#$&* course Mth 271 9/19 8 003. `query 3*********************************************
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Given Solution: `a z^-3 ( 3 z^4) = 3 * z^-3 * z^4 = 3 * z^(4-3) = 3 z. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q0.3.30 (was 0.3.30 simplify(12 s^2 / (9s) ) ^ 3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (4s/3)^3 = (64s^3)/9 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Starting with (12 s^2 / (9s) ) ^ 3 we simplify inside parentheses to get ( 4 s / 3) ^ 3, which is equal to 4^3 * s^3 / 3^3 = 64 s^3 / 27 It is possible to expand the cube without first simplifying inside, but the subsequent simplification is a little more messy and error-prone; however done correctly it gives the same result. It's best to simplify inside the parentheses first. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q0.3.38 (was 0.3.38 simplify ( (3x^2 y^3)^4) ^ (1/3) and (54 x^7) ^ (1/3) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I did not understand this one very well, and I couldn’t understand it by the explanation as well.
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Given Solution: `a ( (3x^2 y^3)^4) ^ (1/3) = ( 3^4 * (x^2)^4 * (y^3)^4) ^ (1/3) = (3^4 x^8 y^12) ^ (1/3). We can take the 1/3 power of y^12, since (y^12)^(1/3) = y^(12 * 1/3) = y^4. We get an integer exponent. The 1/3 power of 3^4 and of x^8, on the other hand, will both result in fractional exponents. We therefore factor out the greatest factor of 3^4 x^8 y^12 of which we can take the 1/3 power and end up with integer exponents. The largest such factor is 3^3 * x^6 * y^12. The 1/3 power of this factor will be 3 x^2 y^4. Beginning with the last step of the original series of equations: (3^4 x^8 y^12) ^ (1/3). We factor the expression inside parentheses to get ( (3^3 x^6 y^12) * (3 x^2) )^(1/3). We take the 1/3 power of each factor, applying the laws of exponents: (3^3 x^6 y^12) ^ (1/3) * (3 x^2)^(1/3). We then take the 1/3 power of each factor in 3^3 x^6 y^12 to get (3^3)^(1/3) * (x^6)^(1/3) * (y^12)^(1/3) * (3 x^2)^(1/3). We apply the laws of exponents to get 3^(1 * 1/3) * x^(6 * 1/3) * y^(12 * 1/3). * (3 x^2)^(1/3). We do the arithmetic to get 3^1 * x^2 * y^4 * (3 x^2)^(1/3), or just 3 x^2 y^4 ( 3 x^2)^(1/3). To simplify (54 x^7)^(1/3) you have to find the maximum factor inside the parentheses which a perfect 3d power. First factor 54 into its prime factors: 54 = 2 * 27 = 2 * 3 * 3 * 3 = 2 * 3^3. Now we have (2 * 3^3 * x^7)^(1/3). 3^3 and x^6 are both perfect 3d Powers. So we factor 3^3 * x^6 out of the expression in parentheses to get ( (3^3 * x^6) * 2x ) ^(1/3). This is equal to (3^3 * x^6)^(1/3) * (2x)^(1/3). Simplifying the perfect cube we end up with 3 x^2 ( 2x ) ^ (1/3) For the second expression: The largest cube contained in 54 is 3^3 = 27 and the largest cube contained in x^7 is x^6. Thus you factor out what's left, which is 2x. Factoring 2x out of (54 x^7)^(1/3) gives you 2x ( 27 x^6) so your expression becomes [ 2x ( 27 x^6) ] ^(1/3) = (2x)^(1/3) * [ 27 x^6 ] ^(1/3) = (2x)^(1/3) * [ (27)^(1/3) (x^6)^(1/3) ] = (2x)^(1/3) * 3 x^2, which in more traditional order is 3 x^2 ( 2x)^(1/3). ** STUDENT COMMENT: This one was a little more difficult than the previous ones. I looked at the answer given and that confused me even more. INSTRUCTOR RESPONSE: Few students can understand the solution by looking at it. Most students will need to rewrite the solution in standard notation in order to understand it. You should refer to the examples at the links included with the Typewriter Notation exercise. You should understand all those examples and should be able to write then in standard notation, given the typewriter notation. You should also be able to write the typewriter notation, given each standard-notation example. A few examples from the current problem: ( (3x^2 y^3)^4) ^ (1/3) is written ( (3^3 x^6 y^12) * (3 x^2) )^(1/3) is written 3^(1 * 1/3) * x^(6 * 1/3) * y^(12 * 1/3). * (3 x^2)^(1/3) is written 3 x^2 y^4 ( 3 x^2)^(1/3) is written &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q0.3.58 (was 0.3.54 factor P(1+r) from expression P(1+r) + P(1+r)^2 + P(1+r)^3 + ... YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution:
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Given Solution: `a Few students get this one. If you didn't you've got a lot of company; if you did congratulations. It's important to understand how this problem illustrates the essence of factoring. It's important also because expressions of this form occur throughout calculus. Factor out P * (1 + r). Divide each term by P ( 1 + r), and your result is P (1 + r) * your quotient. Your quotient would be 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... . The factored form would therefore be P(1+r) [ 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... ]. You can verify that this is identical to the original expression if you multiply it back out. Analogy with different exponents and only three terms: A x^3 + A x^4 + A x^5 can be divided by A x^2 to give quotient x + x^2 + x^3, so the factored expression is A x^2 ( x + x^2 + x^3). ** INSTRUCTOR COMMENTS: This is typically a difficult problem for students. Let's reconstruct these ideas in reverse order: What do you get when you multiply A x^2 ( x + x^2 + x^3) ? Why then do we say that factoring A x^2 our of A x^3 + A x^4 + A x^5 gives us the expression A x^2 ( x + x^2 + x^3) ? Now what is P(1+r) [ 1 + (1+r) + (1+r)^2 + (1+r)^3 ]? Why then would we say that factoring P ( 1 + r) out of P(1+r) + P(1+r)^2 + P(1+r)^3 gives us the expression P(1+r) [ 1 + (1+r) + (1+r)^2 + (1+r)^3 ] ? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Did not understand this one at all! The explanation helped though! STUDENT QUESTION ON TEXT EXAMPLE, WITH INSERTED INSTRUCTOR RESPONSES: Simplify the expression by factoring. a.) 3(x+1)^1/2 (2x - 3)^5/2 + 10(x+1)^3/2 (2x-3)^3/2. The next steps I do not understand how they did. Next step: = (x+1)^1/2(2x-3)^3/2 [3(2x-3) + 10(x+1)] It might be easier to see how this was done if you multiply the last expression out. (x+1)^1/2(2x-3)^3/2 [3(2x-3) + 10(x+1)] = (x+1)^1/2(2x-3)^3/2 * 3(2x-3) + (x+1)^1/2(2x-3)^3/2 * 10(x+1) by the distributive law. (x+1)^1/2(2x-3)^3/2 * 3(2x-3) = 3(x+1)^1/2 (2x - 3)^5/2 and (x+1)^1/2(2x-3)^3/2 * 10(x+1) = 10(x+1)^3/2 (2x-3)^3/2. So (x+1)^1/2(2x-3)^3/2 [3(2x-3) + 10(x+1)] = 3(x+1)^1/2 (2x - 3)^5/2 + 10(x+1)^3/2 (2x-3)^3/2 Now, to factor 3(x+1)^1/2 (2x - 3)^5/2 + 10(x+1)^3/2 (2x-3)^3/2 you use the same ideas, but in reverse order. We first note than both terms contain powers of x+1 and of 2x - 3. Whichever power of x + 1 is the lesser, that power will be common to both terms of our expression. In this case 1/2 is the common power, so (x + 1)^(1/2) is common to both terms. Whichever power of 2x - 3 is the lesser, that power will be common to both terms of our expression. In this case 3/2 is the common power, so (2x - 3)^(3/2) is common to both terms. So we factor (x + 1)^(1/2) * (2x - 3)^(3/2) out of both terms of the original expression. (To figure out what each remaining term will be, just divide each term by (x + 1)^(1/2) * (2x - 3)^(3/2)). "