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21:09:52 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?
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RESPONSE --> 36 x^2 - 9 difference of 2 squares pull the squares out &multiply together (6x-3)(6x+3)
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21:09:56 ** 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 squars. **......!!!!!!!!...................................
RESPONSE --> ok
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21:11:24 R.5.28 (was R.6.24 What do you get when you factor x^2 + 10 x + 1 and how did you get your result?
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RESPONSE --> x^2+10x+1 is prime because no real numbers have a product of 10 and sum of 1
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21:11:29 ** 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 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, 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. **......!!!!!!!!...................................
RESPONSE --> ok
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21:22:08 R.5.34 (was R.6.30). What do you get when you factor x^3 + 125 and how did you get your result?
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RESPONSE --> x^3 + 125=(x+5)(x^2-5x+25)
sum of 2 cubes x^3 + a^3=(x+a)(x^2-ax+a^2).................................................
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21:22:21 ** 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 --> ok
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21:28:58 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?
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RESPONSE --> x^2 - 17 x + 16=(x-16)(x-1)
To factor a second degreee polynomial x^2+Bx+C find integers whose product is C and whose sum is B. that is , if there are numbers a,b, where ab=C and a+b=B.................................................
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21:29:09 ** 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
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22:04:34 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?
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RESPONSE --> 3 x^2 - 3 x + 2 x - 2= 3x(x-1)+2(x-1)= (3x+2)(x-1) factoring trinomials,factor by grouping
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22:04:46 ** 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
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22:14:24 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?
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RESPONSE --> 3 x^2 - 10 x + 8= (3x^2-6x)+(-4x+8)= 3x(x-2)+-4(x-2)= (3x-4)(x-2)
factor by grouping.................................................
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22:14:35 ** 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
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22:25:11 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?
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RESPONSE --> 14 + 6 x - x^2= x^2- 6 x -14= is prime
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22:27:37 ** 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 --> this statement is true ....This is not something you're expected to do at this point. I do not understand
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zwÚYIj assignment #007 ؾ{]Uӂ͛ξ College Algebra 09-21-2005
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23:52:24 Query R.7.10 (was R.7.6). Show how you reduced (x^2 + 4 x + 4) / (x^4 - 16) to lowest terms.
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RESPONSE --> (x^2 + 4 x + 4) / (x^4 - 16)= (x+2)(x+2)/(x-2)(x+2)(x^2+4)= x+2/(x-2)(x^2+4) factor out and then cancel like terms
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23:52:39 ** We factor the denominator to get first (x^2-4)(x^2+4), then (x-2)(x+2)(x^2+4). The numerator factors as (x+2)^2. So the fraction is
(x+2)(x+2)/[(x-2)(x+2)(x^2+4)], which reduces to (x+2)/[(x-2)(x^2+4)]. **......!!!!!!!!...................................
RESPONSE --> ok
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00:15:46 Query R.7.28 (was R.7.24). Show how you simplified[ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ].
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RESPONSE --> ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ]= x-2/4x*12x/x^2-4x+4= x-2/4x*12x/(x-2)(x-2)= 3/x-2
change / to multiply by inverting the 2nd set of the equation factor out cancel out like terms.................................................
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00:15:52 ** [ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ] = (x-2) * / 4x * 12 x / (x^2 - 4x + 4) = (x-2) * 12 x / [ 4x ( x^2 - 4x + 4) ] = 12 x (x-2) / [4x ( x-2) ( x-2) ] = 3/(x - 2) **
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RESPONSE --> ok
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00:25:51 Query R.7.40 (was R.7.36). Show how you found and simplified the sum (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2).
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RESPONSE --> (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2)= 2x-5/(3x+2)+(x+4)*3x+2/1= 2x-5/x+4
invert last denominator with one since it was over 1 to equal itself and factor out then you cancel the like terms to solve.................................................
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00:29:08 ** We have two like terms so we write
(2x-5)/(3x+2) + (x+4)/(3x+2) = [(2x-5)+(x+4)]/(3x+2). Simplifying the numerator we have (3x-1)/(3x+2). **......!!!!!!!!...................................
RESPONSE --> i do see the addition i did not see that in the orig which is how i came up with 2x-5/x+4 i divided, i did not see how 3x+2 was not inverted though
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00:35:50 Query R.7.52 (was R.7.48). Show how you found and simplified the expression(x - 1) / x^3 + x / (x^2 + 1).
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RESPONSE --> (x - 1) / x^3 + x / (x^2 + 1)= x-1/x
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00:38:00 ** Starting with (x-1)/x^3 + x/(x^2+1) we multiply the first term by (x^2 + 1) / (x^2 + 1) and the second by x^3 / x^3 to get a common denominator: [(x-1)/(x^3) * (x^2+1)/(x^2+1)]+[(x)/(x^2+1) * (x^3)/(x^3)], which simplifies to (x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)]. Since the denominator is common to both we combine numerators: (x^3+x-x^2-1+x^4) / ) / [ (x^3)(x^2+1)] . We finally simplify to get (x^4 +x^3 - x^2+x-1) / ) / [ (x^3)(x^2+1)] **
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RESPONSE --> ok i totally do not understand
first term by (x^2 + 1) / (x^2 + 1) and the second by x^3 / x^3 to get a common denominator: [(x-1)/(x^3) * (x^2+1)/(x^2+1)]+[(x)/(x^2+1) * (x^3)/(x^3)], which simplifies to (x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)]. Since the denominator is common to both we combine numerators: (x^3+x-x^2-1+x^4) / ) / [ (x^3)(x^2+1)] . We finally simplify to get (x^4 +x^3 - x^2+x-1) / ) / [ (x^3)(x^2+1)] **.................................................
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00:48:57 Query R.7.58 (was R.7.54). How did you find the LCM of x - 3, x^3 + 3x and x^3 - 9x, and what is your result?
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RESPONSE --> lcm is x - 3, x^3 + 3x and x^3 - 9x= x^3-9x
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00:50:24 ** x-3, x^3+3x and x^3-9x factor into x-3, x(x^2+3) and x(x^2-9) then into (x-3) , x(x^2+3) , x(x-3)(x+3).
The factors x-3, x, x^2 + 3 and x + 3 'cover' all the factors of the three polynomials, and all are needed to do so. The LCM is therefore: x(x-3)(x+3)(x^2+3) **......!!!!!!!!...................................
RESPONSE --> understand now
x-3, x^3+3x and x^3-9x factor into x-3, x(x^2+3) and x(x^2-9) then into (x-3) , x(x^2+3) , x(x-3)(x+3). The factors x-3, x, x^2 + 3 and x + 3 'cover' all the factors of the three polynomials, and all are needed to do so. The LCM is therefore: x(x-3)(x+3)(x^2+3) **.................................................
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01:14:22 Query R.7.64 (was R.7.60). Show how you found and simplified the difference3x / (x-1) - (x - 4) / (x^2 - 2x + 1).
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RESPONSE --> 3x / (x-1) - (x - 4) / (x^2 - 2x + 1)= lcm=(x-1)^2(x-4) 3x(x-1)/(x-1)^2-(x-4)*x-4/(x-1)^2(x-4)= 3x^2-2x-4/(x-1)^2-(x-4)
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01:15:03 ** Starting with 3x / (x-1) - (x-4) / (x^2 - 2x +1) we factor the denominator of the second term to obtain (x - 1)^2. To get a common denominator we multiply the first expression by (x-1) / (x-1) to get
3x(x-1)/(x-1)^2 - (x-4)/(x-1)^2, which gives us (3x^2-3x-x-4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2. DRV**......!!!!!!!!...................................
RESPONSE --> i understand
Starting with 3x / (x-1) - (x-4) / (x^2 - 2x +1) we factor the denominator of the second term to obtain (x - 1)^2. To get a common denominator we multiply the first expression by (x-1) / (x-1) to get 3x(x-1)/(x-1)^2 - (x-4)/(x-1)^2, which gives us (3x^2-3x-x-4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2..................................................
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01:16:24 QUESTION FROM STUDENT: On the practice test I'm having problems with problem #5 I don't know where to start or how to set it up. I'm probably missing something simple and will probably feel stupid by seeing the solution. Could you help with this problem.
A retailer is offering 35% off the purchase price of any pair of shoes during its annual charity sale. The sale price of the shoes pictured in the advertisement is $44.85. Find the original price of the shoes by solving the equation p-.35p = 44.85 for p. INSTRUCTOR RESPONSE: It's very easy to get ahold of the wrong idea on a problem and then have trouble shaking it, or to just fail to look at it the right way. Nothing stupid about it, just human nature. See if the following makes sense. If not let me know. p - .35 p = 44.85. Since p - .35 p = 1 p - .35 p = (1 - .35) p = .65 p we have .65 p = 44.85. Multiplying both sides by 1/.65 we get p = 44.85 / .65 = etc. (you can do the division on your calculator); you'll get something near $67).......!!!!!!!!...................................
RESPONSE --> ok
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