Is it possible for a degree-4 polynomial P(x) to have only complex (no real) roots
work backwards from 4 imaginary roots (remember they come in complex conjugate pairs)
(x-i)(x+i)(x-2i)(x+2i) = 0
(x^2+1)(x^2+4)= 0
x^4 + 5 x^2 + 4 = 0
Is it possible for a degree-4 polynomial P(x) to have only complex (no real) roots?
Yes, it is possible for a degree-4 polynomial to have only complex roots (with no real roots). In fact, there is a simple rule known as the "Fundamental Theorem of Algebra" which states that a polynomial of degree n will have exactly n complex roots (counting multiplicities).
If you want to determine whether a degree-4 polynomial has only complex roots, you can use the discriminant. For a general degree-n polynomial of the form P(x) = aₙ xⁿ + aₙ₋₁ xⁿ⁻¹ + ... + a₁ x + a₀, the discriminant is defined as:
Δ = (aₙ₋₁)² - 4aₙ(aₙ₋₂aₙ₋₃ - aₙ₋₁²) - 4aₙ₋₃aₙ₋₄ + 18aₙ₋₂a²ₙ₋₄ - 27a²ₙ₋₃².
For a degree-4 polynomial, if the discriminant Δ is negative, then all of its roots are complex (no real roots). This is because complex roots come in conjugate pairs, and if the discriminant is negative, it means that the pairs cannot be real.
So, to determine if a degree-4 polynomial has only complex roots, compute the discriminant Δ and check if it is negative. If it is negative, then the polynomial has only complex roots.
Note that if the discriminant is zero or positive, the polynomial may have some combination of real and complex roots.