Expand an equation - sum and product

[Reveille]

Homework Statement

I have been sitting here for the last hour trying to figure it out but I can't seem to be able to find what I'm doing wrong.
I need to expand an equation.

Homework Equations

a² - a - 3

The Attempt at a Solution

a² - 1a - 3
The product is -3 and the sum -1.

-3 * 1 = -3
-1 * 3 = -3.
Both products are -3.
I can't seem to be able to find anything though of which the sum is also -1.
My sums are always either -2 or 2.

 

[Dick]

Homework Statement

I have been sitting here for the last hour trying to figure it out but I can't seem to be able to find what I'm doing wrong.
I need to expand an equation.

Homework Equations

a² - a - 3

The Attempt at a Solution

a² - 1a - 3
The product is -3 and the sum -1.

-3 * 1 = -3
-1 * 3 = -3.
Both products are -3.
I can't seem to be able to find anything though of which the sum is also -1.
My sums are always either -2 or 2.

That's not an equation and you aren't trying to expand it. What you appear to be trying to do is factor a quadratic. That quadratic doesn't factor in integers. Are you trying to solve a^2-a-3=0?

 

[SteamKing]

If you are trying to write the expression a^2 - a - 3 as the product of two monomial factors where everything has nice integer coefficients and such, well, good luck. There aren't any such factors. Sorry you killed an hour on this effort.

In the future, if the factors of the constant term can't produce the coefficient of the linear term, stop, don't waste your time like you did here.

 

[Reveille]

English isn't my native language so I mixed up the 2 things. I am indeed trying to factor a quadratic.
The full question is to simplify it as far as possible and make use of factoring a quadratic.
The full expression is: [itex]\frac{a^2+2a-8}{a^2-a-2}[/itex] / [itex]\frac{a^3+4a^2}{a^2-a-3}[/itex]

I managed to figure the first one out.
[itex]\frac{a^2+2a-8}{a^2-a-2}[/itex]

a²+2a-8
(a + 4)(a - 2)
Which is indeed a²+ 2a - 8

a² - 1a - 2
(a - 2)(a + 1)
a² + 1a - 2a - 2
a² - 1a - 2

[itex]\frac{a^2+2a-8}{a^2-a-2}[/itex] = [itex]\frac{(a+4)(a-2)}{(a-2)(a+1)}[/itex]
Now I can remove the (a-2) because they both have it.
So [itex]\frac{a^2+2a-8}{a^2-a-2}[/itex] = [itex]\frac{(a+4)}{(a+1)}[/itex]

I can't figure out the second part of the expression.

As my first problem is I have no clue how to do the a³+4a²
I decided to instead already solve the bottom. Which I also can't seem to figure out.

 

[Dick]

English isn't my native language so I mixed up the 2 things. I am indeed trying to factor a quadratic.
The full question is to simplify it as far as possible and make use of factoring a quadratic.
The full expression is: [itex]\frac{a^2+2a-8}{a^2-a-2}[/itex] / [itex]\frac{a^3+4a^2}{a^2-a-3}[/itex]

I managed to figure the first one out.
[itex]\frac{a^2+2a-8}{a^2-a-2}[/itex]

a²+2a-8
(a + 4)(a - 2)
Which is indeed a²+ 2a - 8

a² - 1a - 2
(a - 2)(a + 1)
a² + 1a - 2a - 2
a² - 1a - 2

[itex]\frac{a^2+2a-8}{a^2-a-2}[/itex] = [itex]\frac{(a+4)(a-2)}{(a-2)(a+1)}[/itex]
Now I can remove the (a-2) because they both have it.
So [itex]\frac{a^2+2a-8}{a^2-a-2}[/itex] = [itex]\frac{(a+4)}{(a+1)}[/itex]

I can't figure out the second part of the expression.

As my first problem is I have no clue how to do the a³+4a²
I decided to instead already solve the bottom. Which I also can't seem to figure out.

a³+4a² isn't quadratic but it has a simple factorization. Factor the common factor of ##a^2## out. ##a^2-a-3## won't factor. You'll just have to leave that as it is.

 

[HallsofIvy]

The equation [itex]a^2- a- 3= 0[/itex] has roots [tex]a= \frac{1+\sqrt{13}}{2}[/tex] and [itex]a= \frac{1- \sqrt{13}}{2}[/itex]. That means that [itex]a^2- a- 3[/itex] factors as
[tex]\left(a- \frac{1}{2}- \frac{\sqrt{13}}{2}\right)\left(a- \frac{1}{2}+ \frac{\sqrt{13}}{2}\right)[/tex]

That's probably doesn't help you!

 

[Reveille]

The equation [itex]a^2- a- 3= 0[/itex] has roots [tex]a= \frac{1+\sqrt{13}}{2}[/tex] and [itex]a= \frac{1- \sqrt{13}}{2}[/itex]. That means that [itex]a^2- a- 3[/itex] factors as
[tex]\left(a- \frac{1}{2}- \frac{\sqrt{13}}{2}\right)\left(a- \frac{1}{2}+ \frac{\sqrt{13}}{2}\right)[/tex]

That's probably doesn't help you!

I think that
[itex]a^2- a- 3[/itex]
is much simpler than the roots so I'll just leave that as it is but I appreciate your effort to help me.

I came up with:
[itex]\frac{a+4}{a+1}[/itex] / [itex]\frac{a^2(a+4)}{a^2-a-3}[/itex]

 

[Reveille]

a³+4a² isn't quadratic but it has a simple factorization. Factor the common factor of ##a^2## out. ##a^2-a-3## won't factor. You'll just have to leave that as it is.

Thank you! I took a look into that and it seems to work out.

 

[SammyS]

I think that
[itex]a^2- a- 3[/itex]
is much simpler than the roots so I'll just leave that as it is but I appreciate your effort to help me.

I came up with:
[itex]\frac{a+4}{a+1}[/itex] / [itex]\frac{a^2(a+4)}{a^2-a-3}[/itex]

Of course, that's far from being simplified. I hope you realize that.

 

[Reveille]

Of course, that's far from being simplified. I hope you realize that.

I do. :P
I just like to respond to other people so they know I have read their comments and let them know that I appreciate their efforts.

 

[SammyS]

I do. :P
I just like to respond to other people so they know I have read their comments and let them know that I appreciate their efforts.

There's no problem with that, in fact you are to commended for giving the feedback. Some folks just disappear after being helped, leaving no indication that they finally 'get it'.

However, often the person starting the thread gives the final result after being helped.

 

[Reveille]

There's no problem with that, in fact you are to commended for giving the feedback. Some folks just disappear after being helped, leaving no indication that they finally 'get it'.

However, often the person starting the thread gives the final result after being helped.

Woops, I thought I did. I forgot apparently.
My final answer on how to simplify it by factoring is
[itex]\frac{a+4}{a+1}[/itex] / [itex]\frac{a^2(a+4)}{a^2-a-3}[/itex]
The right part of the expression is as a result of common factoring. Both share a²
Instead of factoring the last one out with the roots, I decided to leave it as it was because of having to simplify.
The (a+4)(a+1) explanation is because a²+2a-8 equals (a+4)(a-2) and a²-2-a equals (a-2)(a+1). I can eliminate the (a-2) because they both have it. What remains is (a+4) and (a+1). Therefore a+4 / a+1

 

[SammyS]

Woops, I thought I did. I forgot apparently.
My final answer on how to simplify it by factoring is
[itex]\frac{a+4}{a+1}[/itex] / [itex]\frac{a^2(a+4)}{a^2-a-3}[/itex]
The right part of the expression is as a result of common factoring. Both share a²
Instead of factoring the last one out with the roots, I decided to leave it as it was because of having to simplify.
The (a+4)(a+1) explanation is because a²+2a-8 equals (a+4)(a-2) and a²-2-a equals (a-2)(a+1). I can eliminate the (a-2) because they both have it. What remains is (a+4) and (a+1). Therefore a+4 / a+1

[itex]\displaystyle {\frac{a+4}{a+1}\ }/{\ \frac{a^2(a+4)}{a^2-a-3}}[/itex] can be simplified to a single rational function. I'm quite certain that's what you are supposed to do.

There's still more '"cancelling" to be done after that.

 

[Reveille]

[itex]\displaystyle {\frac{a+4}{a+1}\ }/{\ \frac{a^2(a+4)}{a^2-a-3}}[/itex] can be simplified to a single rational function. I'm quite certain that's what you are supposed to do.

There's still more '"cancelling" to be done after that.

I have tried to do that but I kept failing on how to do so.
I have tried asking other people, they couldn't figure it out either.
As I already had to send my answer in, I just went with what I had.
Mind telling me how to do simplify it further for future purposes?

 

[SammyS]

I have tried to do that but I kept failing on how to do so.
I have tried asking other people, they couldn't figure it out either.
As I already had to send my answer in, I just went with what I had.
Mind telling me how to do simplify it further for future purposes?

I would write it as a compound (a.k.a. complex fraction) first:

[itex]\displaystyle \frac{\displaystyle \frac{a+4}{a+1}}{\displaystyle \ \frac{a^2(a+4)}{a^2-a-3}\ }[/itex]

Then multiply the numerator & denominator of the main fraction by something to cancel both sub-denominators or else by something to cancel the entire denominator of the main fration.