1. **State the problem:** We are given a polynomial $$P(x) = x^5 + x^2 + ax + b$$ and it is divided by $$x^2 - 1$$. The remainder is $$2x + 3$$. We need to find the values of $$a$$ and $$b$$. 2. **Recall the Remainder Theorem for quadratic divisors:** When a polynomial $$P(x)$$ is divided by a quadratic polynomial $$D(x)$$, the remainder $$R(x)$$ is a polynomial of degree less than 2. Here, $$R(x) = 2x + 3$$. 3. Since $$x^2 - 1 = (x-1)(x+1)$$, the remainder satisfies: $$P(1) = R(1)$$ and $$P(-1) = R(-1)$$. 4. Calculate $$P(1)$$: $$P(1) = 1^5 + 1^2 + a(1) + b = 1 + 1 + a + b = 2 + a + b$$ Calculate $$R(1)$$: $$R(1) = 2(1) + 3 = 5$$ Set equal: $$2 + a + b = 5$$ 5. Calculate $$P(-1)$$: $$P(-1) = (-1)^5 + (-1)^2 + a(-1) + b = -1 + 1 - a + b = 0 - a + b = -a + b$$ Calculate $$R(-1)$$: $$R(-1) = 2(-1) + 3 = -2 + 3 = 1$$ Set equal: $$-a + b = 1$$ 6. Solve the system of equations: $$\begin{cases} 2 + a + b = 5 \\ -a + b = 1 \end{cases}$$ From the first equation: $$a + b = 3$$ From the second equation: $$-a + b = 1$$ Add both equations: $$ (a + b) + (-a + b) = 3 + 1 $$ $$ \cancel{a} + b - \cancel{a} + b = 4 $$ $$ 2b = 4 $$ $$ b = 2 $$ Substitute $$b=2$$ into $$a + b = 3$$: $$ a + 2 = 3 $$ $$ a = 1 $$ **Final answer:** $$a = 1$$ and $$b = 2$$.