Finding Imaginary Roots

[barson90]

Find all roots [meaning both real and complex].
f(x)=4x^5 + 2x^3 - 3x

Factor
f(x)=x(4x^4 + 2x^2 - 3)

Quadratic Formula (Modified to fit the fourth degree equation)

x^2 = [-b +/- sqrt (b^2 - 4ac)] / 2a

Let a = 4, b = 2, c = -3


x^2 = {-2 +/- sqrt [4 - 4(4)(-3)]} / 2(4)

x^2 = [-1 + sqrt (13)] / 4

Therefore, the real roots are
x = 0 from the factoring above
x = {+/- sqrt [-1 + sqrt (13)] } / 2 from the quadratic formula
I checked these on a calculator.

Now, here's the real question. Trying to use synthetic division to factor the equation to a quadratic, from which the imaginary conjugate can be found, is a complete mess.

How can I divide 4x^4 + 2x^2 - 3 by {+/- sqrt [-1 + sqrt (13)] } / 2 to "factor" it down to x times x - [sqrt [-1 + sqrt (13)] / 2] times x + [sqrt [-1 + sqrt (13) /2] times a second degree factor?
In other words, do I simply have to put up with the messy synthetic division, or is there a simpler approach?

 

[tkhunny]

Ummm...

1) There is a way to use synthetic (sort of) division with a quadratic factor, but I cannot name a sane individual who has used this method to any value in the last 30 years.

2) You already have all the roots. Why are you now trying to factor it?

3) This nearly is disturbing. Please never write it again. How were you planning to do that division with TWO divisors?

How can I divide 4x^4 + 2x^2 - 3 by {+/- sqrt [-1 + sqrt (13)] } / 2 to

 

[Denis]

3) This nearly is disturbing. Please never write it again. How were you planning to do that division with TWO divisors?

Using 2 pencils?

 

[barson90]

3) This nearly is disturbing. Please never write it again. How were you planning to do that division with TWO divisors?

Using 2 pencils?

Factoring was merely to find real root "zero."

And the two divisors ... the point is to factor the equation by dividing by one of divisors and then dividing by the second divisor, thus factoring the equation. (Is this incorrect?)


Also ... I do not have all the roots. Notice, the problem calls for all roots, both real and imaginary. I found the real roots, as anyone with a graphing calculate can approximate by viewing the graph. I was attempting to use synthetic division to factor out the quadratic term in which the two conjugate imaginary roots can be factored.

It's a fifth degree equation. Therefore, there are five roots. I found the three real roots, therefore, there must be two nonreal (imaginary) roots.

 

[tkhunny]

No, you already found all five. You just don't seem to know it.

From here...

4x^4 + 2x^2 - 3 = 0

You used the quadratic formula to produce x^2 = {two different things}.

Solve each for 'x', rather than x^2. You should find all four remaining roots in there.