Assignment 6 R5

#$&*

course Mth 158

6/20/13 9:00 pm

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?

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Your solution:

I knew that the result of the difference of two squares would result in something like this : [ sqrt( a ) + sort ( b ) ] [ sqrt( a ) - sqrt ( b ) ]

This left me with a result of ( 6x + 3 ) ( 6x - 3 )

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): Factoring it out like that is a bit different than how I learned to do it, but I do see that it is much easier to explain. ??? Is how I am doing it still acceptable?

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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?

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Your solution:

It did not look obviously like any kind of special case, so I decided to use the A, B, C method.

I identified A = 25 B = 10 C = 1 meaning AC = 25

I looked for factors of 25: 25, 1 ; 5, 5 (Because B is positive, I knew that I was not looking for factors that would both be positive)

Then I looked to see if any of those factors would add up to B; the second set, 5 and 5, are equal to 10

25 x^2 + 10 x + 1 =

25 x^2 + 5x + 5x + 1 =

(25 x^2 + 5x )+ (5x + 1) = I used the grouping method

5x (5x + 1) + 1 (5x + 1) = Even though it is understood, I wrote out a factored one so that the next steps would be more clear to see

(5x + 1) (5x + 1) (5x + 1) =

( 5x + 1 )^3

@&

(5x + 1)^3 would include an x^3 term, specifically 125 x^3, so that can't be right.

However you very nearly got the right answer.

You had the expression

5x (5x + 1) + 1 (5x + 1)

Factoring that expression carfully, you can factor out the (5 x + 1). For the moment let's just use ( ### ) to represent the expression (5 x + 1), so your expression will be written

5x * ### + 1 * ###.

You will agree that factoring out ### gives you

(5 x + 1) * ###.

So now you'll have to agree that

5x (5x + 1) + 1 (5x + 1)

factors as

(5 x + 1) ( 5 x + 1),

which is (5x + 1) ^ 2.

The rule you're following, about some numbers adding up to B, works if the coefficient of x^2 is 1, and is works if the constant number in the equation is 1. It doesn't work otherwise, so you might want to look at the following:

The coefficient of x^2 in the expression 25 x^2 + 10 x + 1 is 25. So when the expression is factored into the form (## x + ##) ( ## x + ##), where each ## could stand for any number, the coefficients of x (i.e., the numbers in front of the x's) have to multiply out to 25.

25 can only be expressed as a product of whole numbers are 25 * 1 or 5 # 5. So if the expression can be factored with whole numbers, it has to have one of the following forms:

(25 x + ##) ( x + ##)

or

(5 x + ##) ( 5 x + ##).

Since the constant number in our original expression is 1, the product of the numbers ## in either factorization must be 1. The only way two whole numbers can have a product of 1 is if they are both 1.

So we have only two possible forms for the factored expression:

(25 x + 1) ( x + 1)

or

(5 x + 1) ( 5 x + 1).

If you multiply out both expressions using the distributive law, you will find that only the second one has the desired product.

The solution given here is a sort of last-resort brute-force method, meant to be used when nothing else works. It's also useful for really understanding what's going on with this sort of factorization, but of course there are other ways of looking at this problem. With practice we naturally learn the various ways of thinking about factorization.

*@

@&

Note that the given solution addresses a different expression, namely x^2 + 10 x + 1.

*@

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 am a little confused. Did I use the A, B, C thing incorrectly?

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Self-critique Rating: 1

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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?

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Your solution:

x^3 + 125 is the sum of two cubes

(x + 5) (x^2 - 5x + 25)

confidence rating #$&*: 3

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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): OK

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Self-critique Rating: OK

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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?

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Your solution:

I had to first find all the pairs of integers whose product was 16

Pairs 1, 16 -1, -16 2, 8 -2, -8 4, 4 -4, -4

Sum 17 -17 10 -10 8 -8

The pair -1 and -16 have a product of 16 and a sum of -17 so

x^2 - 17 x + 16 =

(x - 1) (x - 16)

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): OK

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Self-critique Rating: OK

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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?

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Your solution:

For this problem you must factor by grouping

3 x^2 - 3 x + 2 x - 2 =

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

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

(3x + 2) (x - 1)

confidence rating #$&*: 3

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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): OK

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Self-critique Rating: OK

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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?

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Your solution:

Comparing it to Ax^2 + Bx + C we find that A= 3 B = -10 and C = 8

The value of AC = 24

We must find the pair of integers that result in 24 as their product; the possibilities are

1, 24; -1, -24; 2, 12; -2, -12; 3, 8; -3, -8; 4, 6; and -4, -6

We must find which of these pairs adds up to B. -4 and -6 result in a sum of -10

Applying this,

3 x^2 - 10 x + 8 =

3 x^2 - 4x - 6x + 8 =

(3 x^2 - 4x) + (-6x + 8) =

x (3x - 4) -2 (3x - 4) =

(x - 2) (3x - 4)

confidence rating #$&*: 3

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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). **

R.5.88 (was R.6.78). What do you get when you factor 14 + 6 x - x^2 and how did you get your result?

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Your solution:

First I flipped it around so that it read in order of highest degree to lowest degree.

14 + 6 x - x^2 =

- x^2 + 6x + 14 =

-1 (x^2 - 6x -14)

I attempted to see if more factoring was possible by finding pairs of integers that had a product of -14 and added up to -6, but none met both criteria. This seems to be as far as the problem may be factored (prime)

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:

Comparing it to Ax^2 + Bx + C we find that A= 3 B = -10 and C = 8

The value of AC = 24

We must find the pair of integers that result in 24 as their product; the possibilities are

1, 24; -1, -24; 2, 12; -2, -12; 3, 8; -3, -8; 4, 6; and -4, -6

We must find which of these pairs adds up to B. -4 and -6 result in a sum of -10

Applying this,

3 x^2 - 10 x + 8 =

3 x^2 - 4x - 6x + 8 =

(3 x^2 - 4x) + (-6x + 8) =

x (3x - 4) -2 (3x - 4) =

(x - 2) (3x - 4)

confidence rating #$&*: 3

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

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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). **

R.5.88 (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:

First I flipped it around so that it read in order of highest degree to lowest degree.

14 + 6 x - x^2 =

- x^2 + 6x + 14 =

-1 (x^2 - 6x -14)

I attempted to see if more factoring was possible by finding pairs of integers that had a product of -14 and added up to -6, but none met both criteria. This seems to be as far as the problem may be factored (prime)

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:

#*&!

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