1. ## Having trouble factoring

Hey guys..I'm having trouble factoring some functions...I can pretty easily factor and find the solutions for linear and quadratic functions, but when it comes to cubic and..um..higher-powered functions, I'm pretty lost...

The first problem is:

x^3 + 216 = 0

I know to start I should set 216 to its cubic form,

x^3 + 6^3 = 0, but I have no idea what to do after that...

And for the next problem I have

x^4 - x^3 + x - 1 = 0

I worked it out like this:

x^3(x - 1) + 1 (x - 1 )= 0
(x^3+1) (x-1) = 0

so I solve for x

x^3 + 1 = 0
(no solution) and

x-1 = 0
x=1....but the answer says its x = + or - 1...So I definitely did something wrong...

So if anyone could help me I'd greatly appreciate it

2. Originally Posted by jonjon1324
Hey guys..I'm having trouble factoring some functions...I can pretty easily factor and find the solutions for linear and quadratic functions, but when it comes to cubic and..um..higher-powered functions, I'm pretty lost...

The first problem is:

x^3 + 216 = 0

I know to start I should set 216 to its cubic form,

x^3 + 6^3 = 0, but I have no idea what to do after that... make use of the sum of cubes formula!

And for the next problem I have

x^4 - x^3 + x - 1 = 0

I worked it out like this:

x^3(x - 1) + 1 (x - 1 )= 0
(x^3+1) (x-1) = 0

so I solve for x

x^3 + 1 = 0
(no solution) and What's (-1)^3?!

x-1 = 0
x=1....but the answer says its x = + or - 1...So I definitely did something wrong...

So if anyone could help me I'd greatly appreciate it
.

3. "x^3 + 6^3 = 0, but I have no idea what to do after that... make use of the sum of cubes formula!"

Okay so the sum of cubes formula is (x+y)(x^2 - xy + y^2) right?
So if I use it I get (x+6)(x^2-6x+36) ....is it normal for me not to be able to factor the
(x^2-6x+36) part? I should use the quadratic formula for that right?

4. Originally Posted by jonjon1324
"x^3 + 6^3 = 0, but I have no idea what to do after that... make use of the sum of cubes formula!"

Okay so the sum of cubes formula is (x+y)(x^2 - xy + y^2) right?
So if I use it I get (x+6)(x^2-6x+36) ....is it normal for me not to be able to factor the
(x^2-6x+36) part? I should use the quadratic formula for that right?
You won't be able to factorise it any further... check the discriminant, it will be negative...

5. Oh okay..so x is just -6.

and

x^3 + 1 = 0
(no solution) and What's (-1)^3?!

You're right..I forgot you COULD find the cubed root of a negative number...it's the square root you can't find..Okay I get it now...Thanks :-D

Can anyone give me any hints on how to start this problem?

X^4 - 4x^2 + 3 = 0

6. Originally Posted by jonjon1324
Oh okay..so x is just -6.

and

x^3 + 1 = 0
(no solution) and What's (-1)^3?!

You're right..I forgot you COULD find the cubed root of a negative number...it's the square root you can't find..Okay I get it now...Thanks :-D

Can anyone give me any hints on how to start this problem?

X^4 - 4x^2 + 3 = 0
I assume that it's $x^4 - 4x^2 + 3 = 0$.

I would solve this using a dummy variable, $X = x^2$.

So now you get a quadratic

$X^2 - 4X + 3 = 0$

$(X - 3)(X - 1) = 0$

$X - 3 = 0$ or $X - 1 = 0$

$X = 3$ or $X = 1$.

Now solve for $x$, remembering that $X = x^2$.

Case 1:

$x^2 = 3$

$x = \pm \sqrt{3}$.

Case 2:

$x^2 = 1$

$x = \pm 1$.

$x = \{-\sqrt{3}, -1, 1, \sqrt{3}\}$.

7. Oh yeah I had forgotten about the whole substitution thing...this makes more sense now.

So would substitution be appropriate to use in a problem such as
x^3 - 6x^2 - 9x = 0? Because I'm not seeing a way lol

When exactly is it useful to substitute for polynomials and when is it not?

8. Originally Posted by jonjon1324
Oh yeah I had forgotten about the whole substitution thing...this makes more sense now.

So would substitution be appropriate to use in a problem such as
x^3 - 6x^2 - 9x = 0? Because I'm not seeing a way lol

When exactly is it useful to substitute for polynomials and when is it not?
Try taking out a common factor of $x$ first...

9. Okay now I feel really dumb about not seeing that last one lol

I'm trying to also do substitution here but it doesn't seem possible:

3x(x-1)^1/2 + 2 (x-1)^3/2 = 0

It's not coming out right so I'm thinking I'm not supposed to use substitution?

10. Originally Posted by jonjon1324
Okay now I feel really dumb about not seeing that last one lol

I'm trying to also do substitution here but it doesn't seem possible:

3x(x-1)^1/2 + 2 (x-1)^3/2 = 0

It's not coming out right so I'm thinking I'm not supposed to use substitution?
Note that $(x - 1)^{3/2} = (x - 1)^{1/2} (x - 1)$. So take out the common factor of $(x - 1)^{1/2}$.

And please start a new thread for a new question next time.

11. x^3 - 6x^2 - 9x = 0,

x(x^2 - 6x - 9) = 0,

a quadratic formula awaits you . . . .

12. Originally Posted by mr fantastic
Note that $(x - 1)^{3/2} = (x - 1)^{1/2} (x - 1)$. So take out the common factor of $(x - 1)^{1/2}$.

And please start a new thread for a new question next time.

But I don't get what you did...What happened to the 3x? and the 2? I'm confused

13. Originally Posted by pacman
x^3 - 6x^2 - 9x = 0,

x(x^2 - 6x - 9) = 0,

a quadratic formula awaits you . . . .
What did you do? I don't know how you got that

14. Originally Posted by jonjon1324
[snip]
I'm trying to also do substitution here but it doesn't seem possible:

3x(x-1)^1/2 + 2 (x-1)^3/2 = 0

It's not coming out right so I'm thinking I'm not supposed to use substitution?
Originally Posted by mr fantastic
Note that $(x - 1)^{3/2} = (x - 1)^{1/2} (x - 1)$. So take out the common factor of $(x - 1)^{1/2}$.
And please start a new thread for a new question next time. Mr F adds: If you look over this thread I'm sure one of the reasons why this is required is becoming more and more obvious as the thread becomes more and more convoluted and difficult to follow. (And it gets more and more difficult to post replies in such a way as to try and avoid that problem).
Originally Posted by jonjon1324

But I don't get what you did...What happened to the 3x? and the 2? I'm confused
I have told you how to re-write one of the terms in your expression so that it makes it easier for you to see what the common factor is.

15. Originally Posted by mr fantastic
I have told you how to re-write one of the terms in your expression so that it makes it easier for you to see what the common factor is.
Oh okay so that's not the actual problem, that's you showing what to factor out..I get it...So I tried it, and I think I did it correctly..can you let me know if I did?

3x(x-1)^1/2 + 2(x-1)^3/2 = 0
I factor out (x-1)^1/2

(x-1)^1/2 (3+2(x-1)) = 0
(x-1)^1/2 (3+2x-1) = 0
(x-1)^1/2 (1+2x) = 0

I then squared everything to get rid of the ^1/2
and got
(x-1) (1+4x+4x^2) = 0

x-1 = 0, x=1

and factored 4x^2+4x+1 = 0 t get x = -1/2, but I plugged it in the original equation and found that it didn't work out...

Did I do this correctly?

Edit: and would it be better to start a new thread for every question I have, or a thread for similar questions? Because I have a lot of questions but I don't wanna fill the whole first page with my questions

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