factor theorem

**thereddevils** · Jun 10th 2009, 11:44 PM

Find all the real roots of the equation :

$ 2x^6+x^4-8x^2+4=0 $

**Amer** · Jun 11th 2009, 01:05 AM

Originally Posted by thereddevils

Find all the real roots of the equation :

$ 2x^6+x^4-8x^2+4=0 $

the possible root is $\pm 2 , \pm 1,\pm 4 , \pm \frac{1}{2}$

but no one of the possible root is a root of the polynomial that you wrote

there is no real root can be found without calculator see what I have when I put them in the calculator I think there is wrong in your equation

Edited : - correct something yeongil mentioned

**yeongil** · Jun 11th 2009, 01:17 AM

Originally Posted by Amer

the possible root is $\pm 2 , \pm 1,\pm 4$

Minor correction: the possible rational roots are $\pm\frac{1}{2}, \pm 1, \pm 2, \pm 4$ .

And as you discovered, none of the possible rational roots are roots of the polynomial.

01

**thereddevils** · Jun 11th 2009, 05:36 AM

Btw the answer is $\pm\sqrt{2}$

**yeongil** · Jun 11th 2009, 06:37 AM

Originally Posted by thereddevils

Btw the answer is $\pm\sqrt{2}$

No, they are not. If you plug either $\sqrt{2}$ or $-\sqrt{2}$ into the original equation:
$2x^6+x^4-8x^2+4=0$
you get 8 = 0, so neither solution works.

However, if your original equation was supposed to be
$2x^6 + x^4 - 8x^2 {\color{red}-} 4 = 0$
then yes, those two solutions work. In fact, it makes solving this equation much easier, as you can solve it by factoring by grouping:
$\begin{aligned} 2x^6 + x^4 - 8x^2 - 4 &= 0 \\ x^4(2x^2 + 1) - 4(2x^2 + 1) &= 0 \\ (x^4 - 4)(2x^2 + 1) &= 0 \\ (x^2 - 2)(x^2 + 2)(2x^2 + 1) &= 0 \end{aligned}$

Set each factor equal to zero:

$\begin{aligned} x^2 - 2 &= 0 \\ x^2 &= 2 \\ x &= \pm \sqrt{2} \end{aligned}$
(the two real solutions)

$\begin{aligned} x^2 + 2 &= 0 \\ x^2 &= -2 \\ x &= \pm i\sqrt{2} \end{aligned}$
(two non-real solutions)

$\begin{aligned} 2x^2 + 1 &= 0 \\ 2x^2 &= -1 \\ x^2 &= -\frac{1}{2} \\ x &= \pm \frac{\sqrt{2}}{2}i \end{aligned}$
(two more non-real solutions)

So it looks like you copied the problem wrong?

01

**thereddevils** · Jun 11th 2009, 06:50 AM

Originally Posted by yeongil

No, they are not. If you plug either $\sqrt{2}$ or $-\sqrt{2}$ into the original equation:
$2x^6+x^4-8x^2+4=0$
you get 8 = 0, so neither solution works.

However, if your original equation was supposed to be
$2x^6 + x^4 - 8x^2 {\color{red}-} 4 = 0$
then yes, those two solutions work. In fact, it makes solving this equation much easier, as you can solve it by factoring by grouping:
$\begin{aligned} 2x^6 + x^4 - 8x^2 - 4 &= 0 \\ x^4(2x^2 + 1) - 4(2x^2 + 1) &= 0 \\ (x^4 - 4)(2x^2 + 1) &= 0 \\ (x^2 - 2)(x^2 + 2)(2x^2 + 1) &= 0 \end{aligned}$

Set each factor equal to zero:

$\begin{aligned} x^2 - 2 &= 0 \\ x^2 &= 2 \\ x &= \pm \sqrt{2} \end{aligned}$

$\begin{aligned} x^2 + 2 &= 0 \\ x^2 &= -2 \\ x &= \pm i\sqrt{2} \end{aligned}$

$\begin{aligned} 2x^2 + 1 &= 0 \\ 2x^2 &= -1 \\ x^2 &= -\frac{1}{2} \\ x &= \pm \frac{\sqrt{2}}{2}i \end{aligned}$

So it looks like you copied the problem wrong?

01

oh thanks a lot .. nope , i copied it down correctly .. i think its the book's mistake

Btw i hv a question here .. For questions like this where the leading coefficient is not 1 , is it a must for me to reduce the coefficient to 1 before i find the factors : example in this case

$2x^6+x^4-8x^2+4=0$

$x^6+\frac{1}{2}x^4-4x^2+2=0$ , is this step necessary

Before i start substituting the factors of the constant term to find the factors of the equation .

THanks .

**great_math** · Jun 12th 2009, 03:44 AM

Originally Posted by thereddevils

oh thanks a lot .. nope , i copied it down correctly .. i think its the book's mistake

Btw i hv a question here .. For questions like this where the leading coefficient is not 1 , is it a must for me to reduce the coefficient to 1 before i find the factors : example in this case

$2x^6+x^4-8x^2+4=0$

$x^6+\frac{1}{2}x^4-4x^2+2=0$ , is this step necessary

Before i start substituting the factors of the constant term to find the factors of the equation .

THanks .

Absolutely not.. It will just make your work difficult

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