We can always factor a quartic into two quadratics... as long as there are no stipulations on what the coefficients have to be.
We know that the quartic has real coefficients. This means that all complex solutions come in conjugate pairs. The complex roots can be grouped together as the roots of a quadratic with real coefficients. So any real quartic can be factored into quadratics with real coefficients, if you allow the coefficients to be arbitrary real numbers.
Since the lead coefficient is a 1 and since we can ALWAYS factor out a constant, which at worst case turns integer coefficients into rational coefficients, we can easily assume without a loss in generality that the quadratic factors have lead coefficients of 1. This reduces the number of variables we have to deal with. It doesnt lose generality since the other coefficients are arbitrary real numbers anyway.
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We can always factor a quartic into two quadratics... as long as there are no stipulations on what the coefficients have to be.
We know that the quartic has real coefficients. This means that all complex solutions come in conjugate pairs. The complex roots can be grouped together as the roots of a quadratic with real coefficients. So any real quartic can be factored into quadratics with real coefficients, if you allow the coefficients to be arbitrary real numbers.
Since the lead coefficient is a 1 and since we can ALWAYS factor out a constant, which at worst case turns integer coefficients into rational coefficients, we can easily assume without a loss in generality that the quadratic factors have lead coefficients of 1. This reduces the number of variables we have to deal with. It doesnt lose generality since the other coefficients are arbitrary real numbers anyway.
x⁴ + 2x³ - x² - 6x - 3 =
(x² + ax + b)(x² + cx + d)
Distribute it out.
x⁴ + (a + c)x³ + (b + ac + d)x² + (ad + bc)x + bd
So that we can compare coefficients.
We know that a+c = 2, so c = 2-a
x⁴ + 2x³ + (b + 2a - a² + d)x² + (ad + 2b - ab)x + bd
We know that b + 2a - a² + d = -1
a² - 2a - (1 + b + d) = 0
a = 1 ± √(2 + b + d)
x⁴ + 2x³ - x² + (ad + 2b - ab)x + bd
And we know that ad + 2b - ab = -6 and bd = -3
b = -3/d
ad - 6/d + 3a/d = -6
ad² - 6 + 3a + 6d = 0
d = [ -6 ± √(36 - 4a(3a-6) ) ] / 2a
d = [ -6 ± √(36 - 12a² + 24a) ] / 2a
You know a already. Now you can find d. From a and d you can find b and c, and you plug them into the quadratic factorization.
Basically we have four equations
a + c = 2
b + ac + d = -1
ad + bc = -6
bd = -3
In four unknown variables.