Query 3

course Mth 271

9/14 8:35 pm

003. `query 3vvvv

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Question: `q0.3.24 (was 0.3.24 simplify z^-3 (3z^4)

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Your solution:

z^-3(3z^4):

3(z^4)(z-3):

3z^4-3:

3z

confidence rating:

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Given Solution:

`a z^-3 ( 3 z^4) = 3 * z^-3 * z^4 = 3 * z^(4-3) = 3 z. **

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Self-critique (if necessary):

OK

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Self-critique Rating:

OK

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Question: `q0.3.30 (was 0.3.30 simplify(12 s^2 / (9s) ) ^ 3

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Your solution:

(12s^2/(3s)/(9s)/3s))=(4s/3)^3

64s^3/27

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. **

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Self-critique (if necessary):

OK

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Self-critique Rating:

OK

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Question: `q0.3.38 (was 0.3.38 simplify ( (3x^2 y^3)^4) ^ (1/3) and (54 x^7) ^ (1/3)

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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:

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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). **

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Self-critique (if necessary):

OK

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Self-critique Rating:

OK

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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 + ...

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Your solution:

confidence rating:

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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 ] ?

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Self-critique (if necessary):

That one flew over my head but you explanation cleared it all up for me

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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)).

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&#This looks good. Let me know if you have any questions. &#