course Mth 271 9/14 8:35 pm 003. `query 3vvvv
.............................................
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: OK ********************************************* Question: `q0.3.30 (was 0.3.30 simplify(12 s^2 / (9s) ) ^ 3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (12s^2/(3s)/(9s)/3s))=(4s/3)^3 64s^3/27 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: 54= 2*27= 2*3*9 or 2*3*3*3 (2*3^3 x^7)^1/3 (2x(3^3 x^6))^1/3 (3x^2)1/3*2x^1/3 3x^2(2x)^1/3 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): That one flew over my head but you explanation cleared it all up for me ------------------------------------------------ Self-critique Rating: 3 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)). "