# Thread: factor theorem

1. ## factor theorem

Find all the real roots of the equation :

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

2. 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

3. 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

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

5. 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

6. 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 .

7. 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