Assignment 6 R5

#$&*

course Mth 158

2/7/13 around 7:00 PM

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:

 36 x^2 - 9

(6x)^2 - (3)^2

 (6x - 3)(6x + 3)

 

 

confidence rating #$&*: 2

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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): OK

 

 

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Self-critique Rating: OK

 

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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:

 25 x^2 + 10 x + 1

 (5x)^2 + 10x + (1)^2

 Since: 5x * 1 * 2 = 10x, we have a perfect square.

 (5x + 1)^2

 

 

confidence rating #$&*: 3

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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): OK

 

 

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Self-critique Rating: OK

 

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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

 x^3 + (5)^3

 (x + 5)(x^2 - 5x + 5^2)

  (x + 5)(x^2 - 5x + 25)

 

 

confidence rating #$&*: 2

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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?

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

 x^2 - 17 x + 16

 -16 * -1 = +16 and -16 + -1 = -17 so:

 (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): 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?

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

 3x^2 - 3x + 2x - 2

 (3x^2 - 3x) + (2x - 2)

 3x(x - 1) + 2(x - 1)

 (x - 1)(3x + 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): 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?

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

???? Will what I am doing get the correct answer? And if so, what am I doing wrong or missing?

 3x^2 - 10x + 8

 Using Ax^2 + Bx + C to factor a trinomial:

 Since 3 * 8 = 24 and -6 * -4 = 24 and -6 + (-4) = -10

 3x^2 - 4x - 6x + 8

Group:

 (3x^2 - 4x) - (6x + 8)

 x(3x - 4) - 2(3x + 4)

???? I cannot factor out the (3x - 4) and the (3x + 4) because they have different signs. Am using the wrong formula, or way of getting the answer, or did I just do something incorrectly? There were a couple problems in the book like this one where I ended up with opposite signs, but the same numbers and variables in the parentheses.

 

 

confidence rating #$&*: 1

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

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

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). **

@&

You were on the right track.

When all else fails, listing the possibilities systematically is always an option. Sometimes, though, there are an awful lot of possibilities.

*@

 

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:

 My textbook says R.5.82 is: x^2 - 6x + 8

x^2 - 6x + 8

Since -2 * -4 = 8 and -2 + (-4) = -6:

 (x - 2)(x - 4)

 

The textbook says R.5. 88 is: 14 + 6 x - x^2

 I'm not sure how to factor 14 + 6 x - x^2 because when put in a descending order of degree: -x^2 + 6x + 14, the first term is negative and I don't know how that would effect the solution.

 ???? Do I need to put in a descending order of degree? And if not, how would I solve it?

 

@&

This one doesn't factor, at least not as expressed here. The problem you quote from the book does factor, and you've done it correctly.

If you wanted to factor -x^2 + 6 x + 14, not yet knowing that it doesn't factor with rational numbers, you could start by writing it as

- ( x^2 - 6 x - 14 )

and worry about factoring the expression in parentheses first.

However listing your factors with x and -x, as done in the given solution, is also an option.

*@

 

confidence rating #$&*: 1

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

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

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):

------------------------------------------------

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:

???? Will what I am doing get the correct answer? And if so, what am I doing wrong or missing?

 3x^2 - 10x + 8

 Using Ax^2 + Bx + C to factor a trinomial:

 Since 3 * 8 = 24 and -6 * -4 = 24 and -6 + (-4) = -10

 3x^2 - 4x - 6x + 8

Group:

 (3x^2 - 4x) - (6x + 8)

 x(3x - 4) - 2(3x + 4)

???? I cannot factor out the (3x - 4) and the (3x + 4) because they have different signs. Am using the wrong formula, or way of getting the answer, or did I just do something incorrectly? There were a couple problems in the book like this one where I ended up with opposite signs, but the same numbers and variables in the parentheses.

 

 

confidence rating #$&*: 1

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

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

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). **

@&

You were on the right track.

When all else fails, listing the possibilities systematically is always an option. Sometimes, though, there are an awful lot of possibilities.

*@

 

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:

 My textbook says R.5.82 is: x^2 - 6x + 8

x^2 - 6x + 8

Since -2 * -4 = 8 and -2 + (-4) = -6:

 (x - 2)(x - 4)

 

The textbook says R.5. 88 is: 14 + 6 x - x^2

 I'm not sure how to factor 14 + 6 x - x^2 because when put in a descending order of degree: -x^2 + 6x + 14, the first term is negative and I don't know how that would effect the solution.

 ???? Do I need to put in a descending order of degree? And if not, how would I solve it?

 

@&

This one doesn't factor, at least not as expressed here. The problem you quote from the book does factor, and you've done it correctly.

If you wanted to factor -x^2 + 6 x + 14, not yet knowing that it doesn't factor with rational numbers, you could start by writing it as

- ( x^2 - 6 x - 14 )

and worry about factoring the expression in parentheses first.

However listing your factors with x and -x, as done in the given solution, is also an option.

*@

 

confidence rating #$&*: 1

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

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

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):

------------------------------------------------

Self-critique rating:

#*&!

 

*********************************************

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:

???? Will what I am doing get the correct answer? And if so, what am I doing wrong or missing?

 3x^2 - 10x + 8

 Using Ax^2 + Bx + C to factor a trinomial:

 Since 3 * 8 = 24 and -6 * -4 = 24 and -6 + (-4) = -10

 3x^2 - 4x - 6x + 8

Group:

 (3x^2 - 4x) - (6x + 8)

 x(3x - 4) - 2(3x + 4)

???? I cannot factor out the (3x - 4) and the (3x + 4) because they have different signs. Am using the wrong formula, or way of getting the answer, or did I just do something incorrectly? There were a couple problems in the book like this one where I ended up with opposite signs, but the same numbers and variables in the parentheses.

 

 

confidence rating #$&*: 1

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

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

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). **

@&

You were on the right track.

When all else fails, listing the possibilities systematically is always an option. Sometimes, though, there are an awful lot of possibilities.

*@

 

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:

 My textbook says R.5.82 is: x^2 - 6x + 8

x^2 - 6x + 8

Since -2 * -4 = 8 and -2 + (-4) = -6:

 (x - 2)(x - 4)

 

The textbook says R.5. 88 is: 14 + 6 x - x^2

 I'm not sure how to factor 14 + 6 x - x^2 because when put in a descending order of degree: -x^2 + 6x + 14, the first term is negative and I don't know how that would effect the solution.

 ???? Do I need to put in a descending order of degree? And if not, how would I solve it?

 

@&

This one doesn't factor, at least not as expressed here. The problem you quote from the book does factor, and you've done it correctly.

If you wanted to factor -x^2 + 6 x + 14, not yet knowing that it doesn't factor with rational numbers, you could start by writing it as

- ( x^2 - 6 x - 14 )

and worry about factoring the expression in parentheses first.

However listing your factors with x and -x, as done in the given solution, is also an option.

*@

 

confidence rating #$&*: 1

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

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

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):

------------------------------------------------

Self-critique rating:

#*&!#*&!

@&

You're in good shape. See my notes.

*@