Is there an easier way to factor these polynomials?

[Byrgg]

I have two equations, I'm pretty sure I can solve them, but the method I know of is was too long, and I've never had to use very long methods to solve any problems so far. I'm wondering if there's some shortcut method that I don't know, or that I'm forgetting. Here are the equations(I have solve for x):

1. [itex]x^6 - 26x^3 - 27 = 0[/itex]

2. [itex](x^2 + 2x)^2 - (x^2 + 2x) -12 = 0[/itex]

For 1, I thought of doing this:

[itex]x^3(x^3 - 26) - 27 = 0[/itex]

That was all I could come up with, I'm not sure what to do next.

As for 2, I did this:

[itex]x^4 + 4x^3 + 4x^2 - x^2 - 2x - 12 = 0[/itex]
[itex]x^4 + 4x^3 + 3x^2 -2x -12 = 0[/itex]

That's as far as I could get, I figred I could get the factors by continuosly long dividing the polynomial by a factor I could find using the factor theore, but I didn't really want to have do that much work, isn't there a shortcut method? A way of grouping them or something? I'm wondering this about both of the equations. Thanks in advance.

 

[Integral]

1. Let t = x³ then solve for t.

Think about what you just did then apply it to #2

 

[Byrgg]

1. Let t = x3 then solve for t.

Think about what you just did then apply it to #2

What t are you talking about? I don't see a t in either of the problems.

 

[Integral]

It is called a substitution. Let t= x³ then work in t, you introduce a new variable.

 

[Byrgg]

Oh, I was reading it wrong, ok, so I do it like this?

[itex]t(t -26) - 27 = 0[/itex]
[itex]t^2 - 26t - 27 = 0[/itex]

And then what do I do, solve it like a quadratic?

 

[Integral]

Like a 5yr old learning to ride a bike. I have just given you a push... now... PEDAL.

 

[FrogPad]

Like a 5yr old learning to ride a bike. I have just given you a push... now... PEDAL.

(extra characters)

 

[Byrgg]

I guess that means I should continue with my thinking?

[itex]t^2 - 26 - 27 = 0[/itex]
[itex](t - 27)(t + 1) = 0[/itex]

[itex](t - 27) = 0[/itex]
[itex]x^3 - 27 = 0[/itex]
[itex](x - 3)(x^2 + 3x + 9)[/itex]

[itex](t + 1) = 0[/itex]
[itex]x^3 + 1 = 0[/itex]
[itex](x + 1)(x^2 - 1x + 1)[/itex]

Is that right so far? Do I just continue doing what I was doing?

 

[HallsofIvy]

Yes, now put them back together!

 

[Byrgg]

Yes, now put them back together!

What do you mean by "put them back together"? Don't I have to continue breaking them down to solve for the roots? Sorry if I misunderstood what you were saying.

 

[Integral]

Are you looking for ALL solutions or just the ones in the Real numbers?