course Mth 158 9-24 4:11 p.m. 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.
.............................................
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): There were no discrepancies between the two solutions. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: R.5.32 \ 28 (was R.6.24) What do you get when you factor x^2 + 10 x + 1 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X^2 + 10x + 1 is a prime number because there are no two numbers that when multiplied by themselves give you a 1 and a sum of 10 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Well, I see a lot of possibilities here that I would have never thought to do, and still am unsure if I could come to that conclusion on my own. This is an area that I struggle to understand (even with my tutor). ------------------------------------------------ Self-critique Rating: 0 ********************************************* 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: x^3 + 125 is the sum of 2 cubes x*x*x + 5*5*5 The formula is x^3 + a^3 = (x+a) ( x^2 - ax + a^2). So you would write it x^3 + 5^3 = (x+5) (x^2 –5x + 25) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I had to use the book and see the formula for this, but if I had used the formula substituting a b instead of x a, it would have been a little easier to read. The outcome was the same, but the given solution looks much easier to solve. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: x^2 – 17x + 16 First you need to find the integers whose product is 16 Possibilities: 2*8, -2*-8, 4*4, -4*-4, 1*16, -1*-16 And the sum of the two = -17 so x^2 – 17x + 16 = (x-1) (x-16) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I went by the example in the book that showed us how to find the integer whose product is x and sum is the coefficient of the middle term. Even though I can follow the book on some of the problems, I still have trouble trying to solve them without a book. Is this an indication that I am going to have problems taking a test? I am really worried about working these problems without something to guide me.
.............................................
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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After I look at the given solution, I still don’t really understand it.
.............................................
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: 14 + 6x – x^2 = (x +2) (-x +7) confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. ** "