#$&* course Mth 158 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * ** 36x^2-9 is the difference of two squares. We write this as • (6x)^2-3^2 then get • (6x-3)(6x+3), using the special formula difference of two squares. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: R.5.32 \ 28 (was R.6.24) What do you get when you factor 25 x^2 + 10 x + 1 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This problem cannot be factored because there are no factors that would give a product of one and a sum of ten. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** STUDENT SOLUTION: x^2+10x+1 is prime because there are no integers whose product is 10 and sum is 1 INSTRUCTOR COMMENTS: The sum should be 10 and the product 1. I agree that there are no two integers with this property. Furthermore there are no two rational numbers with this property. So you would never be able to find the factors by inspection. However that doesn't mean that there aren't two irrational numbers with the property. For example 10 and 1/10 come close to the criteria, with product 1 and sum 10.1. The quadratic formula tells you in fact that the two numbers are • ( -10 + sqrt( 10^4 - 4 * 1 * 1) ) / (2 * 1) and • ( -10 - sqrt( 10^4 - 4 * 1 * 1) ) / (2 * 1) . Since 10^2 - 4 = 96 is positive, these are real numbers, both irrational. So the polynomial isn't prime. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I made the mistake of assuming that there was no way for this to be factored but it seems that using the quadratic formula would be advantageous to be sure that there are no rational or irrational numbers that would work. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: R.5.34 (was R.6.30). What do you get when you factor x^3 + 125 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this problem we use the format for the sum of cubes which is (a+b)(a^2-ab+b^2) so this becomes (x+5)(x^2-5x+25). So x^3+125 factored is (x+5)(x^2-5x+25) confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: x^3+125 is the sum of two cubes, with 125 = 5^3. We know that a^3 + b^3 = (a+b) ( a^2 - 2 a b + b^2). So we write • x^3+5^3 = (x+5)(x^2-5x+25). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: R.5.46 (was R.6.42). What do you get when you factor x^2 - 17 x + 16 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this problem I split it into two terms with an x and whose numbers would give a sum of -17 and a product of 16. So I began with (x____)(x____) then I knew that no factors of 16 would give me -17 except for -1 and -16 so then it became (x-1)(x-16) which when multiplied back gives the original term of x^2-17x+16. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** x^2-17x+16 is of the form (x + a) ( x + b) = x^2 + (a + b) x + ab, with a+b = -17 and ab = 16. If ab = 16 then, if a and b happen to be integers, we have the following possibilities: • a = 1, b = 16, or • a = 2, b = 8, or • a = -2, b = -8, or • a = 4, b = 4, or • a = -1, b = -16, or • a = -4, b = -4. These are the only possible integer factors of 16. In order to get a + b = -17 we must have at least one negative factor. So the possibilities are reduced to • a = -2, b = -8, or • a = -1, b = -16, or • a = -4, b = -4. The only of the these possibilities that gives us a + b = -17 is a = -1, b = -16. So we conclude that • x^2 - 17 x + 16 = (x-16)(x-1). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: R.5.52 (was R.6.48). What do you get when you factor 3 x^2 - 3 x + 2 x - 2 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To factor this we first need to combine like terms to get a trinomial. So we combine -3x and 2x and this gives 3x^2-1x+2. Now we factor into two terms that contain x and numbers whose sum gives -1x and product give -2. These terms are (3x+2)(x-1). confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** This expression can be factored by grouping: 3x^2-3x+2x-2 = (3x^2-3x)+(2x-2) = 3x(x-1)+2(x-1) = (3x+2)(x-1). ** ADDITIONAL EXPLANATION: To see that (3x^2-3x)+(2x-2) = 3x(x-1)+2(x-1) apply the distributive law to each term in the second expression: 3x ( x - 1) = 3 x^2 - 3x, and 2 ( x - 1) = 2x - 2. To see that 3x(x-1)+2(x-1) = (3x+2)(x-1) apply the distributive law as follows: (3x + 2) ( x - 1) = 3x * (x - 1) + 2 * (x - 1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: R.5.64 (was R.6.60). What do you get when you factor 3 x^2 - 10 x + 8 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Here again we are looking for two terms that will give the original trinomial. So when we break it down we know that 3x^2 can only be broken down into 3x and x giving (3x____)(x_____) now we need to find the factors of 8 that will give us a sum of -10. These numbers are -2 and -4 so the correct factorization would be (3x-4) (x-2). confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Possibilities are • (3x - 8) ( x - 1), • (3x - 1) ( x - 8), • (3x - 2) ( x - 4), • (3x - 4) ( x - 2). The possibility that gives us 3 x^2 - 10 x + 8 is (3x - 4) ( x - 2). ** R.5.82 (was R.6.78). What do you get when you factor 14 + 6 x - x^2 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: After trying to use the method applied previously I have come to the conclusion that this cannot factor. All of the possibilities will not give the original trinomial. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** This expression factors, but not into binomials with integer coefficients. We could list all the possibilities: • (x + 7) ( -x + 2), • (x + 2) ( -x + 7), • (x + 14) ( -x + 1), • (x + 1)(-x + 14) but none of these will give us the desired result. For future reference: You often cannot find the factors by factoring in the usual manner; however it is always possible to find the factors of a second-degree trinomial using the quadratic formula. The quadratic formula applied to this problem tells us that the factors are (x - z1) * (x - z2), were z1 and z2 are the solutions to the equation 14 + 6 x - x^2 0. Using the formula we find that • z1 = ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and • z2 = ( -6 - sqrt(6^2 - 4 * 14 * (-1) ) ) / (2 * -1) . Since sqrt(6^2 - 4 * 14 * (-1) ) = sqrt(36 + 56) = sqrt(92) is a real number these solutions are real numbers but again, as in a previous example, they aren't rational numbers and nobody could ever find them by inspection. This is not something you're expected to do at this point. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: R.5.64 (was R.6.60). What do you get when you factor 3 x^2 - 10 x + 8 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Here again we are looking for two terms that will give the original trinomial. So when we break it down we know that 3x^2 can only be broken down into 3x and x giving (3x____)(x_____) now we need to find the factors of 8 that will give us a sum of -10. These numbers are -2 and -4 so the correct factorization would be (3x-4) (x-2). confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Possibilities are • (3x - 8) ( x - 1), • (3x - 1) ( x - 8), • (3x - 2) ( x - 4), • (3x - 4) ( x - 2). The possibility that gives us 3 x^2 - 10 x + 8 is (3x - 4) ( x - 2). ** R.5.82 (was R.6.78). What do you get when you factor 14 + 6 x - x^2 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: After trying to use the method applied previously I have come to the conclusion that this cannot factor. All of the possibilities will not give the original trinomial. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** This expression factors, but not into binomials with integer coefficients. We could list all the possibilities: • (x + 7) ( -x + 2), • (x + 2) ( -x + 7), • (x + 14) ( -x + 1), • (x + 1)(-x + 14) but none of these will give us the desired result. For future reference: You often cannot find the factors by factoring in the usual manner; however it is always possible to find the factors of a second-degree trinomial using the quadratic formula. The quadratic formula applied to this problem tells us that the factors are (x - z1) * (x - z2), were z1 and z2 are the solutions to the equation 14 + 6 x - x^2 0. Using the formula we find that • z1 = ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and • z2 = ( -6 - sqrt(6^2 - 4 * 14 * (-1) ) ) / (2 * -1) . Since sqrt(6^2 - 4 * 14 * (-1) ) = sqrt(36 + 56) = sqrt(92) is a real number these solutions are real numbers but again, as in a previous example, they aren't rational numbers and nobody could ever find them by inspection. This is not something you're expected to do at this point. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: R.5.64 (was R.6.60). What do you get when you factor 3 x^2 - 10 x + 8 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Here again we are looking for two terms that will give the original trinomial. So when we break it down we know that 3x^2 can only be broken down into 3x and x giving (3x____)(x_____) now we need to find the factors of 8 that will give us a sum of -10. These numbers are -2 and -4 so the correct factorization would be (3x-4) (x-2). confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Possibilities are • (3x - 8) ( x - 1), • (3x - 1) ( x - 8), • (3x - 2) ( x - 4), • (3x - 4) ( x - 2). The possibility that gives us 3 x^2 - 10 x + 8 is (3x - 4) ( x - 2). ** R.5.82 (was R.6.78). What do you get when you factor 14 + 6 x - x^2 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: After trying to use the method applied previously I have come to the conclusion that this cannot factor. All of the possibilities will not give the original trinomial. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** This expression factors, but not into binomials with integer coefficients. We could list all the possibilities: • (x + 7) ( -x + 2), • (x + 2) ( -x + 7), • (x + 14) ( -x + 1), • (x + 1)(-x + 14) but none of these will give us the desired result. For future reference: You often cannot find the factors by factoring in the usual manner; however it is always possible to find the factors of a second-degree trinomial using the quadratic formula. The quadratic formula applied to this problem tells us that the factors are (x - z1) * (x - z2), were z1 and z2 are the solutions to the equation 14 + 6 x - x^2 0. Using the formula we find that • z1 = ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and • z2 = ( -6 - sqrt(6^2 - 4 * 14 * (-1) ) ) / (2 * -1) . Since sqrt(6^2 - 4 * 14 * (-1) ) = sqrt(36 + 56) = sqrt(92) is a real number these solutions are real numbers but again, as in a previous example, they aren't rational numbers and nobody could ever find them by inspection. This is not something you're expected to do at this point. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!