Assignment R5

#$&*

course Mth 158

6/19 11

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

006. `* 6

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

36x^2-9

=6x^2-3^2

Or

(6x-3)^2

confidence rating #$&*: 3

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Given Solution:

* * ** 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 squares. **

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Self-critique (if necessary):

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Self-critique Rating: 3

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Question:

R.5.32 \ 28 (was R.6.24) What do you get when you factor 25 x^2 + 10 x + 1 and how did you get your result?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

25x^2 +10x + 1

This equation is prime because there aren’t any numbers that when multiplied together give you 1 and when added together give you 10.

confidence rating #$&*: 2

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Given Solution:

* * ** 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 be able to 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 to the criteria, 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. **

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Self-critique (if necessary):

??? I still do not understand your explanation of why it is not prime. I understand that you there could be irrational numbers to dispute the fact that it’s prime. I just do not understand how you got those numbers. I got lost when you got to this section:

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. **

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Self-critique Rating:

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Question:

R.5.34 (was R.6.30). What do you get when you factor x^3 + 125 and how did you get your result?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x^3 + 125

125= 5^3

(x + 5) (x^2…..) I don’t understand how to find the next two numbers that fit into the slots.

confidence rating #$&*: 1

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Given Solution:

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).

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Self-critique (if necessary):

All of this makes sense except why is there a 2 in the second slot of the equation?

- We know that a^3 + b^3 = (a+b) ( a^2 - 2 a b + b^2).

If that is there wouldn’t the equation be

(x + 5) (x^2 - 10x + 25) ? since the bolded part would be -2x5 if you were to fill in the slots for 2 a b equaling -10x?

If not, what is the purpose of the 2?

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Self-critique Rating: 1

@&

That was a careless typo on my part. a^2 - 2 a b + b^2 is the square of (a - b), but that is not relevant to this formula.

You are right in the values you get from that incorrect formula.

The correct formula is

a^3 + b^3 = (a+b) ( a^2 - a b + b^2)

and that does lead to the given solution.

*@

@&

That has been there in this given solution for a long time, and lots of very good students have come through this course. You're the first one to notice this.

Thanks for pointing it out. I'l make the correction in the original document as soon as it opens (frontPage appears to be stuck on opening the file right now).

Sorry for the confusion.

*@

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Question:

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x^2 -17x +16

= (x -16) (x -1)

When multiplied, -16 * -1= 16

When added, -16 + -1 = -17

confidence rating #$&*: 3

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Given Solution:

* * ** 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, if a and b happen to be integers, we have the following possibilities:

• 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. So the possibilities are reduced to

• a = -2, b = -8, or

• a = -1, b = -16, or

• a = -4, b = -4.

The only of the these possibilities that gives us a + b = -17 is a = -1, b = -16. So we conclude that

• x^2 - 17 x + 16 = (x-16)(x-1). **

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Self-critique (if necessary):

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Self-critique Rating: 3

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Question:

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3x(x - 1) + 2(x-1) Find what is common in the parentheses and move it outside of the parentheses

Regroup the equation: (3x + 2)(x-1)^2

confidence rating #$&*: 2

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Given Solution:

* * ** 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). **

ADDITIONAL EXPLANATION:

To see that

(3x^2-3x)+(2x-2) =

3x(x-1)+2(x-1)

apply the distributive law to each term in the second expression:

3x ( x - 1) = 3 x^2 - 3x, and

2 ( x - 1) = 2x - 2.

To see that

3x(x-1)+2(x-1) =

(3x+2)(x-1)

apply the distributive law as follows:

(3x + 2) ( x - 1) = 3x * (x - 1) + 2 * (x - 1).

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Self-critique (if necessary):

??? Why in the last step does the equation go from have two sets of (x-1) and then one disappears. Why would the final answer be (3x+2) (x-1)^2 because there are two sets of (x-1)?

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Self-critique Rating: 2

@&

There never is an (x-1)^2.

Among othe things, (3x + 2)(x-1)^2, if multiplied out, would include the term 3 x^3, which is not present in the original expression.

3x(x - 1) + 2(x-1) factors into

(3x + 2) ( x - 1)

by the distributive law.

*@

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Question:

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3x^2 - 10x + 8

(x+4) (3x + 2)

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

* * ** 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). **

??? I know that my answer is not correct and I knew that when I typed it in. How is (3x-4) (x-2) correct? When they are multiplied together it equals the positive 8 but when they are added together it equals 2? I thought it had to add to equal -10?

@&

If you multiply (3x - 4) ( x - 2) using the distributive law you will find that this is correct.

The coefficient of x^2 in this case is not 1. The technique you are trying to use applies only if that coefficient is 1.

*@

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x^2 + 6x +14

The equation is prime because there are not any integers that when multiplied equal 14 and when added equal 6.

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

* * ** This expression factors, but not into binomials 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 often cannot find the factors by factoring in the usual manner; however it is always possible to find the factors of a second-degree trinomial using the quadratic formula.

The quadratic formula applied to this problem tells us that the factors are (x - z1) * (x - z2), were z1 and z2 are the solutions to the equation 14 + 6 x - x^2 0. Using the formula we find that

• z1 = ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and

• z2 = ( -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. **

"

Self-critique (if necessary):

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Self-critique rating:

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Question:

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3x^2 - 10x + 8

(x+4) (3x + 2)

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

* * ** 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). **

??? I know that my answer is not correct and I knew that when I typed it in. How is (3x-4) (x-2) correct? When they are multiplied together it equals the positive 8 but when they are added together it equals 2? I thought it had to add to equal -10?

@&

If you multiply (3x - 4) ( x - 2) using the distributive law you will find that this is correct.

The coefficient of x^2 in this case is not 1. The technique you are trying to use applies only if that coefficient is 1.

*@

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x^2 + 6x +14

The equation is prime because there are not any integers that when multiplied equal 14 and when added equal 6.

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * ** This expression factors, but not into binomials 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 often cannot find the factors by factoring in the usual manner; however it is always possible to find the factors of a second-degree trinomial using the quadratic formula.

The quadratic formula applied to this problem tells us that the factors are (x - z1) * (x - z2), were z1 and z2 are the solutions to the equation 14 + 6 x - x^2 0. Using the formula we find that

• z1 = ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and

• z2 = ( -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. **

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

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Question:

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3x^2 - 10x + 8

(x+4) (3x + 2)

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * ** 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). **

??? I know that my answer is not correct and I knew that when I typed it in. How is (3x-4) (x-2) correct? When they are multiplied together it equals the positive 8 but when they are added together it equals 2? I thought it had to add to equal -10?

@&

If you multiply (3x - 4) ( x - 2) using the distributive law you will find that this is correct.

The coefficient of x^2 in this case is not 1. The technique you are trying to use applies only if that coefficient is 1.

*@

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x^2 + 6x +14

The equation is prime because there are not any integers that when multiplied equal 14 and when added equal 6.

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * ** This expression factors, but not into binomials 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 often cannot find the factors by factoring in the usual manner; however it is always possible to find the factors of a second-degree trinomial using the quadratic formula.

The quadratic formula applied to this problem tells us that the factors are (x - z1) * (x - z2), were z1 and z2 are the solutions to the equation 14 + 6 x - x^2 0. Using the formula we find that

• z1 = ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and

• z2 = ( -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. **

"

Self-critique (if necessary):

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Self-critique rating:

#*&!#*&!

&#Your work looks good. See my notes. Let me know if you have any questions. &#