Solve the quartic equation 2x^4 - 9x^3 + 3x^2 + 11x - 3 = 0 and find all four roots.

Step-by-step solution

1. 1

   List the candidate rational roots. By the Rational Root Theorem any rational root $p/q$ has $p$ dividing the constant term $-3$ and $q$ dividing the leading coefficient $2$. That gives

   $\pm 1,\quad \pm 3,\quad \pm\tfrac12,\quad \pm\tfrac32$

   Eight numbers to test — far better than guessing.

2. 2

   Test them and find the first root. Writing $f(x)=2x^4-9x^3+3x^2+11x-3$:

   $f(1)=2-9+3+11-3=4,\qquad f(3)=162-243+27+33-3=-24$
   $f(-1)=2+9+3-11-3=0$

   So $x=-1$ is a root and $(x+1)$ is a factor. Note that $x=3$ is not a root even though the coefficients tempt you toward it — always evaluate rather than assume.

3. 3

   Divide out $(x+1)$ by synthetic division. Bringing down the coefficients $2,\,-9,\,3,\,11,\,-3$ and using $-1$:

   $2 \quad -11 \quad 14 \quad -3 \quad | \ 0$

   so

   $2x^4-9x^3+3x^2+11x-3=(x+1)\left(2x^3-11x^2+14x-3\right)$

   The remainder $0$ confirms the division; the quartic is now a cubic problem.

4. 4

   Find a root of the cubic from the same candidate list. For $g(x)=2x^3-11x^2+14x-3$:

   $g(1)=2,\qquad g(3)=-6,\qquad g\!\left(\tfrac12\right)=\tfrac32,\qquad g\!\left(\tfrac32\right)=\tfrac{27}{4}-\tfrac{99}{4}+21-3=0$

   So $x=\tfrac32$ is a root, contributing the factor $(2x-3)$. Dividing gives

   $2x^3-11x^2+14x-3=(2x-3)\left(x^2-4x+1\right)$

5. 5

   Solve the remaining quadratic. The last factor has no rational roots, so use the quadratic formula on $x^2-4x+1=0$:

   $x=\frac{4\pm\sqrt{16-4}}{2}=\frac{4\pm 2\sqrt3}{2}=2\pm\sqrt3$

6. 6

   Assemble and verify the factorisation. Altogether

   $2x^4-9x^3+3x^2+11x-3=(x+1)(2x-3)\left(x^2-4x+1\right)$

   Expanding the right-hand side reproduces the original coefficients $2,\,-9,\,3,\,11,\,-3$. A second check uses Vieta: the four roots sum to $-1+\tfrac32+(2-\sqrt3)+(2+\sqrt3)=\tfrac92$, which matches $-\frac{-9}{2}$, and their product is $-1\cdot\tfrac32\cdot(4-3)=-\tfrac32$, matching $\frac{-3}{2}$.