# find all real and imaginary roots

• Sep 29th 2009, 06:29 AM
absvalue
find all real and imaginary roots
I need to find all real and imaginary roots of this equation:

$\displaystyle 2x^7 - 2x^6 + 7x^5 - 7x^4 - 4x^3 + 4x^2 = 0$

Using synthetic division, I found that 1 is a root.

$\displaystyle (x - 1)(2x^6 + 7x^4 - 4x^2) = 0$

I'm not really sure how to proceed from here. Could someone point me in the right direction?
• Sep 29th 2009, 06:56 AM
Soroban
Hello, absvalue!

Quote:

Find all real and imaginary roots: .$\displaystyle 2x^7 - 2x^6 + 7x^5 - 7x^4 - 4x^3 + 4x^2 \:=\: 0$
First, factor out $\displaystyle x^2$ . . .

. . $\displaystyle x^2(2x^5 - 2x^4 + 7x^3 - 7x^2 - 4x + 4) \:=\:0$

Factor by grouping:

. . $\displaystyle x^2\,\bigg[2x^4(x-1) + 7x^2(x-1) - 4(x-1)\bigg] \:=\:0$

. . $\displaystyle x^2(x-1)(2x^4 - 7x^2 - 4) \:=\:0$

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

Then we have:

. . $\displaystyle x^2 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:0}$

. . $\displaystyle x - 1 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:1}$

. . $\displaystyle x^2 - 4 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:4 \quad\Rightarrow\quad\boxed{ x \:=\:\pm2}$

. . . . $\displaystyle 2x^2+1 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:-\frac{1}{2} \quad\Rightarrow\quad\boxed{x \:=\:\pm\frac{i}{\sqrt{2}}}$

• Sep 29th 2009, 07:01 AM
Hello absvalue
Quote:

Originally Posted by absvalue
I need to find all real and imaginary roots of this equation:

$\displaystyle 2x^7 - 2x^6 + 7x^5 - 7x^4 - 4x^3 + 4x^2 = 0$

Using synthetic division, I found that 1 is a root.

$\displaystyle (x - 1)(2x^6 + 7x^4 - 4x^2) = 0$

I'm not really sure how to proceed from here. Could someone point me in the right direction?

Take out a common factor $\displaystyle x^2$ from your second factor; you then have a quadratic in $\displaystyle x^2$, which can be factorised in the usual way:

$\displaystyle (x-1)(2x^6+7x^4-4x^2) = x^2(x-1)(2x^4+7x^2-4)$

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

Can you complete it from here?