Fundamental Polynomial Operations

[Superkevs22]

Hi guys! I'm kind of stuck in my review here in Quantitave on the Polynomial part

Homework Statement

A. The first problem I had is Problem 7 on the link I will provide which has this.
(x + y)³ + (x-y)³ = ?

B. The second problem I had is Problem 8 on the same link I will provide which has this.
x + 2xx-2
1 + 4 x² - 4

Here is the link for proper viewing (sorry I'm still not quite familiar with the tools here in PF ):
http://i.imgur.com/UyJXw2a.jpg?1

Homework Equations

A. (x + y)³ = x³ + y³ = (x + y)(x² - xy + y²)
(x-y)³ = x³ - y³ = (x - y)(x² + xy + y²)

B. I'm sorry I have no idea about the equation on this one

The Attempt at a Solution

When I tried answering it I got:

A. 2x(2x² + 2y²)
And the answer key provided says that the answer is Choice A which is highlighted in yellow

B. I'm sorry I have no idea how to solve this problem

Please help me guys and if it's not too much trouble please show the solution Thanks!

 

[Ray Vickson]

Hi guys! I'm kind of stuck in my review here in Quantitave on the Polynomial part

Homework Statement

A. The first problem I had is Problem 7 on the link I will provide which has this.
(x + y)³ + (x-y)³ = ?

B. The second problem I had is Problem 8 on the same link I will provide which has this.
x + 2xx-2
1 + 4 x² - 4

Here is the link for proper viewing (sorry I'm still not quite familiar with the tools here in PF ):
http://i.imgur.com/UyJXw2a.jpg?1

Homework Equations

A. (x + y)³ = x³ + y³ = (x + y)(x² - xy + y²)
(x-y)³ = x³ - y³ = (x - y)(x² + xy + y²)

B. I'm sorry I have no idea about the equation on this one

The Attempt at a Solution

When I tried answering it I got:

A. 2x(2x² + 2y²)
And the answer key provided says that the answer is Choice A which is highlighted in yellow

B. I'm sorry I have no idea how to solve this problem

Please help me guys and if it's not too much trouble please show the solution Thanks!

No, no, no! You cannot write $(x+y)^3 = x^3 + y^3$ because that is false. Do you think that $(1+3)^3 = 1^3 + 3^3$; that is, do you think that 64 = 28?

To figure out what $(x+y)^3$ is equal to, just expand it out; that is, note that $(x+y)^3 = (x+y)(x+y)^2$ and you can first expand out $(x+y)^2$, then multiply by (x+y and expand out again. There are fast formulas for doing all this more-or-less automatically, but until you really know what you are doing I recommend that you stay away from formulas and just do it from scratch, the long way.

 

[Student100]

For 8 you have a complex fraction, what should you do to begin to simplify it?

 

[Superkevs22]

No, no, no! You cannot write $(x+y)^3 = x^3 + y^3$ because that is false. Do you think that $(1+3)^3 = 1^3 + 3^3$; that is, do you think that 64 = 28?

To figure out what $(x+y)^3$ is equal to, just expand it out; that is, note that $(x+y)^3 = (x+y)(x+y)^2$ and you can first expand out $(x+y)^2$, then multiply by (x+y and expand out again. There are fast formulas for doing all this more-or-less automatically, but until you really know what you are doing I recommend that you stay away from formulas and just do it from scratch, the long way.

Oh! Thank you so much for clearing that one up for me Ray, let me try to solve it again using your advice. I'll get back to you once I'm done.

Thanks too Student 100, I'm thinking of expanding it a bit. Let me get back to you once I'm done also with Problem 8.

Guys thanks a lot I really appreciate it!

 

[Superkevs22]

No, no, no! You cannot write $(x+y)^3 = x^3 + y^3$ because that is false. Do you think that $(1+3)^3 = 1^3 + 3^3$; that is, do you think that 64 = 28?

To figure out what $(x+y)^3$ is equal to, just expand it out; that is, note that $(x+y)^3 = (x+y)(x+y)^2$ and you can first expand out $(x+y)^2$, then multiply by (x+y and expand out again. There are fast formulas for doing all this more-or-less automatically, but until you really know what you are doing I recommend that you stay away from formulas and just do it from scratch, the long way.

Ray thanks a lot! I finally got it!

http://i.imgur.com/rzjJya4.jpg?1

 

[Superkevs22]

For 8 you have a complex fraction, what should you do to begin to simplify it?

It seems I'm having difficulties trying it out

http://i.imgur.com/ZVOA0eO.jpg?1

Am I missing something about simplifying it?

 

[rcgldr]

You can convert the numerator and denominator each into fractions, each with their own common denominator and then multiply the numerator by the inverse of the denominator, or you can multiply the numerator and denominator by the least common multiple of their denominators and simplify from that point.

 

[Student100]

Some of your work got cut off, so I can't see it.

Think about it this way: You have two different mixed fractions, split them up for now x + \frac {2x}{x-2} and 1 + \frac {4}{x^2-4}

The first thing you're going to want to do is put everything in each mixed fraction over a common denominator. So you want something like \frac {x(?what goes here) + 2x} {x-2} and \frac {1(?) +4} {x^2-4}

After you have that done, then you can put it back into complex fraction form\frac {\frac {x(??)+2x}{x-2}}{\frac {1(?)+4}{x^2-4}}

Do you know what to do from there?

Fyi: This link is helpful : http://www.purplemath.com/modules/compfrac.htm

 

[Superkevs22]

Some of your work got cut off, so I can't see it.

Think about it this way: You have two different mixed fractions, split them up for now x + \frac {2x}{x-2} and 1 + \frac {4}{x^2-4}

The first thing you're going to want to do is put everything in each mixed fraction over a common denominator. So you want something like \frac {x(?what goes here) + 2x} {x-2} and \frac {1(?) +4} {x^2-4}

After you have that done, then you can put it back into complex fraction form\frac {\frac {x(??)+2x}{x-2}}{\frac {1(?)+4}{x^2-4}}

Do you know what to do from there?

Fyi: This link is helpful : http://www.purplemath.com/modules/compfrac.htm

Thanks for the response Student100 I really appreciate it! Let me get back to you, I'm going to try solving it again using your advice.

 

[Superkevs22]

You can convert the numerator and denominator each into fractions, each with their own common denominator and then multiply the numerator by the inverse of the denominator, or you can multiply the numerator and denominator by the least common multiple of their denominators and simplify from that point.

Thanks for the reply rcgldr! Let me get back to you too once I finish solving it again using yours and Student100's advice. Thank you so much!

 

[Superkevs22]

Some of your work got cut off, so I can't see it.

Think about it this way: You have two different mixed fractions, split them up for now x + \frac {2x}{x-2} and 1 + \frac {4}{x^2-4}

The first thing you're going to want to do is put everything in each mixed fraction over a common denominator. So you want something like \frac {x(?what goes here) + 2x} {x-2} and \frac {1(?) +4} {x^2-4}

After you have that done, then you can put it back into complex fraction form\frac {\frac {x(??)+2x}{x-2}}{\frac {1(?)+4}{x^2-4}}

Do you know what to do from there?

Fyi: This link is helpful : http://www.purplemath.com/modules/compfrac.htm

I finally did it Student100! Here's what I did:

http://i.imgur.com/P8DgFwQ.jpg

Thank you so much for the advice it really helped me a lot as well as the site you provided.

 

[Student100]

I finally did it Student100! Here's what I did:

http://i.imgur.com/P8DgFwQ.jpg

Thank you so much for the advice it really helped me a lot as well as the site you provided.

Glad we could help.

There is one more thing you should technically do, that may not be required for this problem but it's good practice. That is, what are the restrictions on values for X for which the function is undefined?

 

[Saitama]

You could have started with a well known identity.
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$
In your case a=x+y and b=x-y, substituting
$$(x+y+x-y)((x+y)^2-(x-y)(x+y)+(x-y)^2)=2x(x^2+y^2+2xy-x^2+y^2+x^2+y^2-2xy)$$
$$=2x(x^2+3y^2)$$

 

[Student100]

You could have started with a well known identity.
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$
In your case a=x+y and b=x-y, substituting
$$(x+y+x-y)((x+y)^2-(x-y)(x+y)+(x-y)^2)=2x(x^2+y^2+2xy-x^2+y^2+x^2+y^2-2xy)$$
$$=2x(x^2+3y^2)$$

Excellent Pranav, however, I'm a bit leery to suggest fancy things without a solid understanding of the basics. In this case, just multiplying them out by hand.