如何确定三次/四次实系数多项式的正实根数量？

Great question! Since you're working with real-coefficient cubics and quartics, and can already check the sign of your coefficients, we can use Descartes' Rule of Signs as our core tool, plus some checks for repeated roots and extreme values to narrow down exact counts. Note: Below, we cover counts both with multiplicity (repeated roots count as multiple roots) and distinct positive real roots, since both are common interpretations.

Let’s denote your cubic as ( P(x) = ax^3 + bx^2 + cx + d ) where ( a \neq 0 ). First, a quick simplification: if ( a < 0 ), multiply the entire polynomial by -1—this doesn’t change the roots, just flips the sign of the polynomial, so you can work with a positive leading coefficient to make sign counting easier.

Key Tool: Descartes' Rule of Signs for Cubics #

Count the number of sign changes in the coefficients of ( P(x) ) (skip any zero coefficients, moving left to right). Let this number be ( n_+ ):

• The number of positive real roots (with multiplicity) is either ( n_+ ), or ( n_+ - 2 ) (whichever is non-negative).
• Remember: A cubic always has at least one real root, so this helps eliminate impossible counts.

Exact Positive Root Counts #

0 positive real roots (distinct or with multiplicity)

• If your leading coefficient is positive, all non-zero coefficients ( b, c, d ) are non-negative, and ( d > 0 ): For any ( x > 0 ), ( P(x) = \text{positive} + \text{positive} + \text{positive} + \text{positive} > 0 ), so no positive roots exist.
• If ( d = 0 ) (x=0 is a root), check the remaining quadratic ( ax^2 + bx + c ): if it has no positive roots (e.g., ( b, c \geq 0 )), then you still have 0 positive roots.

1 positive real root (with multiplicity)

• Either: ( P(x) ) has exactly 1 sign change (Descartes’ Rule says this can only correspond to 1 positive root, since 1-2=-1 isn’t valid), and the other two roots are a conjugate complex pair.
• Or: ( P(x) ) has 3 sign changes, but the cubic’s discriminant is negative (meaning only one real root exists, which is positive).

2 positive real roots (with multiplicity)

This only happens when you have a double positive root plus one negative root. Descartes’ Rule will show 2 sign changes (so possible counts are 2 or 0), and checking the cubic’s discriminant will confirm a repeated root. For distinct roots, this counts as 1 unique positive root.

3 positive real roots (with multiplicity)

• ( P(x) ) has 3 sign changes, and the cubic’s discriminant is non-negative (all roots are real). In this case, all three roots are positive (e.g., ( (x-1)(x-2)(x-3) = x^3 -6x^2 +11x -6 ), which has 3 sign changes and three distinct positive roots).

Let your quartic be ( Q(x) = ax^4 + bx^3 + cx^2 + dx + e ) with ( a \neq 0 ). Again, normalize to a positive leading coefficient if needed.

Key Tool: Descartes' Rule of Signs for Quartics #

Count sign changes ( n_+ ) in ( Q(x) ):

• The number of positive real roots (with multiplicity) is ( n_+ ), ( n_+ - 2 ), ( n_+ -4 ), etc., down to 0.
• Quartics can have 0, 2, or 4 real roots total (since complex roots come in pairs), so this limits possible positive root counts.

Exact Positive Root Counts #

0 positive real roots (distinct or with multiplicity)

• Leading coefficient positive, all non-zero coefficients ( b, c, d, e ) non-negative, and ( e > 0 ): For ( x > 0 ), ( Q(x) ) is the sum of positive terms, so no positive roots.
• If ( e = 0 ), check the remaining cubic: if it has 0 positive roots, then you still have 0 positive roots here.

1 positive real root (with multiplicity)

This only occurs with a single positive root plus one negative root and a pair of conjugate complex roots. Descartes’ Rule will show exactly 1 sign change (since 1-2=-1 isn’t valid), confirming 1 positive root.

2 positive real roots (with multiplicity)

Two scenarios here:

1. Two distinct single positive roots plus a pair of conjugate complex roots: Descartes’ Rule will show 2 or 4 sign changes (we can rule out 4 by checking that not all roots are real).
2. One double positive root plus either two negative roots (single or double) or a pair of complex roots: Descartes’ Rule will show 2 sign changes.
   For distinct roots, both scenarios count as 1 or 2 unique positive roots, depending on whether there’s a repeated root.

3 positive real roots (with multiplicity)

This requires a triple positive root plus one negative root. Descartes’ Rule will show 3 sign changes (possible counts 3 or 1), and checking for repeated roots (via discriminant or factoring) confirms the triple root. For distinct roots, this counts as 1 unique positive root.

4 positive real roots (with multiplicity)

• ( Q(x) ) has 4 sign changes, and the quartic’s discriminant is non-negative (all roots are real). All four roots are positive (e.g., ( (x-1)(x-2)(x-3)(x-4) ), which has 4 sign changes and four distinct positive roots).

Extra Check for Exact Counts #

If you need to narrow down from Descartes’ possible counts, calculate the polynomial’s critical points (by taking the derivative and finding its roots) and evaluate ( P(x) ) or ( Q(x) ) at those points. The number of times the function crosses from positive to negative (or vice versa) in ( x > 0 ) will tell you the exact number of distinct positive roots. Since you can judge coefficient signs, you may even be able to deduce the sign of these critical point evaluations without full computation.

内容的提问来源于stack exchange，提问作者AzJ