course mth 158 I am in the process of selling our home and coaching my daughter's volleyball team and feeling way behind in this class. I know you said the dates are flexible, but how behind is too far behind? Volleyball season ends next Saturday and I think that I will have much more free time after that. If I am finding that I cannot catch up, I suppose I will withdraw, as I don't want it to adversely affect my grade. What do you think? assignment #006006. `query 6
......!!!!!!!!...................................
13:46:27 R.5.22 (was R.6.18). What do you get when you factor 36 x^2 - 9 and how did you get your result?
......!!!!!!!!...................................
RESPONSE --> 36 x^2 - 9 since 6^2=36 and 3^2=9 = (6x-3)(6x+3)
.................................................
......!!!!!!!!...................................
13:53:32 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?
......!!!!!!!!...................................
RESPONSE --> x^2 + 10 x + 1 is already a prime factor, as the factors of one are only 1 & 1 -- that sum cannot equal 10. However, for problem number 32: 25x^2 + 10 x + 1 = (5x + 1)^2 Because A+C = 25 and the factors 5,5 equal to the sum of 10 -- this is a perfect square For problem 28: x^2 + 10 x + 25 = (x+5)^2
.................................................
......!!!!!!!!...................................
13:53:56 R.5.34 (was R.6.30). What do you get when you factor x^3 + 125 and how did you get your result?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
13:58:11 ** 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). **
......!!!!!!!!...................................
RESPONSE --> accidentally hit next question -- is there no way to go back? For problem #34 -- I got the correct answer: x^3+5^3 = (x+5)(x^2-5x+25)
.................................................
......!!!!!!!!...................................
14:00:09 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?
......!!!!!!!!...................................
RESPONSE --> x^2 - 17 x + 16 = (x-16)(x-1) The factors of 16 for this problem are -16 & -1. When adding them together, you would get -17
.................................................
......!!!!!!!!...................................
14:00:36 ** 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 we might have 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. The only possibility that gives us a + b = -17 is a = -1, b = -16. So we conclude that x^2 - 17 x + 16 = (x-16)(x-1). **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
14:03:24 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?
......!!!!!!!!...................................
RESPONSE --> (3 x^2 - 3 x) + (2 x - 2) = 3x(x-1) + 2(x-1) = (3x+2)(x-1) To get my answer, I combined each side into a group and factored out the 3x on the left side and the 2 on the right side.
.................................................
......!!!!!!!!...................................
14:03:46 ** 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). **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
14:06:49 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?
......!!!!!!!!...................................
RESPONSE --> 3 x^2 - 10 x + 8 = 3x^2 -6x - 4x + 8 = (3x^2 -6x) - (4x + 8) = (3x^2 -6x) + (4x - 8) = 3x(x-2) + 4(x-2) = (3x+4)(x-2)
.................................................
......!!!!!!!!...................................
14:07:14 ** 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). **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
14:08:21 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?
......!!!!!!!!...................................
RESPONSE --> 14 + 6 x - x^2 is already a prime factor
.................................................
......!!!!!!!!...................................
14:08:46 ** This expression factors, but not into binomtials 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 won't find the factors in the usual manner. The quadratic formula tells us that there are factors ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and ( -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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................