Copyright © 2011 Pearson Education, Inc.

CHAPTER 6 Factoring 6.1 Greatest Common Factor and Factoring by Grouping 6.2 Factoring Trinomials 6.3 Factoring Special Products and Factoring Strategies 6.4 Solving Equations by Factoring Copyright © 2011 Pearson Education, Inc.

Greatest Common Factor and Factoring by Grouping
6.1 Greatest Common Factor and Factoring by Grouping 1. Find the greatest common factor of a set of terms. 2. Factor a monomial GCF out of the terms of a polynomial. 3. Factor polynomials by grouping. Copyright © 2011 Pearson Education, Inc.

Factored form: A number or expression written as a product of factors. Following are some examples of polynomials in factored form: 2x + 8 = 2(x + 4) x2 + 5x + 6 = (x + 2)(x + 3) Factored form Factored form Greatest common factor (GCF) of a set of terms: A monomial with the greatest coefficient and degree that evenly divides all of the given terms. Copyright © 2011 Pearson Education, Inc.

Finding the GCF To find the GCFof two or more monomials, 1. Write the prime factorization in exponential form for each monomial. Treat variables like prime factors. 2. Write the GCF’s factorization by including the prime factors (and variables) common to all the factorizations, each raised to its smallest exponent in the factorizations. 3. Multiply the factors in the factorization created in step 2. Note: If there are no common prime factors, then the GCF is 1. Copyright © 2011 Pearson Education, Inc.

Example 1 Find the GCF of 45a3b and 30a2. Solution Write the prime factorization of each monomial, treating the variables like prime factors. 45a3b = 32 • 5 • a3 • b 30a2 = 2 • 3 • 5 • a2 The common prime factors are 3, 5, and a. GCF = 3 • 5 • a2 = 15a2 Copyright © 2011 Pearson Education, Inc.

Factoring a Monomial GCF Out of a Polynomial To factor a monomial GCF out of the terms of a polynomial, 1. Find the GCF of the terms in the polynomial. 2. Rewrite the polynomial as a product of the GCF and the quotient of the polynomial and the GCF. Polynomial = Copyright © 2011 Pearson Education, Inc.

Example 2a Solution Find the GCF of 36x2 + 42x. 36x2 + 42x = 6x(6x+7)
Factor. 36x2 + 42x Solution Find the GCF of 36x2 + 42x. Rewrite using the form 36x2 + 42x Separate the terms. = 6x(6x+7) Divide the terms by the GCF. It is possible to check your answer by distributing. Copyright © 2011 Pearson Education, Inc.

Example 4 Factor. Solution Notice that this expression is a sum of two products, a(b + 5), and 8(b + 5). Further, note that (b + 5) is the GCF of the two products. Copyright © 2011 Pearson Education, Inc.

Factoring by Grouping To factor a four-term polynomial by grouping, 1. Factor out any monomial GCF (other than 1) that is common to all four terms. 2. Group together pairs of terms and factor the GCF out of each pair. 3. If there is a common binomial factor, then factor it out. 4. If there is no common binomial factor, then interchange the middle two terms and repeat the process. If there is still no common binomial factor, then the polynomial cannot be factored by grouping. Copyright © 2011 Pearson Education, Inc.

Example 5c Factor. Solution There is a monomial GCF, 3p, common to all four terms. Because the polynomial in the parenthesis has four terms, we try to factor by grouping. Copyright © 2011 Pearson Education, Inc.

6.2 Factoring Trinomials 1. Factor trinomials of the form x2 + bx + c. 2. Factor trinomials of the form ax2 + bx + c, where a  1, by trial. 3. Factor trinomials of the form ax2 + bx + c, where a  1, by grouping. 4. Factor trinomials of the form ax2 + bx + c using substitution. Copyright © 2011 Pearson Education, Inc.

First we consider trinomials in the form x2 + bx + c in which the coefficient of the squared term is 1. If the trinomial is factorable, we have two binomial factors. F O I L (x + 2)(x + 3) = x2 + 3x + 2x + 6 = x x Sum of 2 and 3 Product of 2 and 3 Copyright © 2011 Pearson Education, Inc.

Factoring x2 + bx + c To factor a trinomial of the form x2 + bx + c, 1. Find two numbers with a product equal to c and a sum equal to b. 2. The factored trinomial will have the form: (x ) (x ), where the second terms are the numbers found in Step 1. Note: The signs of b and c may cause one or both signs in the binomial factors to be minus signs. Copyright © 2011 Pearson Education, Inc.

Example 1a Factor. x2 – 6x + 8 Solution We must find a pair of numbers whose product is 8 and whose sum is –6. Note that if two numbers have a positive product and negative sum, they must both be negative. List the possible products and sums. Product Sum (–1)(–8) = 8 –1 + (–8) = –9 (–2)(–4) = 8 –2 + (–4) = –6 This is the correct combination. Copyright © 2011 Pearson Education, Inc.

continued Answer x2 – 6x + 8 = (x +(– 2))(x+( – 4)) = (x – 2)(x – 4) Check Multiply the factors to verify that their product is the original polynomial. (x – 2)(x – 4) = x2 – 4x – 2x + 8 = x2 – 6x + 8 It checks. Copyright © 2011 Pearson Education, Inc.

Example 1b Factor. x2 + 2x – 24 Solution: We must find a pair of numbers whose product is –24 and whose sum is 2. Because the product is negative, the two numbers have different signs. Product Sum (–2)(12) = 24 – = 10 (–4)(6) = 24 –4 + 6 = 2 This is the correct combination. Copyright © 2011 Pearson Education, Inc.

continued Answer x2 + 2x – 24 = (x – 4)(x + 6) Check We can check by multiplying the binomial factors to see if their product is the original polynomial. (x – 4)(x + 6) = x2 + 6x – 4x – 24 Multiply the factors using FOIL. = x2 + 2x – 24 The product is the original polynomial. Copyright © 2011 Pearson Education, Inc.

Example 2b Factor. x4 + 2x3 + 5x2 Solution First, factor out the GCF, x2. x2 ( x2 + 2x + 5) Now try to factor the trinomial to two binomials. We must find a pair of numbers whose product is 5 and whose sum is 2. Product Sum (1)(5) = 5 1 + 5 = 6 Copyright © 2011 Pearson Education, Inc.

continued Factor. x4 + 2x3 + 5x2 There is no combination of factors whose product is 5 and sum is 2. The trinomial x2 + 2x + 5 has no binomial factors with integer terms, so x2 (x2 + 2x + 5) is the final factored form. Copyright © 2011 Pearson Education, Inc.

Example 4a Factor. a2 – ab – 20b2 Solution We must find a pair of terms whose product is 20b2 and whose sum is –1b. Product Sum (–1b)(20b) = –20b2 –1b+ 20b= 19b (1b)(–20b) = – 20b2 1b+ (–20b) = –19b (2b)(–10b) = – 20b2 2 b+ (–10b) = –8b (–2b)(10b) = – 20b2 –2b + 10b = 8b (4b)(–5b) = – 20b2 4b + (–5b) = –1b (–4b)(5b) = – 20b2 –4b + 5b = 1b This is the correct combination. Answer a2 – ab – 20b2 = (a – 5b)(a + 4b) Copyright © 2011 Pearson Education, Inc.

Example Factor. 4xy3 + 12xy2 – 72xy Solution This trinomial has a monomial GCF, 4xy. 4xy3 + 12xy2 – 72xy = 4xy(y2 + 3y – 18) We must find a pair of numbers whose product is –18 and whose sum is 3. = 4xy(y – 3)(y + 6) Copyright © 2011 Pearson Education, Inc.

Example 5 Factor. Solution The first terms must multiply to equal 6x2. These could be x and 6x, or 2x and 3x. The last terms must multiply to equal –5. Because –5 is negative, the last terms in the binomials must have different signs. This factor pair must be 1 and 5. Copyright © 2011 Pearson Education, Inc.

continued Now we multiply binomials with various combinations of these first and last terms until we find a combination whose inner and outer products combine to equal 13x. Incorrect combinations. Correct combination. Answer Copyright © 2011 Pearson Education, Inc.

Factoring by Trial and Error To factor a trinomial of the form ax2 + bx + c, where a ≠ 1, by trial and error: 1. Look for a monomial GCF in all of the terms. If there is one, factor it out. 2. Write a pair of first terms whose product is ax2. ax2 Copyright © 2011 Pearson Education, Inc.

3. Write a pair of last terms whose product is c. 4. Verify that the sum of the inner and outer products is bx (the middle term of the trinomial). c + Outer bx Inner Copyright © 2011 Pearson Education, Inc.

If the sum of the inner and outer products is not bx, then try the following: a. Exchange the last terms of the binomials from step 3; then repeat step 4. b. For each additional pair of last terms, repeat steps 3 and 4. c. For each additional pair of first terms, repeat steps 2 – 4. Copyright © 2011 Pearson Education, Inc.

Example 6a Factor. Solution There is no common monomial factor.
The first terms must multiply to equal 6x2. These could be x and 6x, or 3x and 2x. The last terms must multiply to equal 2. Because 2 is a prime number, its factors are 1 and 2. Copyright © 2011 Pearson Education, Inc.

Example 6d Factor. Solution
Factor out the monomial GCF, 3x. Now we factor the trinomial within the parentheses. The first terms must multiply to equal 7x2. These could be x and 7x. The last terms must multiply to equal 3. Because 3 is a prime number, its factors are 1 and 3. Copyright © 2011 Pearson Education, Inc.

Factoring ax2 + bx + c, Where a ≠ 1, by Grouping To factor a trinomial of the form ax2 + bx + c, where a ≠ 1, by grouping, 1. Look for a monomial GCF in all of the terms. If there is one, factor it out. 2. Find two factors of the product ac whose sum is b. 3. Write a four-term polynomial in which bx is written as the sum of two like terms whose coefficients are the two factors you found in step 2. 4. Factor by grouping. Copyright © 2011 Pearson Education, Inc.

Example 7a Factor. Solution For this trinomial, a = 2, b = –15, and c = 7, so ac = (2)(7) = 14. We must find two factors of 14 whose sum is –15. Since 14 is positive and –15 is negative, the two factors must be negative. Factors of ac Sum of Factors of ac (–2)(–7) = 14 –2 + (–7) = –9 (–1)(–14) = 14 –1 + (– 14) = –15 Correct Copyright © 2011 Pearson Education, Inc.

continued –15x –x – 14x Write -15x as –x – 14.
Factor x out 2x2 – x, factor -7 out of -14x+7. Factor out of 2x – 1. Copyright © 2011 Pearson Education, Inc.

Example 7b Factor. Solution Notice that there is a GCF, 2x2, that we can factor out. 2x2(6x2 + 7x – 3) Now we factor the trinomial within the parentheses. a = 6, c = 3; ac = (6)(3) = 18 Find two factors of 18 whose sum is 7. Because the product is negative, the two factors will have different signs. Copyright © 2011 Pearson Education, Inc.

continued Factor. Solution Factors of ac Sum of Factors of ac (1)(18) = 18  = 17 (2)(9) = 18 2 + 9 = 7 (3)(6) = 18 3 + 6 = 3 Correct 2x2(6x2 + 7x – 3) Now write 7x as 2x + 9x and then factor by grouping. Copyright © 2011 Pearson Education, Inc.

Example 8 Factor using substitution. Solution
If we substitute another variable, like u, for y – 5, we can see that the polynomial is actually in the form ax2 + bx + c. Substitute u for y – 5. We can factor by grouping. Copyright © 2011 Pearson Education, Inc.

continued Remember that u was substituted for y – 5, so we must substitute y – 5 back for u to have the factored form of our original polynomial. Substitute y – 5 for u. This resulting factored form can be simplified. Distribute. Combine like terms. Copyright © 2011 Pearson Education, Inc.

Factoring Special Products and Factoring Strategies
6.3 Factoring Special Products and Factoring Strategies 1. Factor perfect square trinomials. 2. Factor a difference of squares. 3. Factor a difference of cubes. 4. Factor a sum of cubes. 5. Use various strategies to factor polynomials. Copyright © 2011 Pearson Education, Inc.

Example 1a Factor. 9a2 + 6a + 1 Solution This trinomial is a perfect square because it is in the form a2 + 2ab + b2, where a = 3a and b = 1. 9a2 + 6a + 1 a2 = (3a)2 = 9a2. b2 = 12 = 1 2ab = (2)(3a)(1) = 6a 9a2 + 6a + 1 = (3a + 1)2 Copyright © 2011 Pearson Education, Inc.

Example 1b Factor. 16x2 – 56x + 49 Solution This trinomial is a perfect square. 16x2 – 56x + 49 = (4x – 7)2 Use a2 – 2ab + b2 = (a – b)2, where a = 4x and b = 7. Copyright © 2011 Pearson Education, Inc.

Example 2a Factor. 9x2 – 16y2 Solution This binomial is a difference of squares because 9x2 – 16y2 = (3x)2 – (4y)2 . To factor it, we use the rule a2 – b2 = (a + b)(a – b). a2 – b2 = (a + b)(a – b) 9x2 – 16y2 = (3x)2 – (4y)2 = (3x + 4y)(3x – 4y) Copyright © 2011 Pearson Education, Inc.

Example 2c Factor. Solution The terms in this binomial have a monomial GCF, 2a7. Factor out the GCF. Factor 36a2 – 49, using a2 – b2 = (a + b)(a – b) with a = 6a and b = 7. Copyright © 2011 Pearson Education, Inc.

Example 2d Factor. n4 – 625 Solution
This binomial is a difference of squares, where a = n2 and b = 25. n4 – 625 = (n2 + 25)(n2 – 25) Use a2 – b2 = (a + b)(a – b). = (n2 + 25)(n + 5)(n – 5) Factor n2 – 25, using a2 – b2 = (a + b)(a – b) with a = n and b = 5. Be sure you have factored completely. Warning: A sum of two squares is prime and cannot be factored. Copyright © 2011 Pearson Education, Inc.

Factoring a Difference of Cubes a3 – b3 = (a – b)(a2 + ab + b2) Warning: The trinomial a2 + ab + b2 is not a perfect square and cannot be factored. Remember a perfect square trinomial has the form a2 + 2ab + b2 . Copyright © 2011 Pearson Education, Inc.

Example 3a Factor. 8x3 – 729 Solution
This binomial is a difference of cubes. a3 – b3 = (a – b) (a a b b2) 8x3 – 729 = (2x)3 – (9)3 = (2x – 9)((2x)2 + (2x)(9) + (9)2) = (2x – 9)(4x2 + 18x + 81) Note: The trinomial may seem like a perfect square. However, to be a perfect square, the middle term should be 2ab. In this trinomial, we only have ab, so it cannot be factored. Copyright © 2011 Pearson Education, Inc.

Example 3b Factor. 216x3 – 64 Solution This binomial is a difference of cubes. This binomial has a GCF, 8. 216x3 – 64 = 8(27x3 – 8) a3 – b3 = (a – b) (a a b b2) 8(27x3 – 8) = 8(3x)3 – (2)3 = 8(3x – 2)[(3x)2 + (3x)(2) + (2)2] = 8(3x – 2)(9x2 + 6x + 4) Copyright © 2011 Pearson Education, Inc.

Example 4a Factor. 125x3 + 64 Solution
This binomial is a sum of cubes. a3 + b3 = (a + b) (a2  a b b2) 125x = (5x)3 + (4)3 = (5x + 4)((5x)2  (5x)(4) + (4)2) = (5x + 4)(25x2  20x + 16) Copyright © 2011 Pearson Education, Inc.

Example 4b Factor. 6x +162xy3 Solution
The terms in this binomial have a monomial GCF, 6x. 6x +162xy3 = 6x(1 + 27y3) Factor out the monomial GCF, 6x. = 6x(1 + 3y)[(1)2 – (1)(3y) + (3y)2] = 6x(1 + 3y)(1 – 3y + 9y2) Simplify. Copyright © 2011 Pearson Education, Inc.

Factoring a Polynomial To factor a polynomial, first factor out any monomial GCF and then consider the number of terms in the polynomial. I. Four terms: Try to factor by grouping. II. Three terms: Determine if the trinomial is a perfect square. A. If the trinomial is not a perfect square, consider its form. 1. If it is in the form x2 + bx + c, find two factors of c whose sum is b, and write the factored form as (x + first number)(x + second number). 2. If it is in the form ax2 + bx + c, where a  1, use trial and error. Or find two factors of ac whose sum is b; write these factors as coefficients of two like terms that, when combined, equal bx; and factor by grouping. Copyright © 2011 Pearson Education, Inc.

Factoring a Polynomial continued III. Two terms: Determine if the binomial is a difference of squares, sum of cubes, or difference of cubes. A. If it is a difference of squares, a2 – b2, the factors are conjugates and the factored form is (a + b)(a – b). Note that a sum of squares cannot be factored. B. If it is a difference of cubes, a3 – b3, the factored form is (a – b)(a2 + ab + b2). C. If it is a sum of cubes, a3 + b3, the factored form is (a + b)(a2 – ab + b2). Note: Always check to see if any of the factors can be factored further. Copyright © 2011 Pearson Education, Inc.

Example 5d Factor completely. 12x2 – 8x – 15 Solution There is no GCF. It is in the form ax2 + bx + c, where a 1, so we use trial and error. After trying various combinations, we find the following correct combination = (6x + 5)(2x – 3 ) Copyright © 2011 Pearson Education, Inc.

Example 5f Factor completely. Solution There is no monomial GCF. Determine whether it is a difference of squares or the sum or difference of cubes. It is the difference of squares. Copyright © 2011 Pearson Education, Inc.

Example 6d Factor. x5 – 2x3 – 27x2 + 54 Solution No common monomial, factor by grouping. (x5 – 2x3) + (– 27x2 + 54) x3(x2 – 2) – 27(x2 – 2) (x2 – 2)(x3 – 27) Difference of cubes (x2 – 2)(x – 3)(x2 + 3x + 9) Copyright © 2011 Pearson Education, Inc.

Solving Equations by Factoring
6.4 Solving Equations by Factoring 1. Use the zero-factor theorem to solve equations by factoring. 2. Solve problems involving quadratic equations. 3. Find the intercepts of quadratic and cubic functions. Copyright © 2011 Pearson Education, Inc.

Example Solve. (x + 4)(x + 5) = 0 Solution Since we have a product of two factors, x + 4 and x + 5, equal to zero, one or both factors must equal 0. x + 4 = or x + 5 = 0 Solve each equation. x =  x = 5 Check Verify that 4 and 5 satisfy the original equation, (x + 4)(x + 5) = 0. For x = 4: For x = 5: (4 + 4)(4 + 5) = 0 (5 + 4)(5 + 5) = 0 0(1) = (1)(0) = 0 Copyright © 2011 Pearson Education, Inc.

Example 2 Solve. 2x2 – 5x = 3 Solution First, we need the equation in standard form. 2x2 – 5x – 3 = 0 (2x + 1)(x – 3) = 0 Factor. 2x + 1 = or x – 3 = 0 Use the zero-factor theorem to solve. The solutions are and To check, we verify that the solutions satisfy the original equations. Copyright © 2011 Pearson Education, Inc.

Example 3 Solve. x3 – 4x2 = 21x Solution x3 – 4x2 – 21x = 0
Write the equation in standard form. x(x2 – 4x – 21) = 0 x(x – 7 )(x + 3) = 0 x = 0 or x – 7 = 0 or x + 3 = 0 The solutions are 0, 7, and 3. Copyright © 2011 Pearson Education, Inc.

Cubic equation in one variable: An equation that can be written in the form ax3 + bx2 + cx + d = 0, where a, b, c, and d, are all real numbers and a  0. Solving Polynomial Equations Using Factoring To solve a polynomial equation using factoring, 1. Write the equation in standard form (Set one side equal to 0 with the other side in descending order of degree when possible). 2. Write the polynomial in factored form. 3. Use the zero-factor theorem to solve. Copyright © 2011 Pearson Education, Inc.

Example 4d Solve 14x2 + 9x + 2 = 10x + 6. Solution x2 + 9x + 2 = 10x + 6 14x2 + 9x  10x + 2  6 = 0 14x2  x  4 = 0 (7x  4)(2x + 1) = 0 7x  4 = 0 or 2x + 1 = 0 7x = 4 or x = 1 x = 4/7 or x = 1/2 The solutions are 4/7 and 1/2. Copyright © 2011 Pearson Education, Inc.

Example 5 The equation h(t) = 16t2 + vot + h0 describes the height, h, in feet of an object t seconds after being thrown upwards with an initial velocity of v0 feet per second from an initial height of h0 feet. Suppose a cannonball is fired from a platform that is 384 feet above the ground . The initial velocity of the object is 160 feet per second. How long will it take for the cannonball to land on the ground? Understand We are given a formula, the initial velocity of 160 feet per second, and the initial height of 384 feet. We are to find the time it takes the cannonball to reach the ground, which is at 0 feet. Copyright © 2011 Pearson Education, Inc.

continued Plan Replace the variables in the formula with given values for h0 and v0, then solve for t. Execute 0 = –16t t + 384 = –16(t2 – 10t – 24) = –16(t – 12)(t + 2) t – 12 = or t + 2 = 0 t = 12 or t = 2 Answer Our answer must describe the amount of time, so only the positive value, 12 makes sense. The cannonball takes seconds to reach the ground. Check h = 16(12)2  160(12)  384 = 2304 – 1920 – 384 = 0 Copyright © 2011 Pearson Education, Inc.

The Pythagorean Theorem Given a right triangle, where a and b represent the lengths of the legs and c represents the length of the hypotenuse, then a2 + b2 = c2. c (hypotenuse) b (leg) a (leg) Copyright © 2011 Pearson Education, Inc.

Example 6 The figure shows a garden to be created. Find the length, in feet, of each side. x + 4 x + 2 x Understand We are given expressions for the lengths of the three sides of a right triangle and we are to find those lengths. Pan and Execute Use the Pythagorean theorem. (x)2 + (x + 2)2 = (x + 4)2 x2 + x2 + 4x + 4 = x2 + 8x + 16 x2 – 4x – 12 = 0 x + 2 = 0 or x – 6 = 0 (x + 2)(x – 6) = 0 x = –2 or x = 6 Copyright © 2011 Pearson Education, Inc.

continued Answer Because x describes a length in feet, only the positive solution is sensible so x must be 6 feet This means x + 2 is 8 feet and x + 4 is 10 feet. Check (6)2 + (8)2 = (10)2 = 100 100 = 100 Copyright © 2011 Pearson Education, Inc.

Example 7a Find the x-intercept(s) and then sketch the graph. f(x) = x2 – 8x + 12 Solution To find the x-intercepts, we let f(x) = 0 and then solve for x. x2 – 8x + 12 = 0 (x – 6)(x – 2) = 0 x – 6 = 0 or x – 2 = 0 x = 6 or x = 2 The x-intercepts are (6,0) and (2,0). Copyright © 2011 Pearson Education, Inc.

continued To sketch the graph, we will also find the
y-intercepts by finding f(0). f(0) = = 12 The y-intercept is (0,12). Finding a few more points is helpful. x y 1 5 3 3 4 4 7 Copyright © 2011 Pearson Education, Inc.

Example 7b Find the x-intercept(s) and then sketch the graph. f(x) = (x + 3)(x – 2)(x – 1) Solution To find the x-intercepts, we let f(x) = 0 and then solve for x. x + 3 = 0 or x – 2 = 0 or x – 1 = 0 x = 1 x = –3 x = 2 The x-intercepts are (3, 0), (2, 0) and (1, 0). Copyright © 2011 Pearson Education, Inc.

continued To sketch the graph, we will also find the
y-intercepts by finding f(0). f(x) = x3 – 7x + 6 f(0) = 03 – 7(0) + 6 = 6 The y-intercept is (0,6). Finding a few more points is helpful. x y 2 12 1 3 Copyright © 2011 Pearson Education, Inc.

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