Solve 2x^3 + 33x^2 + 180x - 300 = 0. How many real roots does it have, and what are they?

Step-by-step solution

1. 1

   Rule out rational roots. Candidates are $\pm\frac{p}{q}$ with $p\mid 300$ and $q\mid 2$. Testing the plausible small ones with $f(x)=2x^{3}+33x^{2}+180x-300$:

   $f(1)=2+33+180-300=-85,\quad f(2)=16+132+360-300=208,$

   $f(-5)=-250+825-900-300=-625,\quad f(-10)=-2000+3300-1800-300=-800.$

   None vanishes, and neither does any other candidate, so the cubic is irreducible over the rationals and cannot be factored by inspection.

2. 2

   Count the real roots from the discriminant. For $ax^{3}+bx^{2}+cx+d$ the discriminant is

   $\Delta=18abcd-4b^{3}d+b^{2}c^{2}-4ac^{3}-27a^{2}d^{2}=-42{,}120{,}000.$

   Because $\Delta<0$ there is exactly one real root and a conjugate pair of complex ones. That settles the counting question before any numerical work — a sweep of $f$ over $[-100,20]$ indeed shows a single sign change, near $x=1.32$.

3. 3

   Bracket and bisect the real root. From $f(1)=-85<0$ and $f(2)=208>0$ the root lies in $(1,2)$. Halving repeatedly and keeping the sign-changing half converges to

   $x\approx 1.321083731.$

   Check: $f(1.321083731)\approx-6\times10^{-14}$ ✓. A rougher quote such as $1.32$ leaves $f\approx-0.30$, so the extra digits matter if the root feeds into later work.

4. 4

   Get the exact value with Cardano. Depressing the cubic by $x=t-\frac{b}{3a}=t-\frac{11}{2}$ gives $t^{3}+pt+q=0$ with

   $p=\frac{3ac-b^{2}}{3a^{2}}=-\frac34,\qquad q=\frac{2b^{3}-9abc+27a^{2}d}{27a^{3}}=-\frac{1249}{4}.$

   Since $\frac{q^{2}}{4}+\frac{p^{3}}{27}=\frac{1560001}{64}-\frac{1}{64}=24375$ and $\sqrt{24375}=25\sqrt{39}$, Cardano gives

   $x=\sqrt[3]{\frac{1249}{8}+25\sqrt{39}}+\sqrt[3]{\frac{1249}{8}-25\sqrt{39}}-\frac{11}{2}.$

5. 5

   Verify the closed form at high precision. Evaluating that expression with 60-digit arithmetic returns

   $x=1.32108373084131024332679\ldots,$

   and substituting it back gives $f(x)\approx 2\times10^{-53}$ ✓. Note the second cube root is of the tiny number $\frac{1249}{8}-25\sqrt{39}\approx 5.004\times10^{-5}$, so evaluating this form in ordinary floating point loses accuracy — the bisection value is the safer one to quote in practice.