1. **State the problem:** We are given the polynomial division $\frac{x^5+4x^2+ax+b}{x^2-1}$ with a remainder $2x+3$. We need to find the values of $a$ and $b$. 2. **Recall the division algorithm for polynomials:** When dividing by $x^2-1$, the remainder must be a polynomial of degree less than 2, which matches the given remainder $2x+3$. 3. **Use the fact that $x^2-1 = (x-1)(x+1)$:** The remainder when dividing by $x^2-1$ is the same as the value of the original polynomial at $x=1$ and $x=-1$ because $x^2-1=0$ at these points. 4. **Evaluate the polynomial at $x=1$:** $$1^5 + 4(1)^2 + a(1) + b = 1 + 4 + a + b = a + b + 5$$ The remainder at $x=1$ is: $$2(1) + 3 = 5$$ Set equal: $$a + b + 5 = 5 \implies a + b = 0$$ 5. **Evaluate the polynomial at $x=-1$:** $$(-1)^5 + 4(-1)^2 + a(-1) + b = -1 + 4 - a + b = b - a + 3$$ The remainder at $x=-1$ is: $$2(-1) + 3 = -2 + 3 = 1$$ Set equal: $$b - a + 3 = 1 \implies b - a = -2$$ 6. **Solve the system of equations:** From step 4: $a + b = 0$ From step 5: $b - a = -2$ Add the two equations: $$ (a + b) + (b - a) = 0 + (-2) \implies 2b = -2 \implies b = -1 $$ Substitute $b = -1$ into $a + b = 0$: $$ a - 1 = 0 \implies a = 1 $$ **Final answer:** $a = 1$, $b = -1$.