Solve 3x^4 + 24x^3 + 82x^2 + 136x + 63 = 0.

Step-by-step solution

1. 1

   Set up the two-quadratic ansatz with the leading 3 in one factor. Since $3$ is prime, any factorisation over the integers must look like

   $3x^{4}+24x^{3}+82x^{2}+136x+63=\left(3x^{2}+ax+b\right)\left(x^{2}+cx+d\right)$

   with $bd=63$. Expanding gives the four matching conditions

   $3c+a=24,\quad 3d+ac+b=82,\quad ad+bc=136,\quad bd=63$

2. 2

   Search the divisor pairs of 63. Trying $b=7,\ d=9$ and solving the first and third conditions gives $a=12,\ c=4$; the second condition then reads $27+48+7=82$ ✓ and the third $108+28=136$ ✓. All four hold simultaneously, so

   $3x^{4}+24x^{3}+82x^{2}+136x+63=\left(3x^{2}+12x+7\right)\left(x^{2}+4x+9\right)$

3. 3

   Always expand a guessed factorisation back before using it. This is not optional bookkeeping: a plausible-looking guess such as $3\left(x^{2}+6x+3\right)\left(x^{2}+6x+7\right)$ expands to $3x^{4}+36x^{3}+138x^{2}+180x+63$, which matches only the first and last coefficients and is therefore wrong. Checking the two middle coefficients is what catches it.

4. 4

   Solve the first quadratic. For $3x^{2}+12x+7=0$:

   $x=\frac{-12\pm\sqrt{144-84}}{6}=\frac{-12\pm\sqrt{60}}{6}=\frac{-12\pm 2\sqrt{15}}{6}=\frac{-6\pm\sqrt{15}}{3}$

   Numerically $x\approx-0.70901$ and $x\approx-3.29099$; both are real since the discriminant $60$ is positive.

5. 5

   Solve the second quadratic. For $x^{2}+4x+9=0$ the discriminant is $16-36=-20<0$, so

   $x=\frac{-4\pm\sqrt{-20}}{2}=-2\pm i\sqrt5$

6. 6

   Collect all four roots and verify numerically.

   $x=\frac{-6+\sqrt{15}}{3},\ \frac{-6-\sqrt{15}}{3},\ -2+i\sqrt5,\ -2-i\sqrt5$

   A numerical root finder on the original quartic returns $-0.70901$, $-3.29099$ and $-2\pm 2.23607i$, matching exactly ✓ ($\sqrt5=2.23607$).