如何求多项式函数的零点？运用代数基本定理求解$f(x)=x^3-5x^2-4x-20$

Hey Kate, great job working through the Rational Root Theorem first—since none of the constant term's factors worked, we know this cubic has no rational zeros. That means we're dealing with either one irrational real zero plus a pair of complex conjugate zeros, or three irrational real zeros (the discriminant will clarify which). Let's tie this to the Fundamental Theorem of Algebra (FTA) and solve it properly:

First, Recap the Fundamental Theorem of Algebra #

FTA tells us every degree-$n$ polynomial has exactly $n$ complex zeros (counting multiplicities). For your cubic, that means 3 total zeros—we just need to find their exact forms.

Step 1: Eliminate the Quadratic Term (Depress the Cubic) #

Cubic equations are easier to solve when we get rid of the $x^2$ term. For a general cubic $x^3 + bx^2 + cx + d = 0$, use the substitution:
x = y - b/3
Here, $b=-5$, so:
x = y + 5/3

Plug this into $f(x)$ and simplify to get a "depressed cubic" of the form $y^3 + py + q = 0$:
$$
\left(y+\frac{5}{3}\right)^3 -5\left(y+\frac{5}{3}\right)^2 -4\left(y+\frac{5}{3}\right) -20 = 0
$$

Expanding and combining like terms:

• The $y^2$ terms cancel out perfectly (that's the point of the substitution!)
• We end up with: $y^3 - \frac{37}{3}y - \frac{970}{27} = 0$
  Multiply through by 27 to eliminate fractions: $27y^3 - 333y - 970 = 0$

Step 2: Use the Cubic Discriminant to Classify Zeros #

For a depressed cubic $y^3 + py + q = 0$, the discriminant is:
$$
\Delta = \left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3
$$

Plugging in $p=-\frac{37}{3}$ and $q=-\frac{970}{27}$:
$$
\Delta = \left(-\frac{485}{27}\right)^2 + \left(-\frac{37}{9}\right)^3 = \frac{235225}{729} - \frac{50653}{729} = \frac{184572}{729} > 0
$$

Since $\Delta > 0$, FTA confirms we have one real zero and two complex conjugate zeros (no repeated roots here).

Step 3: Calculate the Real Zero with Cardano's Formula #

Cardano's formula gives the real root for a depressed cubic when $\Delta > 0$:
$$
y = \sqrt[3]{-\frac{q}{2} + \sqrt{\Delta}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\Delta}}
$$

Substitute our values:

• $-\frac{q}{2} = \frac{485}{27}$
• $\sqrt{\Delta} = \frac{\sqrt{184572}}{27} = \frac{2\sqrt{5127}}{9}$

So:
$$
y = \frac{1}{3}\left( \sqrt[3]{485 + 6\sqrt{5127}} + \sqrt[3]{485 - 6\sqrt{5127}} \right)
$$

Convert back to $x$ using our substitution $x = y + 5/3$:
$$
x = \frac{5}{3} + \frac{1}{3}\left( \sqrt[3]{485 + 6\sqrt{5127}} + \sqrt[3]{485 - 6\sqrt{5127}} \right)
$$

Step 4: Find the Complex Conjugate Zeros #

Once we have the real zero $r$, we can factor $f(x)$ as $(x - r)(x^2 + mx + n)$ where $m$ and $n$ are real numbers. You can do this via polynomial long division, or use Vieta's formulas (which pair nicely with FTA):

• The sum of all zeros equals $5$ (for $x^3 + bx^2 + cx + d$, sum of zeros is $-b$)
• The product of all zeros equals $20$ (product is $-d$)

Let the complex zeros be $s$ and $\overline{s}$ (their conjugate). Then:
$r + s + \overline{s} = 5$ → $s + \overline{s} = 5 - r$ (this is $-m$, the coefficient of $x$ in the quadratic)
$r \cdot s \cdot \overline{s} = 20$ → $r \cdot |s|^2 = 20$ → $|s|^2 = 20/r$ (this is $n$, the constant term)

Solve the quadratic $x^2 - (5 - r)x + (20/r) = 0$ using the quadratic formula:
$$
x = \frac{(5 - r) \pm \sqrt{(5 - r)^2 - 80/r}}{2}
$$

Since $(5 - r)^2 - 80/r$ will be negative (we know there are complex zeros), the square root will give an imaginary number, resulting in our pair of conjugate complex zeros.

Wrapping Up with FTA #

Notice we ended up with exactly 3 zeros (1 real, 2 complex), which aligns perfectly with the Fundamental Theorem of Algebra's guarantee for a cubic polynomial.

内容的提问来源于stack exchange，提问作者Kate