R6

course Mth 158

f 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:

(6x)^2 – 3^2

(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):

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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 x^2 + 10 x + 1 and how did you get your result?

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Your solution:

Can not do this problem because there is no answer for to get ten because the only way you could even attempt the problem would be to take 10 mult by 1 but then you have to add 10 to 1 and you get 11 so that don’t’ work either.

There is no answer, polynomial is prime.

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):

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Self-critique rating #$&*:3

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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 + 5^3

found least common multiple that would be ^3

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):

Wasn’t thinking about that formula.

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Self-critique rating #$&*:3

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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:

Intergers that will work for –17 is –16 and –1 so problem will be

(x-16)(x-1)

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?

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Your solution:

Group

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

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

(3x+2)(x-1)

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):

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Self-critique rating #$&*:3

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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:

Can not find any number that will equal 10 but can add together to get 8.

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). **

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?

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Your solution:

Not sure how to do this one.

confidence rating #$&*:

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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. **

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