Solve the cubic equation 2x³ + 3x² − 8x + 3 = 0.

Step-by-step solution

1. 1

   List the candidate rational roots. For $a_nx^n + \dots + a_0$ any rational root is $\pm\frac{p}{q}$ with $p \mid a_0 = 3$ and $q \mid a_n = 2$. So the full candidate list is

   $\pm 1,\ \pm 3,\ \pm\frac12,\ \pm\frac32$

   Because the leading coefficient is $2$ and not $1$, the halves must be tested — skipping them is how this cubic gets mistaken for irreducible.

2. 2

   Test the candidates and find one that works. Trying $x = \tfrac12$:

   $2\left(\tfrac18\right) + 3\left(\tfrac14\right) - 8\left(\tfrac12\right) + 3 = \tfrac14 + \tfrac34 - 4 + 3 = 0$

   So $x = \tfrac12$ is a root, and by the factor theorem $\left(x - \tfrac12\right)$ — equivalently the integer factor $(2x - 1)$ — divides the cubic.

3. 3

   Divide out the known factor. Polynomial (or synthetic) division by $2x - 1$ gives a quadratic quotient with no remainder:

   $2x^3 + 3x^2 - 8x + 3 = (2x - 1)\left(x^2 + 2x - 3\right)$

   A zero remainder is the confirmation that the division was done correctly.

4. 4

   Factor the quadratic quotient. Two numbers multiplying to $-3$ and summing to $+2$ are $+3$ and $-1$:

   $x^2 + 2x - 3 = (x + 3)(x - 1)$

   $\Longrightarrow \quad 2x^3 + 3x^2 - 8x + 3 = (2x - 1)(x + 3)(x - 1)$

5. 5

   Set each factor to zero and check.

   $2x - 1 = 0 \Rightarrow x = \tfrac12, \qquad x + 3 = 0 \Rightarrow x = -3, \qquad x - 1 = 0 \Rightarrow x = 1$

   Substituting back: $x=-3$ gives $-54 + 27 + 24 + 3 = 0$, and $x=1$ gives $2 + 3 - 8 + 3 = 0$. The sum of the roots is $-3 + \tfrac12 + 1 = -\tfrac32 = -\tfrac{a_2}{a_3}$ and their product is $-\tfrac32 = -\tfrac{a_0}{a_3}$, both matching Vieta's formulas.