1. **State the problem:** We have a quadratic polynomial $$f(x) = x^2 + ax + b$$ and two conditions based on remainders when divided by linear factors: - When divided by $$x+1$$, the remainder is 2, so $$f(-1) = 2$$. - When divided by $$x-3$$, the remainder is -10, so $$f(3) = -10$$. We need to find the values of $$a$$ and $$b$$. 2. **Write the equations from the conditions:** From $$f(-1) = 2$$: $$(-1)^2 + a(-1) + b = 2 \implies 1 - a + b = 2$$ Simplify: $$-a + b = 1 \quad \text{(Equation 1)}$$ From $$f(3) = -10$$: $$3^2 + 3a + b = -10 \implies 9 + 3a + b = -10$$ Simplify: $$3a + b = -19 \quad \text{(Equation 2)}$$ 3. **Solve the system of linear equations:** Equations: $$-a + b = 1$$ $$3a + b = -19$$ Subtract Equation 1 from Equation 2 to eliminate $$b$$: $$ (3a + b) - (-a + b) = -19 - 1$$ $$3a + b + a - b = -20$$ $$4a = -20$$ $$a = -5$$ Substitute $$a = -5$$ into Equation 1: $$-(-5) + b = 1$$ $$5 + b = 1$$ $$b = 1 - 5 = -4$$ 4. **Final answer:** $$a = -5, \quad b = -4$$ Thus, the quadratic polynomial is: $$f(x) = x^2 - 5x - 4$$