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# Factoring

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This was a project for math, but I hope others might find it useful f they need help with factoring.

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### Factoring

1. 1. Factoring By Hayley McMillon
2. 2. Greatest Common Factor In this section you will learn how to extract the greatest common factor or GCF from an expression. There are two great ways to find the GCF of an expression one method is to create a list of allthe possible factors and the other is a factor tree. Both of these methods will show you whetherthe coefficients of the expression all have a factor in common and if they have multiple factors in common, which one is the biggest. After using one of these methods to find the common factor of the coefficients all you have to do is divide each coefficient by the factor and then place thequotients inside a set of parentheses, and the factor outside the parentheses. Another cool thing about extracting GCF’s is that it applies to variables such as x,y, and z. To extract the variable all terms within the expression must have the specified variable whether it be x, y, z, etc. Then you look at the exponents of the variables and choose the smallest exponent in the expression and subtract that from each of the variables and place it outside of the parentheses. After this you will have completely extracted the GCF’s out of the entire expression. * Examples of how to extract the greatest common factor will be seen on the next pages
3. 3. Greatest Common Factor Example 1 Example of A List *The GCF is highlighted 25x^3 + 50x^2 + 5x + 1525x^3/5 + 50x^2/5 + 5x/5 + 15/3 25 50 5 15 5(5x^3 + 10x^2 + x + 3) 5*5 10 * 5 1*5 3*5 Don’t Forget To Check Your Work 5(5x^3 + 10x^2 + x + 3) 25 * 1 25 * 2 1 * 15 5 * 5x^3 + 5 * 10x^2 + 5 * 5x + 5 *3 50 * 1 25x^3 + 50x^2 + 5x + 15 Step 1 : Find the factors of each coefficient using a list. Step 2: If there is a GCF divide it out of each coefficient. Step 3: Then place the expression in a set of parentheses and the GCF outside of the expression. Step 4: Make sure to check your work by using the distributive property to multiply each term by 5, and if the answer matches the original expression the answer is correct.
4. 4. Greatest Common Factor Example 2 Factor Tree * GCF is highlighted 4x^5 + 8x^4 + 12x^34x^3/4x^5 + 8x^4/4x^3 + 12x^3/4x^3 4 8 12 4x^3(x^2 + 2x + 3) ⤩ 4 2 4 3 2 2 ⤩ ⤩ Don’t Forget Check Your Work 2 2 2 2 4x^3(x^2 + 2x + 3) 4x^3 * x^2 + 4x^3 * 2x + 4x^3 * 3 4x^5 + 8x^4 + 12x^3 Step 1 : Find the factors of each coefficient using a factor tree. Step 2: If there is a GCF divide it out of each coefficient. Then look at the variables of each constant and decide if all terms have the same variable which can be divided it out of the expression. Since “x” can be divided out of each term you must look for the smallest exponent attached to “x” and then use this degree of x to divide out the rest of the variables within the expression. Step 3: Then place the expression in a set of parentheses and the GCF 4x^3 outside of the expression. Step 4: Make sure to check your work by using the distributive property to multiply each term by 4x^3, and if the answer matches the original expression the answer is correct.
5. 5. Special Products In this section you will learn about....Difference of Perfect Squares.Perfect Square Trinomials
6. 6. Difference of Perfect Squares In this section we will learn about the difference of perfect squares. To be a difference of perfect squares binomial a perfect square term must be subtracted from another perfect square term. An example of this is 25a^2 - 36. The formula for the difference of perfect squares binomial is a^2 - b^2 and the fully factored form is (a + b)(a - b). To get the formula from a^2-b^2 to (a + b)(a - b), you have to determine the square roots of the terms and and then place them in two sets of parentheses which have differentsigns. This will look like (a + b)(a - b) and be the fully factored form of a^2 - b^2 . A great way to check the work ofyour fully factored form of the difference of perfect squares binomial is to use the FOIL method which stands for the First terms, Outside terms, Inside terms, and Last terms. To use the Foil method you would first multiply the First two terms together for example you would multiply 5a, and 5a in (5a + 6)(5a - 6) to get 25a^2 from our earlierexample 25a^2 - 36. Next you would multiply the Outside terms 5a and -6 in (5a + 6)(5a - 6)to get -30a. Then you would multiply the Inside terms 6 and 5a to get 30a from (5a + 6)(5a - 6). Last but certainly not least you would multiply the Last terms 6 and -6 from (5a + 6)(5a - 6) to get -36. Next you would gather the products from each step to make the expression 25a^2 - 30a + 30a - 36 then you would simplify the expression and would be left with 25a^2 - 36. Using the FOIL method allows us to easily multiply without making things to messy whether it be with difference of perfect squares binomials or any other type of factoring. * Examples can be seen on the next page.
7. 7. Difference of Perfect Squares Example Step 1: To factor 9a^2 - 64^2 completely you should first look for Factor 9a^2 - 64b^2 the GCF of the expression. As none of the constants or variables of the expression share a GCF we can skip the step of extracting 9a^2 - 64b^2 the GCF. (3a + 8b)(3a - 8b) Step 2: As this is a difference of perfect squares problem we can assume that the constants are squares of a another number. The square roots of the constants and variables are 3a and 8b. Don’t Forget To Check Your Work! Step 3 : Now that we know the square roots of each term we can (3a+8b)(3a-8b) place them in two sets of parentheses with (3a + 8b) in one and First Terms: (3a + 8b)(3a-8b) = 3a * 3a = 9a^2 (3a-b) in the other.Outside Terms: (3a + 8b)(3a - 8b) = 3a * -8b = -24ab Step 4: Once all these steps are done you will have the fully Inside Terms: (3a + 8b)(3a - 8b) = 8b * 3a = 24ab factored form of 9a^2 - 64b^2 which is (3a + 8b)(3a - 8b). Last Terms: (3a + 8b)(3a - 8b) = 8b * -8b = 64b^2 Step 5: To check your work all you have to do is use the FOIL 9a^2 - 24ab + 24ab - 64b^2 method explained earlier to multiply the factored expression and 9a^2 - 64b^2 then simplify the new expression. Once this is done you will be left with the original form of 9a^2 - 64b^2 and will assure you that your answer is correct.
8. 8. Perfect Square Trinomials In this section we will learn about how to factor the perfect square trinomial. To be classified as a perfectsquare trinomial it must be a polynomial that has three terms where the first and least terms are perfectsquares and the middle term is twice the product of the square roots of those terms. The formula for theperfect square trinomial is a^2 + 2ab + b^2 = (a + b)^2 or a^2 -2ab + b^2 = (a - b)^2. * Examples with steps can be seen on the next page
9. 9. Perfect Square Trinomial Example 9a^2 + 36ab + 36b^2 Step 1: To factor this perfect square trinomial you should first find the GCF which is 9 and then divide 9a^2/ 9 + 36ab/ 9 + 36b^2/ 9 it out out of each term in the expression. 9(a^2 + 4ab + 4b^2) Step 2: Then you should take the three remaining 9(a + 2b)(a + 2b) terms in the expression and place it within two sets 9(a + 2b)^2 of parentheses with the 9 still outside both sets of parentheses and the other two terms separated as 2b in each set of parentheses, which will look like Don’t Forget To Check Your Work! 9(a + 2b)^2 9(a + 2b)(a + 2b), and is the fully factored form of 9(a + 2b)(a + 2b) 9a^2 + 36ab + 36b^2. First Terms: 9(a + 2b)(a + 2b) = a * a = a^2 Step 3: To check your work use the FOIL method toOutside Terms: 9(a + 2b)(a + 2b) = a * 2b = 2ab multiply the expression out and then simplify the Inside Terms: 9(a + 2b)(a + 2b) = 2b * a = 2ab expression created after using the FOIL method.Last Terms: 9(a + 2b)(a + 2b) = 2b * 2b = 4b^2 Next distribute the 9 outside of the expression to 9(a^2 + 2ab + 2ab + 4b^2) each of the terms within the parentheses and if your 9(a^2 + 4ab + 4b^2) answer comes out as the original polynomial 9a^2 + a^2 * 9 + 4ab * 9 + 4b^2 * 9 36ab + 36b^2 then your answer is correct. 9a^2 + 36ab + 36b^2
10. 10. Factoring Trinomials In this section we will learn about how to factor trinomials. A trinomial is a polynomial that has three terms. Factoring trinomials is similar to factoring all other polynomials except that after finding the GCFyou must find the binomial factors (will be explained in greater detail below) that will go into the last spots in the 2 sets of parentheses: (x ?)(x ?), after doing this you should have the fully factored form of the trinomial and all you have to do is check your work. Binomial FactorsTo find the binomial factors that will satisfy the last spots in the two sets of parentheses you have to take the expression (x^2 + 3 - 10) (just used as an example) and look at the middle and last term along with their signs and set up a list that has on one side the factors whose product will equal -10, and on theother side the sum of those same factors. The correct binomial factors will be the two that not only equal -10 but will also add up to positive 3. In this case the binomial factors will be 5 and -2 because when multiplied they equal -10 and when added amount to a positive three. You will then take 5 and -2 and place them in the last two spots in the two sets of parentheses so that it will look like (x + 5)(x - 2) when finished, and will be the fully factored form of the trinomial.
11. 11. Factoring Trinomials * binomial factors are highlighted Example Don’t Forget To Check Your Work! Factors of - 10 Sum of Factors 3(x + 5)(x - 2) 3x^2 + 9x - 30 First Terms: 3(x + 5)(x - 2) = x * x = x^2 5 * -2 5 + -2 = 3 Outside Terms: 3(x + 5)(x - 2) = x * -2 = -2x3x^2/ 3 + 9x/ 3 - 30/ 3 -5 * 2 -5 + 2 = -3 Inside Terms: 3(x + 5)(x - 2) = 5 * x = 5x 3(x^2 + 3x - 10) Last Terms: 3(x + 5)(x - 2) = 5 * -2 = -10 3(x + 5)(x - 2) 3(x^2 - 2x + 5x - 10) -10 * 1 -10 + 1 = -9 3(x^2 + 3x - 10) 3x^2 + 9x - 30 10 * -1 10 + -1 = 9 Step 1: First find the GCF and divide it out. In this case the GCF is 3. After dividing out the GCF you should be left with 3(x^2 + 3x - 10). Step 2: First set up two sets of parentheses as 3(x )(x ), and then find the binomial factors explained earlier to find the last term within each set of parentheses. Remember that the factors must be equal to to the sum of the middle coefficient. In this example the factors that satisfy the requirement stated before are 5 and -2. Step 3: Next place these numbers into their respective set of parentheses which when done should look like this 3(x + 5)(x - 2). Step 4: Then once all the steps are completed you should check your work by using the FOIL method, then simplify the expression created after using the FOIl method, and then distribute the GCF of 3 to all the terms of the expression and you should be left with the original trinomial of 3x^2 + 9x - 10.
12. 12. Factoring By Grouping In this section you will learn about ....Factoring trinomials by grouping4 - term Polynomials
13. 13. Factoring Trinomials By Grouping In this section we will learn about how to factor a trinomial by grouping. To factor a trinomial by grouping you must first find the GCF if one exists and extract it. Thenyou take the first term and last term’s coefficients and multiply them together to get the product. Then you take this product and create a list that has the factors of the product and on the other side a list of the sum of all those factors. Next you take the factors that add up to the middleterm and multiply to the product of the first and last term. Next you do what is called splitting the middle term and replace the middle term with the factors of the product earlier. After doing this you split the 4 term polynomial into two groups one that has the first to terms and the second group with the last two terms so that an expression like a^2 + 4a - 2a - 8 becomes grouped as a^2 + 4a - 2a + 8. Then you take the first group and extract the GCF which in this case is a, and in the second group is 2. after this is done you should be left with a(a + 4) - 2(a + 4). Then you should see that there is a common binomial which is a + 4. You, then take one of these common binomials and put a and -2 in another set of parentheses to create the fully factored form as (a + 4)(a - 2) of the original trinomial a^2 + 2a - 8.
14. 14. Factoring Trinomials By Grouping Example Don’t Forget To Check Your Work! Factors of 6 Sum of Factors (x + 1)(2x + 3) (x + 1)(2x + 3) = 2x^2 2*3 2+3=5 2X^2 + 5x+ 3 (x + 1)(2x + 3) = 3x (2x^2 + 5x + 3) -3 * -2 -3 + -2 = -5 (x + 1)(2x + 3) = 2x 2*3=6 (x + 1)(2x + 3) = 3 1*6 1+6=7 2x^2 + 2x + 3x + 3 2x^2 + 3x + 2x + 3 2x^2 + 5x + 3 -1 * -6 -1 + -6 = -7 (2x^2 + 2x + 3x + 3) (2(x)(x) + 2(x) + 3x + 3) [2x(x + 1) + 3x + 3] Step 1: First look to see if there is a GCF. As there is not we can skip this step. Step 2: Next take the first and last terms coefficients and multiply them together to get 6. After doing this set up a table or [(2x(x + 1) + 3x + 3] list and on one side have the factors of six and on the other side the sum of these same factors. Step 3: Next look to see which factors multiply to a positive 6 and add up to positive 5. In this example the factors that [(2x(x + 1) + 3(x + 1)] satisfy both requirements are 2 and 3. Step 4: Next split the middle term by replacing 5x with 2x and 3x, so the expression will look like 2x^2 + 2x + 3x + 3. Common binomial = (x + 1) Step 5: Then split the expression into two groups the first with the first two terms 2x^2 + 2x, and then the second group* Remember 2x and 3 make up the other with the last two terms 3x + 3. Next look at the first group and take out the variable, x and coefficient, 2 they share in set of parentheses common so that it will look like 2x(x + 1), repeat these same steps for the second group which should look like 3(x + 1) (x + 1)(2x + 3) when finished. Step 6: Next you should see that there is a common binomial in [2x(x + 1) + 3(x + 1)] which is (x + 1). Next you can take this common binomial and then create a second set of parentheses to house 2x, and 3; once this is all done the fully factored form of 2x^2 + 5x + 3 is (x + 1)(2x + 3). Step 7: Don’t forget to check your work by using the foil method.
15. 15. 4-Term Polynomials In this section we will learn about factoring 4 - term polynomials through grouping. Factoring four term polynomials are very similar to factoring trinomials through grouping except in this case you get to skip the step of splitting the middle term into two terms because a 4- termpolynomial already has 4 terms like its name implies. Other than that the process is the same as in factoring trinomials through grouping. * An example can be seen on the next page.
16. 16. 4-Term Polynomials Example 2x^3 - 4x^2 + 8x - 162x^3/ 2 - 4x^2/ 2 + 8x/ 2 - 16/ 2 2(x^3 - 2x^2 + 4x - 8) 2(x^3 - 2x^2 + 4x - 8) 2(x^3 - 2x^2) + (4x - 8) Step 1: Factor out the GCF which is 2 in this example. 2[x^2(x - 2) + 4(x - 2)] Step 2: Group the terms by putting the first two terms in one group and the last two terms in the second group. Step 3: Then factor out the GCF from each group which in this case Common Binomial: (x - 2) is x^2 for the first group and 4 for the second group. 2(x - 2)(x^2 + 4) Step 4: Next look to see if their is a common binomial. The common binomial in this example is (x - 2). Then take the GCF’s factored outDon’t Forget To Check Your Work! of each group and place them in a set of parentheses together to get 2(x - 2)(x^2 + 4) (x^2 + 4) in one set of parentheses and (x - 2) in the other set of 2(x - 2)(x^2 + 4) = x^3 parentheses which will leave the fully factored form of 2x^3 - 4x^2 + 2(x - 2)(x^2 + 4) = 4x 8x - 16 as 2(x - 2)(x^2 + 4). 2(x - 2)(x^2 + 4) = -2x^2 Step 5: Don’t forget to check your work by using the FOIL method 2(x - 2)(x^2 + 4) = -8 and then distribute the two to get the original polynomial 2x^3 - 4x^2 + 2(x^3 + 4x - 2x^2 - 8) 8x - 16. 2x^3 - 2x^2 + 4x - 8
17. 17. Sum and Difference of Perfect Cubes In this section you will learn about ....The sum and difference of cubes
18. 18. Sum and Difference of Perfect Cubes The sum and difference of perfect cubes can be summed up as a binomial whose termsare perfect cubes. A sum of perfect cubes looks like a^3 + b^3, and a difference of perfect cubes looks like a^3 - b^3. To factor a sum or difference of perfect cubes you must use the sum and difference cube patterns. These patterns are as follows: Sum of Perfect Cubes: a^3 + b^3 = (a + b)(a^2 - ab + b^2) Difference of Perfect Cubes: a^3 - b^3 = (a - b)(a^2 + ab + b^2) * Examples can be seen on the following pages.
19. 19. Sum of Perfect Cubes Example Find the cube root of each term x^3 = x * x * x 8x ^3 + 64 cube root = x 8(x^3 + 8) 8c^3 = 2 * 2 * 2 Step 1: Begin by extracting the GCF. The GCF is 8 in this a=x b= 2 cube root = 2 example. a^3 + b^3 = (a + b)(a^2 - ab + b^2) Step 2: Then take the first term x^3 and find the cube root.x^3 + 2^3 = (x + 2)(x^2 - x * 2 + 2^2) The cube root of x^3 is x. Then repeat this step for 8 the 8(x^3 + 8) = 8(x + 2)(x^2 - 2x + 4) second term. The cube root of 8 is 2. After doing this you will know that a in the sum of cubes pattern will equal x, and b in Don’t forget to check your work! the sum of cubes pattern will equal 2 8(x + 2)(x^2 - 2x + 4) Step 3: Next set up the sum of cubes pattern a^3 + b^3 = (a 8(x + 2)(x^2 - 2x + 4) = x ^3 + b)(a^2 - ab + b^2) and plug in the values of a and b, and 8(x + 2)(x^2 - 2x + 4) = -2x^2 then simplify the expressions. Don’t forget to put the GCF of 8(x + 2)(x^2 - 2x + 4) = 4x 8 in front of (x + 2)(x^2 - 2x + 4). This will be the fully 8(x + 2)(x^2 - 2x + 4) = 2x^2 factored form of 8x^3 + 64. 8(x + 2)(x^2 -2x + 4) = -4x Step 4: After doing all this you should check your work by 8(x + 2)(x^2 - 2x + 4) = 8 multiplying each term in (x + 2) by each term in (x^2 - 2x + 4), 8(x^3 - 2x^2 + 4x + 2x^2 - 4x + 8) and then simplify. Then don’t forget to distribute the 8 to all 8(x^3 + 8) terms in the expression to be sure that the answer of 8(x + 2) x^3 * 8 + 8 * 8 (x^2 - 2x + 4) is correct. 8x^3 + 64
20. 20. Difference of Perfect Cubes Example Find the cube root of each term 8x^3 = 2x * 2x * 2x cube root = 2x 8x^3 - 27d^3 a = 2x b = 3d 27d^3 = 3d * 3d * 3d a^3 - b^3 = (a - b)(a^2 + ab + b^2) cube root = 3d(2x)^3 - (3d)^3 = (2x - 3d)((2x)^2 + 2x * 3d + (3d)^2) (2x)^3 - (3d)^3 = (2x - 3d)(4x^2 + 6dx + 9d^2) Don’t Forget To Check Your Work! Step 1: Since there is no GCF we can skip this step. (2x - 3d)(4x^2 + 6dx + 9d^2) Step 2: Next find the cube root of each term. In this example the cube (2x - 3d)(4x^2 + 6dx + 9d^2) = 8x^3 root of 8x^3 is 2x, and the cube root of 27d^3 is 3d. In the difference of (2x - 3d)(4x^2 + 6dx + 9d^2) = 12dx^2 cubes pattern “a” will equal 2x and “b” will equal 3d. (2x - 3d)(4x^2 + 6dx + 9d^2) = 18d^2x Step 3: Next substitute the “a” and “b” values into a^3 -b^3 = (a - b)(a^2 (2x - 3d)(4x^2 + 6dx + 9d^2) = -12dx^2 + ab + b^2). After doing this simplify what can be simplified. What you (2x - 3d)(4x^2 + 6dx + 9d^2) = -18d^2x are left with is the fully factored form of 8x^3 - 27d^3. (2x - 3d)(4x^2 + 6dx + 9d^2) = -27d^3 Step 4: Don’t forget to check your work by multiplying each term in (2x -8x^3 + 12dx^2 - 12dx^2 + 18d^2x - 18d^2 - 27d^3 3d) by every term in (4x^2 + 6dx + 9d^2) To make sure that the fully 8x^3 -27d^3 factored form 8x^3 - 27 d^3 is indeed (2x - 3d) (4x^2 + 6dx + 9d^2).
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