Review of Factoring Methods: GCF, Grouping, Trinomials Section 8.6 Review of Factoring Methods: GCF, Grouping, Trinomials

Objectives Factor out the greatest common factor Factor by grouping Use factoring to solve formulas for a specified variable Factor trinomials Use substitution to factor trinomials Use the grouping method to factor trinomials

Objective 1: Factor Out the Greatest Common Factor When we factor a polynomial, we write a sum of terms as a product of factors. To perform the most basic type of factoring, we determine whether the terms of the given polynomial have any common factors. This process, called factoring out the greatest common factor, is based on the distributive property. A polynomial that cannot be factored is called a prime polynomial or an irreducible polynomial.

EXAMPLE 1 Factor: 3xy2z3 + 6xz2 − 9xyz4 Strategy We will determine the GCF of the terms of the polynomial. Then we will write each term of the polynomial as the product of the GCF and one other factor. Why We can then use the distributive property to factor out the GCF.

EXAMPLE 1 Solution We begin by factoring each term: Factor: 3xy2z3 + 6xz2 − 9xyz4 Solution We begin by factoring each term: Since each term has one factor of 3, one factor of x, and two factors of z, and there are no other common factors, 3xz2 is the greatest common factor of the three terms. We write each term as the product of the GCF, 3xz2, and one other factor and proceed as follows: We can check the factorization using multiplication.

Objective 2: Factor By Grouping Although the terms of many polynomials don’t have a common factor, other than 1, it is possible to factor some of them by arranging their terms in convenient groups. This method is called factoring by grouping. To factor a polynomial, it is often necessary to factor more than once. When factoring a polynomial, always look for a common factor first.

EXAMPLE 4 Factor: 2c − 2d + cd − d2 Strategy Since the four terms of the polynomial do not have a common factor (other than 1), we will attempt to factor the polynomial by grouping. We will factor out a common factor from the first two terms and from the last two terms. Why This will produce a common binomial factor that can be factored out.

EXAMPLE 4 Factor: 2c − 2d + cd − d2 Solution The first two terms have a common factor, 2, and the last two terms have a common factor, d. When we factor out the common factor from each group, a common binomial factor c − d appears.

Objective 3: Use Factoring to Solve Formulas for a Specified Variable Factoring is often required to solve a formula for one of its variables.

EXAMPLE 6 Electronics. The formula r1r2 = rr2 + rr1 is used in electronics to relate the combined resistance, r, of two resistors wired in parallel. The variable r1 represents the resistance of the first resistor, and the variable r2 represents the resistance of the second. Solve for r2. Strategy To isolate r2 on one side of the equation, we will get all the terms involving r2 on the left side and all the terms not involving r2 on the right side. Why To solve a formula for a specified variable means to isolate that variable on one side of the equation, with all other variables and constants on the opposite side.

EXAMPLE 6 Electronics. The formula r1r2 = rr2 + rr1 is used in electronics to relate the combined resistance, r, of two resistors wired in parallel. The variable r1 represents the resistance of the first resistor, and the variable r2 represents the resistance of the second. Solve for r2. Solution Simplify the left side by removing the common factor r1 - r from the numerator and denominator.

Objective 4: Factor Trinomials Many trinomials factor as the product of two binomials.

EXAMPLE 7 Factor: x2 – 6x + 8 Strategy We will assume that this trinomial is the product of two binomials. We must find the terms of the binomials. Why Since the terms of x2 – 6x + 8 do not have a common factor (other than 1), the only option is to try to factor it as the product of two binomials.

EXAMPLE 7 Factor: x2 – 6x + 8 Solution We represent the binomials using two sets of parentheses. Since the first term of the trinomial is x2, we enter x and x as the first terms of the binomial factors. The second terms of the binomials must be two integers whose product is 8 and whose sum is −6. We list all possible integer-pair factors of 8 in the table.

EXAMPLE 7 Solution, Contd. Factor: x2 – 6x + 8 Solution, Contd. The fourth row of the table contains the correct pair of integers −2 and −4, whose product is 8 and whose sum is −6. To complete the factorization, we enter −2 and −4 as the second terms of the binomial factors. x2 − 6x + 8 = (x − 2)(x − 4) Check: We can verify the factorization by multiplication:

Objective 5: Use Substitution to Factor Trinomials For more complicated expressions, especially those involving a quantity within parentheses, a substitution sometimes helps to simplify the factoring process.

EXAMPLE 11 Factor: (x + y)2 + 7(x + y) + 12 Strategy We will use a substitution where we will replace each expression x + y with the variable z and factor the resulting trinomial. Why The resulting trinomial will be easier to factor because it will be in only one variable, z.

EXAMPLE 11 Solution Factor: (x + y)2 + 7(x + y) + 12 Factor the trinomial.

Objective 6: Use the Grouping Method to Factor Trinomials Another way to factor trinomials is to write them as equivalent four-termed polynomials and factor by grouping.

EXAMPLE 12 Factor by grouping: a. x2 + 8x + 15 b. 10x2 + 13xy – 3y2 Strategy In each case, we will express the middle term of the trinomial as the sum of two terms. Why We want to produce an equivalent four-termed polynomial that can be factored by grouping.

EXAMPLE 12 Factor by grouping: a. x2 + 8x + 15 b. 10x2 + 13xy – 3y2 Solution a. Since x2 + 8x + 15 = 1x2 + 8x + 15, we identify a as 1, b as 8, and c as 15. The key number is ac =1(15) = 15. We must find two integers whose product is the key number 15 and whose sum is b = 8. Since the integers must have a positive product and a positive sum, we consider only positive factors of 15. The second row of the table contains the correct pair of integers 3 and 5, whose product is 15 and whose sum is 8. We can express the middle term, 8x, of the trinomial as the sum of two terms, using the integers 3 and 5 as coefficients of the two terms and factor the equivalent four-termed polynomial by grouping: Check the factorization by multiplying.

EXAMPLE 12 Solution, Contd. Factor by grouping: a. x2 + 8x + 15 b. 10x2 + 13xy – 3y2 Solution, Contd. b. In 10x2 + 13xy – 3y2, we have a = 10, b = 13, and c = −3. The key number is ac = 10(−3) = −30. We must find a factorization of −30 such that the sum of the factors is b = 13. Since the factors must have a negative product, their signs must be different. The possible factor pairs are listed in the table. The seventh row contains the correct pair of numbers 15 and –2, whose product is −30 and whose sum is 13. They serve as the coefficients of two terms, 15xy and −2xy, that we place between 10x2 and −3y2.

EXAMPLE 12 Solution, Contd. Factor by grouping: a. x2 + 8x + 15 b. 10x2 + 13xy – 3y2 Solution, Contd.