1. **Stating the problem:** We have a polynomial $$P(x) = x^4 - 6x^3 + ax^2 + bx + c$$ with conditions: - $$P(p) = 0$$ and $$p(2) = 0$$ (meaning $$x=2$$ is a root). - $$P(x)$$ is divisible by $$x^2 - 4x + 3 = (x-1)(x-3)$$. - Define $$Q(x) = x^2 + dx + e$$. - Define $$R(x) = \frac{P(x)}{(x-1)(x-2)(x^2+1)}$$. - Given the equation $$ (x-1)y^2 + (x+1)y - (x^2 - 3x + 2) = 0 $$. - Partial fraction decomposition: $$R(x) = \frac{m}{x-1} + \frac{n}{x-2} + \frac{rx + s}{x^2 + 1}$$. 2. **Using divisibility:** Since $$P(x)$$ is divisible by $$(x-1)(x-3)$$ and also has root at $$x=2$$, the roots include $$1, 2, 3$$. Thus, $$P(x)$$ can be written as: $$ P(x) = (x-1)(x-2)(x-3)(x + k) $$ for some constant $$k$$ because degree is 4. 3. **Expanding $$P(x)$$:** First, expand $$(x-1)(x-2)(x-3)$$: $$ (x-1)(x-2) = x^2 - 3x + 2 $$ Then: $$ (x^2 - 3x + 2)(x-3) = x^3 - 3x^2 - 3x^2 + 9x + 2x - 6 = x^3 - 6x^2 + 11x - 6 $$ So: $$ P(x) = (x^3 - 6x^2 + 11x - 6)(x + k) = x^4 + kx^3 - 6x^3 - 6kx^2 + 11x^2 + 11kx - 6x - 6k $$ Simplify: $$ P(x) = x^4 + (k - 6)x^3 + (-6k + 11)x^2 + (11k - 6)x - 6k $$ 4. **Matching coefficients with given $$P(x)$$:** Given: $$ P(x) = x^4 - 6x^3 + ax^2 + bx + c $$ Equate coefficients: - Coefficient of $$x^3$$: $$k - 6 = -6 \Rightarrow k = 0$$ - Coefficient of $$x^2$$: $$-6k + 11 = a \Rightarrow a = 11$$ - Coefficient of $$x$$: $$11k - 6 = b \Rightarrow b = -6$$ - Constant term: $$-6k = c \Rightarrow c = 0$$ 5. **Final polynomial:** $$ P(x) = x^4 - 6x^3 + 11x^2 - 6x $$ 6. **Partial fraction decomposition of $$R(x)$$:** Recall: $$ R(x) = \frac{P(x)}{(x-1)(x-2)(x^2 + 1)} $$ Since $$P(x) = (x-1)(x-2)(x-3)x$$ (because $$k=0$$ means factor is $$x$$), rewrite: $$ P(x) = (x-1)(x-2)(x)(x-3) $$ So: $$ R(x) = \frac{(x-1)(x-2)(x)(x-3)}{(x-1)(x-2)(x^2 + 1)} = \frac{x(x-3)}{x^2 + 1} $$ 7. **Expressing $$R(x)$$ in partial fractions:** We want: $$ \frac{x(x-3)}{x^2 + 1} = \frac{m}{x-1} + \frac{n}{x-2} + \frac{rx + s}{x^2 + 1} $$ Multiply both sides by $$(x-1)(x-2)(x^2 + 1)$$: $$ x(x-3)(x-1)(x-2) = m(x-2)(x^2 + 1) + n(x-1)(x^2 + 1) + (rx + s)(x-1)(x-2) $$ Since $$R(x) = \frac{x(x-3)}{x^2 + 1}$$, the left side is actually: $$ x(x-3)(x-1)(x-2)$$ which is $$P(x)$$, so the decomposition is consistent. 8. **Summary:** - $$a = 11$$, $$b = -6$$, $$c = 0$$. - $$P(x) = x^4 - 6x^3 + 11x^2 - 6x$$. - $$R(x) = \frac{x(x-3)}{x^2 + 1}$$. **Final answers:** $$ a = 11, \quad b = -6, \quad c = 0 $$ $$ R(x) = \frac{x(x-3)}{x^2 + 1} $$