# Is there an easier way to factor these polynomials?

1. Sep 19, 2006

### 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. $x^6 - 26x^3 - 27 = 0$

2. $(x^2 + 2x)^2 - (x^2 + 2x) -12 = 0$

For 1, I thought of doing this:

$x^3(x^3 - 26) - 27 = 0$

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

As for 2, I did this:

$x^4 + 4x^3 + 4x^2 - x^2 - 2x - 12 = 0$
$x^4 + 4x^3 + 3x^2 -2x -12 = 0$

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.

Last edited: Sep 19, 2006
2. Sep 19, 2006

### Integral

Staff Emeritus
1. Let t = x3 then solve for t.

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

3. Sep 19, 2006

### Byrgg

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

4. Sep 19, 2006

### Integral

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

5. Sep 19, 2006

### Byrgg

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

$t(t -26) - 27 = 0$
$t^2 - 26t - 27 = 0$

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

6. Sep 19, 2006

### Integral

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

7. Sep 19, 2006

:rofl:

(extra characters)

8. Sep 19, 2006

### Byrgg

I guess that means I should continue with my thinking?

$t^2 - 26 - 27 = 0$
$(t - 27)(t + 1) = 0$

$(t - 27) = 0$
$x^3 - 27 = 0$
$(x - 3)(x^2 + 3x + 9)$

$(t + 1) = 0$
$x^3 + 1 = 0$
$(x + 1)(x^2 - 1x + 1)$

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

9. Sep 20, 2006

### HallsofIvy

Yes, now put them back together!

10. Sep 20, 2006

### Byrgg

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.

11. Sep 20, 2006

### Integral

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