1. **State the problem:** We are given a cubic polynomial $p(x) = 2x^3 + ax^2 + bx + c$ with roots $x=\frac{1}{2}$, $x=n$, and $x=-n$, where $a,b,c,n$ are integers. The $y$-intercept is 4, meaning $p(0) = c = 4$. We need to find $p(x)$ with simplified coefficients. 2. **Use the root-factor relationship:** Since the roots are $\frac{1}{2}$, $n$, and $-n$, the polynomial can be expressed as $$p(x) = k(x - \frac{1}{2})(x - n)(x + n)$$ where $k$ is the leading coefficient. Given the leading term is $2x^3$, we have $k=2$. 3. **Simplify the factors:** Note that $(x - n)(x + n) = x^2 - n^2$. So, $$p(x) = 2(x - \frac{1}{2})(x^2 - n^2)$$ 4. **Expand the polynomial:** $$p(x) = 2\left(x(x^2 - n^2) - \frac{1}{2}(x^2 - n^2)\right) = 2\left(x^3 - n^2 x - \frac{1}{2}x^2 + \frac{1}{2}n^2\right)$$ $$= 2x^3 - 2n^2 x - x^2 + n^2$$ 5. **Rewrite in standard form:** $$p(x) = 2x^3 - x^2 - 2n^2 x + n^2$$ 6. **Use the $y$-intercept condition:** Since $p(0) = c = 4$, substitute $x=0$: $$p(0) = n^2 = 4 \implies n^2 = 4$$ 7. **Solve for $n$:** $$n = \pm 2$$ 8. **Substitute $n^2=4$ back into $p(x)$:** $$p(x) = 2x^3 - x^2 - 8x + 4$$ 9. **Final polynomial:** $$\boxed{p(x) = 2x^3 - x^2 - 8x + 4}$$ This polynomial has roots $\frac{1}{2}$, $2$, and $-2$, leading coefficient 2, and $y$-intercept 4.