Finding Solutions (Polynomials)

[adr8]

I am asked to find all the solutions for the following problem:

2x^6-x^4-2x^2+1=0


I used the cubic formula to factor this equation. I so far have (x+1) (2x^5-2x^4+x^3-x^2-x+1). This part came out right, but what was making me wonder can I factor out (2x^5-2x^4+x^3-x^2-x+1)? The reason why I ask is because I was using the cubic formula.

[INDENT][U][SIZE=3] 2x^4-4x^3+5x^2-6x

[/SIZE][/U][/INDENT]
[SIZE=3]x+1   2x^5-2x^4+x^3-x^2-x+1
[/SIZE]
      [SIZE=3]-[/SIZE][SIZE=3][U](2x^5+2x^4)  
[/U]              -4x^4+x^3
            [/SIZE][U][SIZE=3]-(-4x^4-4x^3)[/SIZE][/U][SIZE=3]

                       5x^3-x^2 [/SIZE]
[SIZE=3]                     [U]-(5x^3+5x^2)[/U][/SIZE][SIZE=3] 
                                  -6x^2+x
                                  [U]-(6x^2+6x)[/U][/SIZE][INDENT][SIZE=3]                                   -5x+1[/SIZE][/INDENT]

I tried making the division sign but I needed the little bar to complete the sign.​

​

 

[Deleted member 4993]

I am asked to find all the solutions for the following problem:

2x^6-x^4-2x^2+1=0


I used the cubic formula to factor this equation. I so far have (x+1) (2x^5-2x^4+x^3-x^2-x+1). This part came out right, but what was making me wonder can I factor out (2x^5-2x^4+x^3-x^2-x+1)? The reason why I ask is because I was using the cubic formula

[INDENT][SIZE=3][U]2x^4-4x^3+5x^2-6x[/U][/SIZE][/INDENT]
[SIZE=3]x+1                                  2x^5-2x^4+x^3-x^2-x+1[/SIZE][INDENT][SIZE=3]-[/SIZE][SIZE=3][U](2x^5+2x^4)  
[/U]               -4x^4+x^3
              [/SIZE][U][SIZE=3]-(-4x^4-4x^3)[/SIZE][/U]
[SIZE=3]                                                         5x^3-x^2 [/SIZE]
[SIZE=3]                                                     [U]-(5x^3+5x^2)[/U][/SIZE][SIZE=3] 
                                            -6x^2+x
                                            [U]-(6x^2+6x)[/U][/SIZE][INDENT][SIZE=3]                                                                                                       -5x+1[/SIZE]
[/INDENT]
[/INDENT]

I tried making the division sign but I needed the little bar to complete the sign.​

​

First I would make the sbstitution:

u = x²

Then the equation (2x^6-x^4-2x^2+1=0) becomes:

2u³ - u² - 2u + 1 = 0

By observation, u = 1 is a solution.

(u-1)(2u² + u - 1) = 0

(u-1)(u+1)(2u -1) = 0

Now continue......

 

[adr8]

Ok I got it, I know where I messed up. Thanks

 

[lookagain]

Here's another approach:

I am asked to find all the solutions for the following problem:

2x^6 - x^4 - 2x^2 + 1 = 0

Alternative
-----------


Try seeing if the "factoring by grouping" method may be applied.


Factor the greatest common factor from the first two terms:


\(\displaystyle x^4(2x^2 - 1) - 2x^2 + 1 = 0\)


If this factor by grouping is to work, then the \(\displaystyle (2x^2 - 1)\)
factor must also be present in the last two terms. Leave spaces for
an addition or subtraction sign, coefficients and variables as part
of another g.c.f:


\(\displaystyle x^4(2x^2 - 1) \ ????? \ (2x^2 - 1) = 0\)


What multiplied by \(\displaystyle 2x^2 \ gives \ -2x^2 \ ? \ \ Answer: \ \ -1\)


Check this: Does -1 multiplied by the -1 in this alleged common factor
equal +1, the last term in the original equation? Answer: Yes.


Write this minus one as a subtraction sign next to a "1."
Then follow this up with the factor, \(\displaystyle (2x^2 - 1):\)


Then the factorization up to that point is:


\(\displaystyle x^4(2x^2 - 1) - 1(2x^2 - 1) = 0\)


Factor out this common factor:


\(\displaystyle (2x^2 - 1)(x^4 - 1) = 0\)


The first binomial factor does not factor over the integers.
The second factor is a difference of two squares, and they
can be further factored as this:


\(\displaystyle (2x^2 - 1)(x^2 - 1)(x^2 + 1) = 0\)


The second factor from this result is also a difference
of two squares, but the third factor is prime
(over the integers).


\(\displaystyle (2x^2 - 1)(x - 1)(x + 1)(x^2 + 1) = 0\)


\(\displaystyle \text{Hint:}\)

Set each of these four factors to zero and solve for the x-values.
Consider the root method for a couple of these, as well as there
being imaginary solutions to at least one of the factors set equal
to zero.

 

[adr8]

Thanks, I will write down your method because sometimes this cubic formula can be quite confusing. I got -1,+1, and squrt 1/2 as my solutions .

 

[lookagain]

Thanks, I will write down your method because sometimes this cubic formula can be quite confusing. I got -1,+1, and squrt 1/2 as my solutions .

adr8, the instructions state to find all solutions, so that would include
imaginary ones also.

\(\displaystyle 2x^2 - 1 = 0 \ \ becomes\)

\(\displaystyle 2x^2 = 1 \)

\(\displaystyle x^2 = \frac{1}{2}\)

\(\displaystyle x = \pm\sqrt{\frac{1}{2}} \ \ becomes\)

\(\displaystyle x = \pm\frac{1}{\sqrt{2}} \ \ becomes\)

\(\displaystyle x = \pm\frac{\sqrt{2}}{2}\)

----------------------------------------------------


\(\displaystyle x^2 + 1 = 0\)

\(\displaystyle x^2 = -1\)

\(\displaystyle x = \pm\sqrt{-1}\)

\(\displaystyle x = \pm i\)

----------------------------------------------------

Along with x = -1. x = 1, the solutions above
make a total of six solutions for this 6th degree
polynomial set equal to zero.