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# 5 3 factoring trinomial ii

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• 1. Factoring Trinomials II
• 2. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c.
• 3. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods.
• 4. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable.
• 5. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers.
• 6. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method
• 7. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method For this method, we need to find four numbers that fit certain descriptions.
• 8. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished.
• 9. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished. Example A. Let {1, 3} and {1, 2} be two pairs of numbers. Is it possible to split the {1, 2 }, put them in the boxes that makes the equality true?
• 10. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished. Example A. Let {1, 3} and {1, 2} be two pairs of numbers. Is it possible to split the {1, 2 }, put them in the boxes that makes the equality true? a. 1* (&#xB1; ) + 3*(&#xB1; ) = 5.
• 11. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished. Example A. Let {1, 3} and {1, 2} be two pairs of numbers. Is it possible to split the {1, 2 }, put them in the boxes that makes the equality true? a. 1* (&#xB1; ) + 3*(&#xB1; ) = 5. Yes, 1* (2) + 3 * (1) = 5
• 12. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished. Example A. Let {1, 3} and {1, 2} be two pairs of numbers. Is it possible to split the {1, 2 }, put them in the boxes that makes the equality true? a. 1* (&#xB1; ) + 3*(&#xB1; ) = 5. Yes, 1* (2) + 3 * (1) = 5 b. 1* (&#xB1; ) + 3* (&#xB1; ) = &#x2013;5.
• 13. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished. Example A. Let {1, 3} and {1, 2} be two pairs of numbers. Is it possible to split the {1, 2 }, put them in the boxes that makes the equality true? a. 1* (&#xB1; ) + 3*(&#xB1; ) = 5. Yes, 1* (2) + 3 * (1) = 5 b. 1* (&#xB1; ) + 3* (&#xB1; ) = &#x2013;5. Yes, 1* (1) + 3* (&#x2013;2) = &#x2013;5
• 14. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished. Example A. Let {1, 3} and {1, 2} be two pairs of numbers. Is it possible to split the {1, 2 }, put them in the boxes that makes the equality true? a. 1* (&#xB1; ) + 3*(&#xB1; ) = 5. Yes, 1* (2) + 3 * (1) = 5 b. 1* (&#xB1; ) + 3* (&#xB1; ) = &#x2013;5. Yes, 1* (1) + 3* (&#x2013;2) = &#x2013;5 or 1* (&#x2013;2) + 3* (&#x2013;1) = &#x2013;5
• 15. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished. Example A. Let {1, 3} and {1, 2} be two pairs of numbers. Is it possible to split the {1, 2 }, put them in the boxes that makes the equality true? a. 1* (&#xB1; ) + 3*(&#xB1; ) = 5. Yes, 1* (2) + 3 * (1) = 5 b. 1* (&#xB1; ) + 3* (&#xB1; ) = &#x2013;5. Yes, 1* (1) + 3* (&#x2013;2) = &#x2013;5 or 1* (&#x2013;2) + 3* (&#x2013;1) = &#x2013;5 c. 1* (&#xB1; ) + 3* (&#xB1; ) = 8.
• 16. Factoring Trinomials II Now let&#x2019;s try to factor trinomials of the form ax2 + bx + c. We&#x2019;ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished. Example A. Let {1, 3} and {1, 2} be two pairs of numbers. Is it possible to split the {1, 2 }, put them in the boxes that makes the equality true? a. 1* (&#xB1; ) + 3*(&#xB1; ) = 5. Yes, 1* (2) + 3 * (1) = 5 b. 1* (&#xB1; ) + 3* (&#xB1; ) = &#x2013;5. Yes, 1* (1) + 3* (&#x2013;2) = &#x2013;5 or 1* (&#x2013;2) + 3* (&#x2013;1) = &#x2013;5 c. 1* (&#xB1; ) + 3* (&#xB1; ) = 8. No, since the most we can obtain is 1* (1) + 3* (2) = 7.
• 17. Factoring Trinomials II (Reversed FOIL Method)
• 18. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring.
• 19. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2.
• 20. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2. The only way to get 3x2 is (3x &#xB1; #)(1x &#xB1; #).
• 21. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2. The only way to get 3x2 is (3x &#xB1; #)(1x &#xB1; #). The #&#x2019;s must be 1 and 2
• 22. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2. The only way to get 3x2 is (3x &#xB1; #)(1x &#xB1; #). The #&#x2019;s must be 1 and 2 to get the constant term +2.
• 23. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2. The only way to get 3x2 is (3x &#xB1; #)(1x &#xB1; #). The #&#x2019;s must be 1 and 2 to get the constant term +2. We need to place 1 and 2 as the #'s so the product will yield the correct middle term +5x.
• 24. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2. The only way to get 3x2 is (3x &#xB1; #)(1x &#xB1; #). The #&#x2019;s must be 1 and 2 to get the constant term +2. We need to place 1 and 2 as the #'s so the product will yield the correct middle term +5x. That is, (3x &#xB1; #)(1x &#xB1; #) must yields +5x,
• 25. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2. The only way to get 3x2 is (3x &#xB1; #)(1x &#xB1; #). The #&#x2019;s must be 1 and 2 to get the constant term +2. We need to place 1 and 2 as the #'s so the product will yield the correct middle term +5x. That is, (3x &#xB1; #)(1x &#xB1; #) must yields +5x, or that 3(&#xB1; # ) +1(&#xB1; #) = 5 where the #&#x2019;s are 1 and 2.
• 26. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2. The only way to get 3x2 is (3x &#xB1; #)(1x &#xB1; #). The #&#x2019;s must be 1 and 2 to get the constant term +2. We need to place 1 and 2 as the #'s so the product will yield the correct middle term +5x. That is, (3x &#xB1; #)(1x &#xB1; #) must yields +5x, or that 3(&#xB1; # ) +1(&#xB1; #) = 5 where the #&#x2019;s are 1 and 2. Since 3(1) +1(2) = 5,
• 27. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2. The only way to get 3x2 is (3x &#xB1; #)(1x &#xB1; #). The #&#x2019;s must be 1 and 2 to get the constant term +2. We need to place 1 and 2 as the #'s so the product will yield the correct middle term +5x. That is, (3x &#xB1; #)(1x &#xB1; #) must yields +5x, or that 3(&#xB1; # ) +1(&#xB1; #) = 5 where the #&#x2019;s are 1 and 2. Since 3(1) +1(2) = 5,
• 28. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2. The only way to get 3x2 is (3x &#xB1; #)(1x &#xB1; #). The #&#x2019;s must be 1 and 2 to get the constant term +2. We need to place 1 and 2 as the #'s so the product will yield the correct middle term +5x. That is, (3x &#xB1; #)(1x &#xB1; #) must yields +5x, or that 3(&#xB1; # ) +1(&#xB1; #) = 5 where the #&#x2019;s are 1 and 2. Since 3(1) +1(2) = 5,
• 29. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2. The only way to get 3x2 is (3x &#xB1; #)(1x &#xB1; #). The #&#x2019;s must be 1 and 2 to get the constant term +2. We need to place 1 and 2 as the #'s so the product will yield the correct middle term +5x. That is, (3x &#xB1; #)(1x &#xB1; #) must yields +5x, or that 3(&#xB1; # ) +1(&#xB1; #) = 5 where the #&#x2019;s are 1 and 2. Since 3(1) +1(2) = 5, we see that 3x2 + 5x + 2 = (3x + 2)(1x + 1).
• 30. Factoring Trinomials II (Reversed FOIL Method) Let&#x2019;s see how the above examples are related to factoring. Example B. Factor 3x2 + 5x + 2. The only way to get 3x2 is (3x &#xB1; #)(1x &#xB1; #). The #&#x2019;s must be 1 and 2 to get the constant term +2. We need to place 1 and 2 as the #'s so the product will yield the correct middle term +5x. That is, (3x &#xB1; #)(1x &#xB1; #) must yields +5x, or that 3(&#xB1; # ) +1(&#xB1; #) = 5 where the #&#x2019;s are 1 and 2. Since 3(1) +1(2) = 5, we see that 3x2 + 5x + 2 = (3x + 2)(1x + 1). 5x
• 31. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2.
• 32. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #).
• 33. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s
• 34. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = &#x2013;7.
• 35. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = &#x2013;7. It's 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7.
• 36. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = &#x2013;7. It's 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. So 3x2 &#x2013; 7x + 2 = (3x &#x2013;1)(1x &#x2013; 2)
• 37. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = &#x2013;7. It's 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. So 3x2 &#x2013; 7x + 2 = (3x &#x2013;1)(1x &#x2013; 2) Example D. Factor 3x2 + 5x &#x2013; 2.
• 38. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = &#x2013;7. It's 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. So 3x2 &#x2013; 7x + 2 = (3x &#x2013;1)(1x &#x2013; 2) Example D. Factor 3x2 + 5x &#x2013; 2. We start with (3x &#xB1; #)(1x &#xB1; #).
• 39. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = &#x2013;7. It's 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. So 3x2 &#x2013; 7x + 2 = (3x &#x2013;1)(1x &#x2013; 2) Example D. Factor 3x2 + 5x &#x2013; 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s
• 40. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = &#x2013;7. It's 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. So 3x2 &#x2013; 7x + 2 = (3x &#x2013;1)(1x &#x2013; 2) Example D. Factor 3x2 + 5x &#x2013; 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = +5.
• 41. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = &#x2013;7. It's 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. So 3x2 &#x2013; 7x + 2 = (3x &#x2013;1)(1x &#x2013; 2) Example D. Factor 3x2 + 5x &#x2013; 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = +5. Since c is negative, they must have opposite signs .
• 42. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = &#x2013;7. It's 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. So 3x2 &#x2013; 7x + 2 = (3x &#x2013;1)(1x &#x2013; 2) Example D. Factor 3x2 + 5x &#x2013; 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = +5. Since c is negative, they must have opposite signs . It is 3(+2) + 1(&#x2013;1) = +5.
• 43. Factoring Trinomials II Example C. Factor 3x2 &#x2013; 7x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = &#x2013;7. It's 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. So 3x2 &#x2013; 7x + 2 = (3x &#x2013;1)(1x &#x2013; 2) Example D. Factor 3x2 + 5x &#x2013; 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1 and 2 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = +5. Since c is negative, they must have opposite signs . It is 3(+2) + 1(&#x2013;1) = +5. So 3x2 + 5x + 2 = (3x &#x2013;1)(1x + 2)
• 44. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2.
• 45. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #).
• 46. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8.
• 47. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8. This is impossible.
• 48. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8. This is impossible. Hence the expression is prime.
• 49. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8. This is impossible. Hence the expression is prime. If both the numbers a and c in ax2 + bx + c have many factors then there are many possibilities to check.
• 50. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8. This is impossible. Hence the expression is prime. If both the numbers a and c in ax2 + bx + c have many factors then there are many possibilities to check. Example F. Factor 3x2 + 11x &#x2013; 4.
• 51. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8. This is impossible. Hence the expression is prime. If both the numbers a and c in ax2 + bx + c have many factors then there are many possibilities to check. Example F. Factor 3x2 + 11x &#x2013; 4. We start with (3x &#xB1; #)(1x &#xB1; #).
• 52. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8. This is impossible. Hence the expression is prime. If both the numbers a and c in ax2 + bx + c have many factors then there are many possibilities to check. Example F. Factor 3x2 + 11x &#x2013; 4. We start with (3x &#xB1; #)(1x &#xB1; #). Since 4 = 2(2) = 1(4),
• 53. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8. This is impossible. Hence the expression is prime. If both the numbers a and c in ax2 + bx + c have many factors then there are many possibilities to check. Example F. Factor 3x2 + 11x &#x2013; 4. We start with (3x &#xB1; #)(1x &#xB1; #). Since 4 = 2(2) = 1(4), we need to fill in 2&amp;2 or 1&amp;4 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = +11.
• 54. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8. This is impossible. Hence the expression is prime. If both the numbers a and c in ax2 + bx + c have many factors then there are many possibilities to check. Example F. Factor 3x2 + 11x &#x2013; 4. We start with (3x &#xB1; #)(1x &#xB1; #). Since 4 = 2(2) = 1(4), we need to fill in 2&amp;2 or 1&amp;4 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = +11. It can't be 2&amp;2.
• 55. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8. This is impossible. Hence the expression is prime. If both the numbers a and c in ax2 + bx + c have many factors then there are many possibilities to check. Example F. Factor 3x2 + 11x &#x2013; 4. We start with (3x &#xB1; #)(1x &#xB1; #). Since 4 = 2(2) = 1(4), we need to fill in 2&amp;2 or 1&amp;4 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = +11. It can't be 2&amp;2. Try 1&amp;4,
• 56. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8. This is impossible. Hence the expression is prime. If both the numbers a and c in ax2 + bx + c have many factors then there are many possibilities to check. Example F. Factor 3x2 + 11x &#x2013; 4. We start with (3x &#xB1; #)(1x &#xB1; #). Since 4 = 2(2) = 1(4), we need to fill in 2&amp;2 or 1&amp;4 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = +11. It can't be 2&amp;2. Try 1&amp;4, it is 3(+4) + 1(&#x2013;1) = +11.
• 57. Factoring Trinomials II Example E. Factor 3x2 + 8x + 2. We start with (3x &#xB1; #)(1x &#xB1; #). We need to fill in 1&amp;2 so that 3(&#xB1; # ) + 1(&#xB1; # ) = +8. This is impossible. Hence the expression is prime. If both the numbers a and c in ax2 + bx + c have many factors then there are many possibilities to check. Example F. Factor 3x2 + 11x &#x2013; 4. We start with (3x &#xB1; #)(1x &#xB1; #). Since 4 = 2(2) = 1(4), we need to fill in 2&amp;2 or 1&amp;4 as #'s so that 3(&#xB1; # ) + 1(&#xB1; # ) = +11. It can't be 2&amp;2. Try 1&amp;4, it is 3(+4) + 1(&#x2013;1) = +11. So 3x2 + 11x &#x2013; 4 = (3x &#x2013; 1)(1x + 4).
• 58. Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c.
• 59. Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c. Example G. Factor 12x2 &#x2013; 5x &#x2013; 3.
• 60. Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c. Example G. Factor 12x2 &#x2013; 5x &#x2013; 3. Since 3 must be 3(1),
• 61. Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c. Example G. Factor 12x2 &#x2013; 5x &#x2013; 3. Since 3 must be 3(1), we need to find two numbers such that (#)(#) = 12
• 62. Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c. Example G. Factor 12x2 &#x2013; 5x &#x2013; 3. Since 3 must be 3(1), we need to find two numbers such that (#)(#) = 12 and that (&#xB1; #)(&#xB1; 3) + (&#xB1; #)(&#xB1;1) = &#x2013; 5.
• 63. Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c. Example G. Factor 12x2 &#x2013; 5x &#x2013; 3. Since 3 must be 3(1), we need to find two numbers such that (#)(#) = 12 and that (&#xB1; #)(&#xB1; 3) + (&#xB1; #)(&#xB1;1) = &#x2013; 5. 12 = 1(12) = 2(6) = 3(4)
• 64. Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c. Example G. Factor 12x2 &#x2013; 5x &#x2013; 3. Since 3 must be 3(1), we need to find two numbers such that (#)(#) = 12 and that (&#xB1; #)(&#xB1; 3) + (&#xB1; #)(&#xB1;1) = &#x2013; 5. 12 = 1(12) = 2(6) = 3(4) 1&amp;12 and 2&amp;6 can be quickly eliminated.
• 65. Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c. Example G. Factor 12x2 &#x2013; 5x &#x2013; 3. Since 3 must be 3(1), we need to find two numbers such that (#)(#) = 12 and that (&#xB1; #)(&#xB1; 3) + (&#xB1; #)(&#xB1;1) = &#x2013; 5. 12 = 1(12) = 2(6) = 3(4) 1&amp;12 and 2&amp;6 can be quickly eliminated. We get (3)(&#x2013;3) + (4)(+1) = &#x2013; 5.
• 66. Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c. Example G. Factor 12x2 &#x2013; 5x &#x2013; 3. Since 3 must be 3(1), we need to find two numbers such that (#)(#) = 12 and that (&#xB1; #)(&#xB1; 3) + (&#xB1; #)(&#xB1;1) = &#x2013; 5. 12 = 1(12) = 2(6) = 3(4) 1&amp;12 and 2&amp;6 can be quickly eliminated. We get (3)(&#x2013;3) + (4)(+1) = &#x2013; 5. So 12x2 &#x2013; 5x &#x2013; 3 = (3x + 1)(4x &#x2013; 3).
• 67. Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c. Example G. Factor 12x2 &#x2013; 5x &#x2013; 3. Since 3 must be 3(1), we need to find two numbers such that (#)(#) = 12 and that (&#xB1; #)(&#xB1; 3) + (&#xB1; #)(&#xB1;1) = &#x2013; 5. 12 = 1(12) = 2(6) = 3(4) 1&amp;12 and 2&amp;6 can be quickly eliminated. We get (3)(&#x2013;3) + (4)(+1) = &#x2013; 5. So 12x2 &#x2013; 5x &#x2013; 3 = (3x + 1)(4x &#x2013; 3). Remark: In the above method, finding (#)(&#xB1; #) + (#)( &#xB1; #) = b does not guarantee that the trinomial will factor.
• 68. Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c. Example G. Factor 12x2 &#x2013; 5x &#x2013; 3. Since 3 must be 3(1), we need to find two numbers such that (#)(#) = 12 and that (&#xB1; #)(&#xB1; 3) + (&#xB1; #)(&#xB1;1) = &#x2013; 5. 12 = 1(12) = 2(6) = 3(4) 1&amp;12 and 2&amp;6 can be quickly eliminated. We get (3)(&#x2013;3) + (4)(+1) = &#x2013; 5. So 12x2 &#x2013; 5x &#x2013; 3 = (3x + 1)(4x &#x2013; 3). Remark: In the above method, finding (#)(&#xB1; #) + (#)( &#xB1; #) = b does not guarantee that the trinomial will factor. We have to match the sign of c also.
• 69. Factoring Trinomials II Example H. Factor 3x2 &#x2013; 7x &#x2013; 2 .
• 70. Factoring Trinomials II Example H. Factor 3x2 &#x2013; 7x &#x2013; 2 . We start with (3x &#xB1; #)(1x &#xB1; #).
• 71. Factoring Trinomials II Example H. Factor 3x2 &#x2013; 7x &#x2013; 2 . We start with (3x &#xB1; #)(1x &#xB1; #). We find that: 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7.
• 72. Factoring Trinomials II Example H. Factor 3x2 &#x2013; 7x &#x2013; 2 . We start with (3x &#xB1; #)(1x &#xB1; #). We find that: 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. But this won't work since (&#x2013;2)(&#x2013;1) = 2 = c.
• 73. Factoring Trinomials II Example H. Factor 3x2 &#x2013; 7x &#x2013; 2 . We start with (3x &#xB1; #)(1x &#xB1; #). We find that: 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. But this won't work since (&#x2013;2)(&#x2013;1) = 2 = c. In fact this trinomial is prime.
• 74. Factoring Trinomials II Example H. Factor 3x2 &#x2013; 7x &#x2013; 2 . We start with (3x &#xB1; #)(1x &#xB1; #). We find that: 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. But this won't work since (&#x2013;2)(&#x2013;1) = 2 = c. In fact this trinomial is prime. There might be multiple matchings for (#)(&#xB1; #) + (#)( &#xB1; #) = b make sure you chose the correct one, if any.
• 75. Factoring Trinomials II Example H. Factor 3x2 &#x2013; 7x &#x2013; 2 . We start with (3x &#xB1; #)(1x &#xB1; #). We find that: 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. But this won't work since (&#x2013;2)(&#x2013;1) = 2 = c. In fact this trinomial is prime. There might be multiple matchings for (#)(&#xB1; #) + (#)( &#xB1; #) = b make sure you chose the correct one, if any. Example I: Factor 1x2 + 5x &#x2013; 6 .
• 76. Factoring Trinomials II Example H. Factor 3x2 &#x2013; 7x &#x2013; 2 . We start with (3x &#xB1; #)(1x &#xB1; #). We find that: 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. But this won't work since (&#x2013;2)(&#x2013;1) = 2 = c. In fact this trinomial is prime. There might be multiple matchings for (#)(&#xB1; #) + (#)( &#xB1; #) = b make sure you chose the correct one, if any. Example I: Factor 1x2 + 5x &#x2013; 6 . We have: 1(+3) + 1(+2) = +5
• 77. Factoring Trinomials II Example H. Factor 3x2 &#x2013; 7x &#x2013; 2 . We start with (3x &#xB1; #)(1x &#xB1; #). We find that: 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. But this won't work since (&#x2013;2)(&#x2013;1) = 2 = c. In fact this trinomial is prime. There might be multiple matchings for (#)(&#xB1; #) + (#)( &#xB1; #) = b make sure you chose the correct one, if any. Example I: Factor 1x2 + 5x &#x2013; 6 . We have: 1(+3) + 1(+2) = +5 1(+6) + 1(&#x2013;1) = +5
• 78. Factoring Trinomials II Example H. Factor 3x2 &#x2013; 7x &#x2013; 2 . We start with (3x &#xB1; #)(1x &#xB1; #). We find that: 3(&#x2013;2) + 1(&#x2013;1) = &#x2013;7. But this won't work since (&#x2013;2)(&#x2013;1) = 2 = c. In fact this trinomial is prime. There might be multiple matchings for (#)(&#xB1; #) + (#)( &#xB1; #) = b make sure you chose the correct one, if any. Example I: Factor 1x2 + 5x &#x2013; 6 . We have: 1(+3) + 1(+2) = +5 1(+6) + 1(&#x2013;1) = +5 The one that works is x2 + 5x &#x2013; 6 = (x + 6)(x &#x2013; 1).
• 79. Factoring Trinomials II Finally, before starting the reverse-FOIL procedure
• 80. Factoring Trinomials II Finally, before starting the reverse-FOIL procedure 1. make sure the terms are arranged in order.
• 81. Factoring Trinomials II Finally, before starting the reverse-FOIL procedure 1. make sure the terms are arranged in order. 2. if there is any common factor, pull out the GCF first.
• 82. Factoring Trinomials II Finally, before starting the reverse-FOIL procedure 1. make sure the terms are arranged in order. 2. if there is any common factor, pull out the GCF first. 3. make sure that x2 is positive, if not, factor out the negative sign first.
• 83. Factoring Trinomials II Finally, before starting the reverse-FOIL procedure 1. make sure the terms are arranged in order. 2. if there is any common factor, pull out the GCF first. 3. make sure that x2 is positive, if not, factor out the negative sign first. Example J. Factor &#x2013;x3 + 3x + 2x2
• 84. Factoring Trinomials II Finally, before starting the reverse-FOIL procedure 1. make sure the terms are arranged in order. 2. if there is any common factor, pull out the GCF first. 3. make sure that x2 is positive, if not, factor out the negative sign first. Example J. Factor &#x2013;x3 + 3x + 2x2 &#x2013;x3 + 3x + 2x2 Arrange the terms in order = &#x2013;x3 + 2x2 + 3x
• 85. Factoring Trinomials II Finally, before starting the reverse-FOIL procedure 1. make sure the terms are arranged in order. 2. if there is any common factor, pull out the GCF first. 3. make sure that x2 is positive, if not, factor out the negative sign first. Example J. Factor &#x2013;x3 + 3x + 2x2 &#x2013;x3 + 3x + 2x2 Arrange the terms in order = &#x2013;x3 + 2x2 + 3x Factor out the GCF = &#x2013; x(x2 &#x2013; 2x &#x2013; 3)
• 86. Factoring Trinomials II Finally, before starting the reverse-FOIL procedure 1. make sure the terms are arranged in order. 2. if there is any common factor, pull out the GCF first. 3. make sure that x2 is positive, if not, factor out the negative sign first. Example J. Factor &#x2013;x3 + 3x + 2x2 &#x2013;x3 + 3x + 2x2 Arrange the terms in order = &#x2013;x3 + 2x2 + 3x Factor out the GCF = &#x2013; x(x2 &#x2013; 2x &#x2013; 3) = &#x2013; x(x &#x2013; 3)(x + 1)
• 87. Factoring Trinomials II Finally, before starting the reverse-FOIL procedure 1. make sure the terms are arranged in order. 2. if there is any common factor, pull out the GCF first. 3. make sure that x2 is positive, if not, factor out the negative sign first. Example J. Factor &#x2013;x3 + 3x + 2x2 &#x2013;x3 + 3x + 2x2 Arrange the terms in order = &#x2013;x3 + 2x2 + 3x Factor out the GCF = &#x2013; x(x2 &#x2013; 2x &#x2013; 3) = &#x2013; x(x &#x2013; 3)(x + 1)
• 88. Factoring Trinomials II Ex. A. Factor the following trinomials. If it&#x2019;s prime, state so. 1. 3x2 &#x2013; x &#x2013; 2 2. 3x2 + x &#x2013; 2 3. 3x2 &#x2013; 2x &#x2013; 1 4. 3x2 + 2x &#x2013; 1 5. 2x2 &#x2013; 3x + 1 6. 2x2 + 3x &#x2013; 1 7. 2x2 + 3x &#x2013; 2 8. 2x2 &#x2013; 3x &#x2013; 2 9. 5x2 &#x2013; 3x &#x2013; 2 11. 3x2 + 5x + 2 12. 3x2 &#x2013; 5x &#x2013; 2 15. 6x2 + 5x &#x2013; 6 10. 5x2 + 9x &#x2013; 2 13. 3x2 &#x2013; 5x + 2 14. 6x2 &#x2013; 5x &#x2013; 6 16. 6x2 &#x2013; x &#x2013; 2 17. 6x2 &#x2013; 13x + 2 18. 6x2 &#x2013; 13x + 2 19. 6x2 + 7x + 2 20. 6x2 &#x2013; 7x + 2 21. 6x2 &#x2013; 13x + 6 22. 6x2 + 13x + 6 23. 6x2 &#x2013; 5x &#x2013; 4 24. 6x2 &#x2013; 13x + 8 25. 6x2 &#x2013; 13x &#x2013; 8 25. 4x2 &#x2013; 9 26. 4x2 &#x2013; 49 27. 25x2 &#x2013; 4 28. 4x2 + 9 29. 25x2 + 9 B. Factor. Factor out the GCF, the &#x201C;&#x2013;&#x201D;, and arrange the terms in order first. 30. &#x2013; 6x2 &#x2013; 5xy + 6y2 31. &#x2013; 3x2 + 2x3&#x2013; 2x 32. &#x2013;6x3 &#x2013; x2 + 2x 33. &#x2013;15x2 &#x2013; 25x2 &#x2013; 10x 34. 12x2y2 &#x2013;14x2y2 + 4xy2