1. **State the problem:** Given the cubic function $f(x) = 4x^3 - 8x^2 + ax + b$, where $a$ and $b$ are constants, we know that $2x - 1$ is a factor of $f(x)$ and that the remainder when $f(x)$ is divided by $x + 2$ is 20. We need to find the remainder when $f(x)$ is divided by $x - 1$. 2. **Use the factor condition:** Since $2x - 1$ is a factor, then $f\left(\frac{1}{2}\right) = 0$ because setting $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$. Calculate: $$f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 8\left(\frac{1}{2}\right)^2 + a\left(\frac{1}{2}\right) + b = 4\cdot\frac{1}{8} - 8\cdot\frac{1}{4} + \frac{a}{2} + b = \frac{1}{2} - 2 + \frac{a}{2} + b = -\frac{3}{2} + \frac{a}{2} + b.$$ Set this equal to zero: $$-\frac{3}{2} + \frac{a}{2} + b = 0 \Rightarrow \frac{a}{2} + b = \frac{3}{2} \Rightarrow a + 2b = 3.$$ 3. **Use the remainder condition:** The remainder when dividing by $x + 2$ is $f(-2) = 20$ Calculate: $$f(-2) = 4(-2)^3 - 8(-2)^2 + a(-2) + b = 4(-8) - 8(4) - 2a + b = -32 - 32 - 2a + b = -64 - 2a + b.$$ Set equal to 20: $$-64 - 2a + b = 20 \Rightarrow -2a + b = 84.$$ 4. **Solve the system of linear equations:** $$\begin{cases} a + 2b = 3 \\ -2a + b = 84 \end{cases}$$ Multiply first equation by 2: $$2a + 4b = 6$$ Add to second equation: $$2a + 4b + (-2a + b) = 6 + 84 \Rightarrow 5b = 90 \Rightarrow b = 18.$$ Substitute $b=18$ into first equation: $$a + 2(18) = 3 \Rightarrow a + 36 =3 \Rightarrow a = 3 - 36 = -33.$$ 5. **Find remainder when dividing by $x - 1$:** The remainder is $f(1)$. Calculate: $$f(1) = 4(1)^3 - 8(1)^2 + a(1) + b = 4 - 8 - 33 + 18 = (4 - 8) + (-33 + 18) = -4 - 15 = -19.$$ **Final answer:** The remainder when $f(x)$ is divided by $x - 1$ is **$-19$**.