Factor 2x^3 - x^2 - 5x - 2 completely.

Step-by-step solution

1. 1

   Try grouping first, and see it fail. Pairing gives $\left(2x^3-x^2\right)+(-5x-2)=x^2(2x-1)-(5x+2)$. The two brackets $(2x-1)$ and $(5x+2)$ are different, so grouping does not apply here — which is the signal to switch to the Rational Root Theorem.

2. 2

   List the candidate rational roots. A rational root $\tfrac{p}{q}$ needs $p\mid2$ (the constant) and $q\mid2$ (the leading coefficient), so the candidates are

   $\pm1,\ \pm2,\ \pm\tfrac12$

   The halves appear precisely because the leading coefficient is $2$ rather than $1$ — a step often skipped.

3. 3

   Test the candidates.

   $f(2)=2(8)-4-10-2=16-16=0\ \checkmark$

   so $x=2$ is a root and $(x-2)$ is a factor.

4. 4

   Divide by (x - 2) with synthetic division. Using the root $2$ on the coefficients $2,\ -1,\ -5,\ -2$:

   $2\ \to\ -1+4=3\ \to\ -5+6=1\ \to\ -2+2=0$

   Zero remainder ✓, and the quotient is $2x^2+3x+1$:

   $2x^3-x^2-5x-2=(x-2)\left(2x^2+3x+1\right)$

5. 5

   Factor the quadratic with the AC method. We need two numbers with product $2\times1=2$ and sum $3$: they are $1$ and $2$. Splitting and grouping:

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

6. 6

   Write the complete factorisation and roots.

   $2x^3-x^2-5x-2=(x-2)(2x+1)(x+1)$

   The roots are $x=2$, $x=-\tfrac12$ and $x=-1$ — note that $-\tfrac12$ is one of the fractional candidates, which is why they had to be on the list.

7. 7

   Verify. Comparing the original with the triple product at every integer from $-30$ to $29$ gives exact agreement ✓. Vieta also checks: the roots sum to $2-\tfrac12-1=\tfrac12=-b/a$ ✓ and their product is $2\cdot\left(-\tfrac12\right)\cdot(-1)=1=-d/a$ ✓.