Calculator Function – How to take the cube root of a number To take the cube root of a number, press MATH, then select option 4. Example: What is 3 13824 24 ? The Greatest Common Factor Factors (either numbers or polynomials) When an integer is written as a product of integers, each of the integers in the product is a factor of the original number. When a polynomial is written as a product of polynomials, each of the polynomials in the product is a factor of the original polynomial. Factoring – writing a polynomial as a product of polynomials. Greatest common factor – largest quantity that is a factor of all the integers or polynomials involved. Finding the GCF of a List of Integers or Terms 1) Prime factor the numbers. 2) Identify common prime factors. 3) Take the product of all common prime factors. • If there are no common prime factors, GCF is 1. Example Find the GCF of each list of numbers. 1) 12 and 8 2) 7 and 20 Example Find the GCF of each list of numbers. 1) 12 and 8 12 = 2 · 2 · 3 8=2·2·2 So the GCF is 2 · 2 = 4. 2) 7 and 20 7=1·7 20 = 2 · 2 · 5 There are no common prime factors so the GCF is 1. Example Find the GCF of each list of numbers. 1) 6, 8 and 46 2) 144, 256 and 300 . Example Find the GCF of each list of numbers. 1) 6, 8 and 46 6=2·3 8=2·2·2 46 = 2 · 23 So the GCF is 2. 2) 144, 256 and 300 144 = 2 · 2 · 2 · 3 · 3 256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 300 = 2 · 2 · 3 · 5 · 5 So the GCF is 2 · 2 = 4. Example Find the GCF of each list of terms. 1) x3 and x7 2) 6x5 and 4x3 Example Find the GCF of each list of terms. 1) x3 and x7 x3 = x · x · x x7 = x · x · x · x · x · x · x So the GCF is x · x · x = x3 2) 6x5 and 4x3 6x5 = 2 · 3 · x · x · x 4x3 = 2 · 2 · x · x · x So the GCF is 2 · x · x · x = 2x3 Example Find the GCF of the following list of terms. a3b2, a2b5 and a4b7 Example Find the GCF of the following list of terms. a3b2, a2b5 and a4b7 a3b2 = a · a · a · b · b a2b5 = a · a · b · b · b · b · b a4b7 = a · a · a · a · b · b · b · b · b · b · b So the GCF is a · a · b · b = a2b2 Notice that the GCF of terms containing variables will use the smallest exponent found amongst the individual terms for each variable. The first step in factoring a polynomial is to find the GCF of all its terms. Then we write the polynomial as a product by factoring out the GCF from all the terms. The remaining factors in each term will form a polynomial. Example Factor out the GCF in each of the following polynomials. 1) 6x3 – 9x2 + 12x = 2) 14x3y + 7x2y – 7xy = Example Factor out the GCF in each of the following polynomials. 1) 6x3 – 9x2 + 12x = 3 · x · 2 · x2 – 3 · x · 3 · x + 3 · x · 4 = 3x(2x2 – 3x + 4) 2) 14x3y + 7x2y – 7xy = 7 · x · y · 2 · x2 + 7 · x · y · x – 7 · x · y · 1 = 7xy(2x2 + x – 1) Example Factor out the GCF in each of the following polynomials. 3) 6(x + 2) – y(x + 2) = 4) xy(y + 1) – (y + 1) = Example Factor out the GCF in each of the following polynomials. 3) 6(x + 2) – y(x + 2) = 6 · (x + 2) – y · (x + 2) = (x + 2)(6 – y) 4) xy(y + 1) – (y + 1) = xy · (y + 1) – 1 · (y + 1) = (y + 1)(xy – 1) Remember that factoring out the GCF from the terms of a polynomial should always be the first step in factoring a polynomial. This will usually be followed by additional steps in the process. Example Factor 90 + 15y2 – 18x – 3xy2. 90 + 15y2 – 18x – 3xy2 = 3(30 + 5y2 – 6x – xy2) = 3(5 · 6 + 5 · y2 – 6 · x – x · y2) = 3(5(6 + y2) – x (6 + y2)) = 3(6 + y2)(5 – x) Example 90 + 15y2 – 18x – 3xy2 = 3(30 + 5y2 – 6x – xy2) = 3(5 · 6 + 5 · y2 – 6 · x – x · y2) = 3(5(6 + y2) – x (6 + y2)) = 3(6 + y2)(5 – x) Factoring Trinomials of the 2 Form x + bx + c Recall by using the FOIL method that F O I L (x + 2)(x + 4) = x2 + 4x + 2x + 8 = x2 + 6x + 8 To factor x2 + bx + c into (x + one #)(x + another #), note that b is the sum of the two numbers and c is the product of the two numbers. So we’ll be looking for 2 numbers whose product is c and whose sum is b. Note: there are fewer choices for the product, so that’s why we start there first. Example Factor the polynomial x2 + 13x + 30. Since our two numbers must have a product of 30 and a sum of 13, the two numbers must both be positive. Positive factors of 30 Sum of Factors 1, 30 31 2, 15 17 3, 10 13 Note, there are other factors, but once we find a pair that works, we do not have to continue searching. So x2 + 13x + 30 = (x + 3)(x + 10). Example Factor the polynomial x2 – 11x + 24. Since our two numbers must have a product of 24 and a sum of -11, the two numbers must both be negative. Negative factors of 24 Sum of Factors – 1, – 24 – 25 – 2, – 12 – 14 – 3, – 8 So x2 – 11x + 24 = (x – 3)(x – 8). – 11 Prime Polynomials Example Factor the polynomial x2 – 6x + 10. Since our two numbers must have a product of 10 and a sum of – 6, the two numbers will have to both be negative. Negative factors of 10 Sum of Factors – 1, – 10 – 11 – 2, – 5 –7 Since there is not a factor pair whose sum is – 6, x2 – 6x +10 is not factorable and we call it a prime polynomial. When polynomials contain four terms, it is sometimes easier to group like terms in order to factor. Your goal is to create a common factor. You can also move terms around in the polynomial to create a common factor. Practice makes you better in recognizing common factors. » FACTOR: 3xy - 21y + 5x – 35 Factor the first two terms: 3xy - 21y = 3y (x – 7) Factor the last two terms: + 5x - 35 = 5 (x – 7) The green parentheses are the same so it’s the common factor Now you have a common factor (x - 7) (3y + 5) » FACTOR: 6mx – 4m + 3rx – 2r Factor the first two terms: 6mx – 4m = 2m (3x - 2) Factor the last two terms: + 3rx – 2r = r (3x - 2) The green parentheses are the same so it’s the common factor Now you have a common factor (3x - 2) (2m + r) » FACTOR: 15x – 3xy + 4y –20 Factor the first two terms: 15x – 3xy = 3x (5 – y) Factor the last two terms: + 4y –20 = 4 (y – 5) The green parentheses are opposites so change the sign on the 4 ; - 4 (-y + 5) or – 4 (5 - y) Now you have a common factor (5 – y) (3x – 4) When trinomials have a degree of “2”, they are known as quadratics. We learned earlier to use the “diamond” to factor trinomials that had a “1” in front of the squared term. x2 + 12x + 35 (x + 7)(x + 5) When there is a coefficient larger than “1” in front of the squared term, we can use a modified diamond or square to find the factors. Always remember to look for a GCF before you do ANY other factoring. Let’s try this example 3x2 + 13x + 4 Make a box Write the factors of the first term. Write the factors of the last term. Multiply on the diagonal and add to see if you get the middle term of the trinomial. If so, you’re done! When factoring using a difference of squares, look for the following three things: ˃ only 2 terms ˃ minus sign between them ˃ both terms must be perfect squares If all 3 of the above are true, write two ( ), one with a + sign and one with a – sign : ( + ) ( - ). » 1. » 2. » 3. » 4. » 5. » 6. a2 – 16 x2 – 25 4y2 – 16 9y2 – 25 3r2 – 81 2a2 + 16 When factoring using perfect square trinomials, look for the following three things: ˃ 3 terms ˃ last term must be positive ˃ first and last terms must be perfect squares If all three of the above are true, write one ( )2 using the sign of the middle term. » 1. » 2. » 3. » 4. » 5. » 6. a2 – 8a + 16 x2 + 10x + 25 4y2 + 16y + 16 9y2 + 30y + 25 3r2 – 18r + 27 2a2 + 8a - 8 Factoring Trinomials of the 2 Form ax + bx + c Returning to the FOIL method, F O I L (3x + 2)(x + 4) = 3x2 + 12x + 2x + 8 = 3x2 + 14x + 8 To factor ax2 + bx + c into (#1·x + #2)(#3·x + #4), note that a is the product of the two first coefficients, c is the product of the two last coefficients and b is the sum of the products of the outside coefficients and inside coefficients. Note that b is the sum of 2 products, not just 2 numbers, as in the last section. Example Factor the polynomial 25x2 + 20x + 4. Possible factors of 25x2 are {x, 25x} or {5x, 5x}. Possible factors of 4 are {1, 4} or {2, 2}. We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors. Keep in mind that, because some of our pairs are not identical factors, we may have to exchange some pairs of factors and make 2 attempts before we can definitely decide a particular pair of factors will not work. Example Factor the polynomial 25x2 + 20x + 4. We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to 20x. Factors of Factors of 25x2 4 Resulting Binomials Product of Outside Terms Product of Inside Terms Sum of Products {x, 25x} {1, 4} (x + 1)(25x + 4) 4x 25x 29x (x + 4)(25x + 1) x 100x 101x {x, 25x} {2, 2} (x + 2)(25x + 2) 2x 50x 52x {5x, 5x} {2, 2} (5x + 2)(5x + 2) 10x 10x 20x Example Continued Check the resulting factorization using the FOIL method. F O I L (5x + 2)(5x + 2) = 5x(5x) + 5x(2) + 2(5x) + 2(2) = 25x2 + 10x + 10x + 4 = 25x2 + 20x + 4 So our final answer when asked to factor 25x2 + 20x + 4 will be (5x + 2)(5x + 2) or (5x + 2)2. Example Factor the polynomial 21x2 – 41x + 10. Since the middle term is negative, possible factors of 10 must both be negative: {-1, -10} or {-2, -5}. We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors. Example Factor the polynomial 21x2 – 41x + 10. Factors Factors Resulting Product of Product of Sum of of 21x2 of 10 Binomials Outside Terms Inside Terms Products {3x, 7x}{1, 10}(3x – 1)(7x – 10) 30x 7x 37x (3x – 10)(7x – 1) 3x 70x 73x {3x, 7x} {2, 5} (3x – 2)(7x – 5) 15x 14x 29x (3x – 5)(7x – 2) 6x 35x 41x Example Continued Check the resulting factorization using the FOIL method. F (3x – 5)(7x – 2) = O I L 3x(7x) + 3x(-2) - 5(7x) - 5(-2) = 21x2 – 6x – 35x + 10 = 21x2 – 41x + 10 So our final answer when asked to factor 21x2 – 41x + 10 will be (3x – 5)(7x – 2). Now that we’ve learned all the types of factoring, we need to remember to use them all. Whenever it says to factor, you must break down the expression into the smallest possible factors. Let’s review all the ways to factor. 1. Look for GCF first. 2. Count the number of terms: a) 4 terms – factor by grouping b) 3 terms – look for perfect square trinomial – if not, try diamond or box c) 2 terms – look for difference of squares If any ( ) still has an exponent of 2 or more, see if you can factor again. We know that an equation must be solved for the unknown. Up to now, most of what we have only solved were equations with a degree of 1. 2x + 8 = 4x +6 -2x + 8 = 6 -2x = -2 x=1 If an equation has a degree of 2 or higher, we cannot solve it until it has been factored. You must first get “0” on one side of the = sign before you try any factoring. Once you have “0” on one side, use all your rules for factoring to make 2 ( ) or factors. Next, set each factor = 0 and solve for the unknown. x2 + 12x = 0 Factor GCF x(x + 12) = 0 (set each factor = 0, & solve) x=0 x + 12 = 0 x = - 12 You now have 2 answers, x = 0 and x = -12. Q. Factor the polynomial x 3 – 400. If the polynomial cannot be factored, write prime. Answer: The first term is a perfect cube, but the second term is not. It is a prime polynomial. As with some quadratic equations, factoring a polynomial equation is one way to find its real roots. Recall the Zero Product Property. You can find the roots, or solutions, of the polynomial equation P(x) = 0 by setting each factor equal to 0 and solving for x. Every polynomial function with degree n greater than or equal to 1 has exactly n complex zeros, including multiplicities The polynomial 3x5 + 18x4 + 27x3 = 0 has two multiple roots, 0 and –3. The root 0 is a factor three times because 3x3 = 0. The multiplicity of root r is the number of times that x – r is a factor of P(x). When a real root has even multiplicity, the graph of y = P(x) touches the x-axis but does not cross it. When a real root has odd multiplicity greater than 1, the graph “bends” as it crosses the x-axis. You cannot always determine the multiplicity of a root from a graph. It is easiest to determine multiplicity when the polynomial is in factored form. Solve the polynomial equation by factoring. 4x6 + 4x5 – 24x4 = 0 4x6 + 4x5 – 24x4 = 0 Factor out the GCF, 4x4. 4x4(x2 + x – 6) = 0 Factor the quadratic. 4x4(x + 3)(x – 2) = 0 Set each factor equal to 0. 4x4 = 0 or (x + 3) = 0 or (x – 2) = 0 Solve for x. x = 0, x = –3, x = 2 The roots are 0, –3, and 2. Check Use a graph. The roots appear to be located at x = 0, x = –3, and x = 2. Solve the polynomial equation by factoring. x4 + 25 = 26x2 x4 + 25 = 26x2 Set the equation equal to 0. x4 – 26 x2 + 25 = 0 Factor the trinomial in quadratic form. (x2 – 25)(x2 – 1) = 0 Factor the difference of two squares. (x – 5)(x + 5)(x – 1)(x + 1) Solve for x. x – 5 = 0, x + 5 = 0, x – 1 = 0, or x + 1 =0 x = 5, x = –5, x = 1 or x = –1 The roots are 5, –5, 1, and –1. Identify the roots of each equation. State the multiplicity of each root. x3 + 6x2 + 12x + 8 = 0 Since this is difficult to factor, use a graph. A calculator graph shows a bend near (–2, 0). The root –2 has a multiplicity of 3. Therefore x + 2 is a factor three times. Check to see if that is true! (x + 2)(x + 2)(x + 2)= x3 + 6x2 + 12x + 8 Identify the roots of each equation. State the multiplicity of each root. x4 + 8x3 + 18x2 – 27 = 0 x4 + 8x3 + 18x2 – 27 = 0 A calculator graph shows a bend near (–3, 0) and crosses at (1, 0). The root 1 has a multiplicity of 1. The root –3 has a multiplicity of 3. Therefore (x – 1) is a factor once, and (x + 3) is a factor three times. The root 1 has a multiplicity of 1. The root –3 has a multiplicity of 3. Therefore (x – 1) is a factor once, and (x + 3) is a factor three times. (x – 1)(x + 3)(x + 3)(x + 3)=x4 + 8x3 + 18x2 – 27 Location Principle If P is a polynomial function and P(x1) and P(x2) have opposite signs, then there is a real number r between x1 and x2 that is a zero of P, that is, P(r) = 0 P( x) x 4 3x 3 5x 2 13x 6 Lesson Quiz Identify the roots of each equation. State the multiplicity of each root. 1. 5x4 – 20x3 + 20x2 = 0 2. x3 – 12x2 + 48x – 64 = 0 3. x3 + 9 = x2 + 9x Lesson Quiz Identify the roots of each equation. State the multiplicity of each root. 1. 5x4 – 20x3 + 20x2 = 0 0 and 2 each with multiplicity 2 2. x3 – 12x2 + 48x – 64 = 0 4 with multiplicity 3 3. x3 + 9 = x2 + 9x –3, 3, 1

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