Solve 3x^4 + 18x^3 + 38x^2 + 32x + 12 = 0. Does it have any real solutions?

Step-by-step solution

1. 1

   Test the obvious candidates first — and reject them. The coefficients look as if the quartic might be $3(x+1)^{2}(x+2)^{2}$, and that guess is easy to check by substitution:

   $f(-1)=3-18+38-32+12=3,\qquad f(-2)=48-144+152-64+12=4.$

   Neither is zero, so $x=-1$ and $x=-2$ are not roots. Expanding the guess confirms why: $3(x+1)^{2}(x+2)^{2}=3x^{4}+18x^{3}+39x^{2}+36x+12$, which differs from the given polynomial in the $x^{2}$ and $x$ terms.

2. 2

   Rule out every rational root. By the rational root theorem any rational root is $\pm\frac{p}{q}$ with $p\mid 12$ and $q\mid 3$, i.e. one of $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12,\pm\tfrac13,\pm\tfrac23,\pm\tfrac43$. All coefficients are positive, so $f(x)>0$ for every $x\ge 0$ and only the negative candidates need testing — and each of them gives a nonzero value. There is no rational root, and in fact no rational quadratic factorisation either.

3. 3

   Complete the square at the quartic level. Look for a constant $a$ making $f(x)-3(x^{2}+3x+a)^{2}$ a positive-definite quadratic. Since

   $3(x^{2}+3x+a)^{2}=3x^{4}+18x^{3}+(27+6a)x^{2}+18ax+3a^{2},$

   the choice $a=\tfrac32$ leaves exactly

   $f(x)=3\left(x^{2}+3x+\tfrac32\right)^{2}+2x^{2}+5x+\tfrac{21}{4}.$

4. 4

   Show the leftover quadratic is always positive. For $2x^{2}+5x+\tfrac{21}{4}$ the discriminant is

   $25-4(2)\left(\tfrac{21}{4}\right)=25-42=-17<0,$

   and its minimum value is $\tfrac{21}{4}-\tfrac{25}{8}=\tfrac{17}{8}$. Since the first term is a square multiplied by $3$, it is never negative, so

   $f(x)\ge \tfrac{17}{8}>0\quad\text{for every real }x.$

   A quartic that never touches zero has no real roots.

5. 5

   Locate the four complex roots numerically. Because the coefficients are real the roots come in conjugate pairs. Solving numerically,

   $x\approx-2.384922\pm 0.678668i,\qquad x\approx-0.615078\pm 0.521776i.$

   Check against Vieta: the four roots sum to $-6=-\tfrac{18}{3}$ ✓ and their product is $4=\tfrac{12}{3}$ ✓. The corresponding real factorisation is $3(x^{2}+4.76984x+6.14844)(x^{2}+1.23016x+0.65057)$, neither factor having rational coefficients.