Is there an easier way to factor these polynomials?

  • Thread starter Thread starter Byrgg
  • Start date Start date
  • Tags Tags
    Polynomials
AI Thread Summary
The discussion revolves around finding easier methods to factor two polynomial equations. The first equation, x^6 - 26x^3 - 27 = 0, can be simplified by substituting t = x^3, transforming it into a quadratic form, t^2 - 26t - 27 = 0. The second equation, (x^2 + 2x)^2 - (x^2 + 2x) - 12 = 0, can also benefit from a similar substitution approach. Participants emphasize the importance of using substitution to simplify the equations and encourage continuing the process to find all solutions, including real roots. The conversation highlights the effectiveness of substitution as a shortcut method in polynomial factoring.
Byrgg
Messages
335
Reaction score
0
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:
Physics news on Phys.org
1. Let t = x3 then solve for t.

Think about what you just did then apply it to #2
 
Integral said:
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.
 
It is called a substitution. Let t= x3 then work in t, you introduce a new variable.
 
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?
 
Like a 5yr old learning to ride a bike. I have just given you a push... now... PEDAL.
 
Integral said:
Like a 5yr old learning to ride a bike. I have just given you a push... now... PEDAL.

:smile:


(extra characters)
 
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?
 
Yes, now put them back together!
 
  • #10
HallsofIVy said:
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.
 
  • #11
Are you looking for ALL solutions or just the ones in the Real numbers?
 
Back
Top