course Mth 158 007. `* 7
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Given Solution: * * ** 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)]. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): You have to make sure you factor everything completely like in the denominator Self-critique Rating: ********************************************* Question: * R.7.28 (was R.7.24). Show how you simplified[ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ]. ********************************************* Your solution: Bring the denominators to the numerator so x-2 * 4x / x^2 -4x +4 * 12x then x^2-4x+4 factors out to (x-2)(x-2) then cancel like factors which are (x-2) / (x-2) cancel out and then 4x/12x cancels out and leaves 3 leaving 1/ (x-2)*3 Confidence Assessment:
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Given Solution: [ ( 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) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The 12x went to the numerator of the first expression and when I solved the 12x goes to the numerator of the second expression in the entire division problem. That is why my answer the simplified 3 is in the denominator and in the given solution the 3 is the numerator Self-critique Rating: ********************************************* Question: * R.7.40 (was R.7.36). Show how you found and simplified the sum (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2). ********************************************* Your solution: Group so [(2x-5)+(x+4)]/(3x+2) then add like terms giving (3x-1)/(3x+2) Confidence Assessment:
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Given Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * R.7.52 (was R.7.48). Show how you found and simplified the expression (x - 1) / x^3 + x / (x^2 + 1). ********************************************* Your solution: Get a common denominator multiplying by x^2 + 1 and x^3 so [(x-1)/(x^3) (x^2+1)/(x^2+1)]+[(x)/(x^2+1) (x^3)/(x^3)] simplifying to (x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)] and further simplifying to (x^3+x-x^2-1+x^4) / ) / [ (x^3)(x^2+1)] and again simplify (x^4 +x^3 - x^2+x-1) / ) / [ (x^3)(x^2+1)] Confidence Assessment:
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Given Solution: 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)] &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It is tough when you have to keep simplifying and factor correctly and with division and it’s rules but being careful and not rushing works for me. Self-critique Rating: ********************************************* Question: * 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? ********************************************* Your solution: Factor each expression so x(x^2+3) and (x-3) and x(x+3)(x-3) so the LCM is x(x-3)(x+3)(x^2+3) Confidence Assessment:
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Given Solution: 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: • LCM = x(x-3)(x+3)(x^2+3) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): You have to make sure to simplify/factor completely to see the common Self-critique Rating: ********************************************* Question: * R.7.64 (was R.7.60). Show how you found and simplified the difference 3x / (x-1) - (x - 4) / (x^2 - 2x + 1). ********************************************* Your solution: I’m not sure b/c it gets confusing to know where each expression is with the way written. Confidence Assessment:
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Given Solution: * * ** Starting with 3x / (x-1) - (x-4) / (x^2 - 2x +1) we factor the denominator of the second term to obtain (x - 1)^2. Since the first denominator (x - 1) is already a factor of the second, our common denominator is (x - 1)^2. To express the given expression in terms of the common denominator we then 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok it makes sense once I looked at the solution given. First thing is to factor and then see if anything cancels out and grouping like terms. And keep simplifying until you can’t anymore.