{2} and use the above method to factor. Example 2 a. 9x^{2}  4 = (3x)^{2}  2^{2} As before, we always factor out a common monomial first whenever possible. Example 3 a. x^{3}  x^{5} = x^{3}(l  x^{2}) 4.8 EQUATIONS INVOLVING PARENTHESESOften we must solve equations in which the variable occurs within parentheses. We can solve these equations in the usual manner after we have simplified them by applying the distributive law to remove the parentheses. Example 1 Solve 4(5  y) + 3(2y  1) = 3. Solution 20  4y + 6y  3 = 3 Now combining like terms and solving for y yields 2y + 17 = 3 2y = 14 y=l The same method can be applied to equations involving binomial products. Example 2 Solve (x + 5)(x + 3)  x = x^{2} + 1. Solution x^{2} + 8x + 15  x = x^{2} + 1 Now, combining like terms and solving for x yields x^{2} + 7x + 15 = x^{2} + 1 7x = 14 x = 2 4.9 WORD PROBLEMS INVOLVING NUMBERSParentheses are useful in representing products in which the variable is contained in one or more terms in any factor. Example 1 One integer is three more than another. If x represents the smaller integer, represent in terms of x a. The larger integer. Solution Let us say we know the sum of two numbers is 10. If we represent one number by x, 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 terms of X a. The larger integer. Solution 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 represents the smaller integer, represent in terms of x a. The larger integer Solution a. x + 2 Sometimes, the mathematical models (equations) for word problems involve parentheses. We can use the approach outlined on page 115 to obtain the equation. Then, we proceed to solve the equation by first writing equivalently the equation without parentheses. Example 4 One integer is five more than a second integer. Three times the smaller integer plus twice the larger equals 45. Find the integers. Solution Steps 12 Step 3 Step 4 3x + 2(x + 5) = 45 Step 5 Step 6 4.10 APPLICATIONSIn this section, we will examine several applications of word problems that lead to equations 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 number of 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 next example is helpful in solving coin problems. Example 1 A collection of coins consisting of dimes and quarters has a value of $5.80. There are 16 more dimes than quarters. How many dimes and quarters are in the col lection? Solution Steps 12 We first write what we want to find as word phrases. Then, we
represent each phrase in terms of a variable. 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 earned in one year at simple interest equals the product of the rate of interest r and the amount of money p invested (i = r * p). For example, $1000 invested for one year at 9% yields i = (0.09)(1000) = $90. A table like the one shown in the next example is helpful in solving interest problems. Example 2 Two investments produce an annual interest of $320. $1000 more is invested at 11% than at 10%. How much is invested at each rate? Solution Steps 12 We first write what we want to find as word phrases. Then, we
represent each phrase in terms of a variable. Step 3 Next, we make a table showing the amount of money invested, the rates of interest, and the amounts of interest. Step 4 Now, we can write an equation relating the interest from each in vestment and the total interest received. Step 5 To solve for x, first multiply each member by 100 to obtain Step 6 $1000 is invested at 10%; $1000 + $1000, or $2000, is invested at 11%. MIXTURE PROBLEMS The basic idea of solving mixture problems is that the amount (or value) of the substances being mixed must equal the amount (or value) of the final mixture. A table like the ones shown in the following examples is helpful in solving mixture problems. Example 3 How much candy worth 80c a kilogram (kg) must a grocer blend with 60 kg of candy worth $1 a kilogram to make a mixture worth 900 a kilogram? Solution Steps 12 We first write what we want to find as a word phrase. Then, we
represent the phrase in terms of a variable. 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 to obtain a 25% solution? Solution Steps 12 We first write what we want to find as a word phrase. Then, we represent 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 each solution. Step 4 We can now write an equation relating the amounts of pure acid before and after combining the solutions. Step 5 To solve for x, first multiply each member by 100 to obtain 20x + 30(10) = 25(x + 10) Step 6 Add 10 quarts of 20% solution to produce the desired solution. CHAPTER SUMMARY
