Math help i need to factor the symmetric quartic equation for x. x is a complex number. could u please help me

i need to factor the symmetric quartic equation for x. x is a complex number. could u please help me?
x^4-3x^3-2x^2-3x+1=0

Hello,

f(x) = x⁴ - 3x³ - 2x² - 3x + 1

If x=0,
f(0) = 0 - 0 - 0 - 0 + 1 = 1
Hence zero is not a root of this polynomial.

Since zero is not a root, we can divide the whole equation by x²:
x⁴ - 3x³ - 2x² - 3x + 1 = 0
x² - 3x - 2 - 3/x + 1/x² = 0

Let X=x+(1/x), then:
X = x + 1/x
X² = (x + 1/x)² = x² + 2 + 1/x²
x² + 1/x² = X² - 2

Then:
x² - 3x - 2 - 3/x + 1/x² = 0
(x² + 1/x²) - 3(x + 1/x) - 2 = 0
(X² - 2) - 3X - 2 = 0
X² - 3X - 4 = 0
X² - 4X + X - 4 = 0
X(X - 4) + (X - 4) = 0
(X + 1)(X - 4) = 0
X=-1 and X=4

Let's then solve for x:
X = x + 1/x = -1
x² + x + 1 = 0
∆ = 1-4 = -3 = (i√3)²
x = ½(-1 ± i√3) = exp(±2iπ/3)

X = x + 1/x = 4
x² - 4x + 1 = 0
∆ = 16 - 4 = 12 = (2√3)²
x = 2 ± √3

Hence the four complex roots of the quartic equation:
x₀ = ½(-1 + i√3) = exp(2iπ/3)
x₁ = ½(-1 - i√3) = exp(-2iπ/3)
x₂ = 2 - √3
x₃ = 2 + √3

And the resulting factorisation:
f(x) = x⁴ - 3x³ - 2x² - 3x + 1 = [x - exp(2iπ/3)][x - exp(-2iπ/3)](x - 2 + √3)(x - 2 - √3)

Methodically,
Dragon.Jade :-)

Hello, Guy. I am sorry. I cannot contact anyone by mail. You should ask your question here, and I might solve it for you if I come around to find it. Or better yet. Just apply the reasonning to your own equation. It's actually the same.

Regards,
Dragon.Jade :-)

Report Abuse

Hey Dragon,

I have a similar problem. Could you please help me?


ax^4+bx^3+cx^2+bx+a=0