A cubic polynomial is defined by f(x) = 4x3 − 7x − 3. Find the remainder when f(x) is divided by (x-2). Show that (2x+1) is a factor of f(x) and hence factorize f(x) completely.

Here, $a=2$.

The remainder is $\boxed{15}$.

If $(2x+1)$ is a factor, then $f\left(-\frac{1}{2}\right)$ must be 0.

Since $f\left(-\frac{1}{2}\right) = 0$, $(2x+1)$ is a factor of $f(x)$.

Dividing $4x^3 - 7x - 3$ by $(2x+1)$ gives $2x^2 - x - 3$.
So, $f(x) = (2x+1)(2x^2 - x - 3)$.

Therefore, the complete factorization of $f(x)$ is $\boxed{(2x+1)(x+1)(2x-3)}$.

The equation becomes $4x^3 - 7x - 3 = 0$, which is $f(x)=0$.
From part b), we know $f(x) = (2x+1)(x+1)(2x-3)$.
So, $(2x+1)(x+1)(2x-3) = 0$.
This gives the solutions for $x$:

• $2x+1 = 0 \implies x = -\frac{1}{2}$
• $x+1 = 0 \implies x = -1$
• $2x-3 = 0 \implies x = \frac{3}{2}$

• $\cos\theta = -\frac{1}{2}$ The reference angle is $\alpha = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ$. Since $\cos\theta$ is negative, $\theta$ is in the second or third quadrant. $\theta = 180^\circ - 60^\circ = 120^\circ$ $\theta = 180^\circ + 60^\circ = 240^\circ$

• $\cos\theta = -1$ $\theta = 180^\circ$

• $\cos\theta = \frac{3}{2}$ There are no solutions for this case, as the range of $\cos\theta$ is $[-1, 1]$.

The solutions for $\theta$ are $\boxed{120^\circ, 180^\circ, 240^\circ}$.