A polynomial f(x) = (p-1)x3 + px2 + qx + r gives remainders 4, 0, and 5 when divided by (x-1), (x+1), and (x+2) respectively. Find the values of p, q, r.

The Remainder Theorem states that if a polynomial $f(x)$ is divided by $(x-a)$, the remainder is $f(a)$.
Given $f(x) = (p-1)x^3 + px^2 + qx + r$.

• When $f(x)$ is divided by $(x-1)$, the remainder is 4: $f(1) = (p-1)(1)^3 + p(1)^2 + q(1) + r = 4$ $p-1 + p + q + r = 4$

  $$2p + q + r = 5 \quad \text{(Equation 1)}$$

• When $f(x)$ is divided by $(x+1)$, the remainder is 0: $f(-1) = (p-1)(-1)^3 + p(-1)^2 + q(-1) + r = 0$ $-(p-1) + p - q + r = 0$ $-p + 1 + p - q + r = 0$

  $$1 - q + r = 0 \implies q - r = 1 \quad \text{(Equation 2)}$$

• When $f(x)$ is divided by $(x+2)$, the remainder is 5: $f(-2) = (p-1)(-2)^3 + p(-2)^2 + q(-2) + r = 5$ $-8(p-1) + 4p - 2q + r = 5$ $-8p + 8 + 4p - 2q + r = 5$

  $$-4p - 2q + r = -3 \quad \text{(Equation 3)}$$

From Equation 2, express $q$ in terms of $r$:
$q = r + 1$

Substitute $q = r+1$ into Equation 1:
$2p + (r+1) + r = 5$
$2p + 2r + 1 = 5$

$$2p + 2r = 4 \implies p + r = 2 \quad \text{(Equation 4)}$$

Substitute $q = r+1$ into Equation 3:
$-4p - 2(r+1) + r = -3$
$-4p - 2r - 2 + r = -3$
$-4p - r - 2 = -3$

$$-4p - r = -1 \quad \text{(Equation 5)}$$

Add Equation 4 and Equation 5:
$(p + r) + (-4p - r) = 2 + (-1)$
$-3p = 1$
$p = -\frac{1}{3}$

Substitute $p = -\frac{1}{3}$ into Equation 4:
$-\frac{1}{3} + r = 2$
$r = 2 + \frac{1}{3} = \frac{