When the polynomial x^4 + bx^3 + 5x^2 + dx + 6 is divided by x - 2 the remainder is 16. When it is divided by x + 1 the remainder is 10. Find the value of constant d. Date of Exam May 2008 Subject Mathematics, Surveying and Transportation Engineering Problem When the polynomial $x^4 + bx^3 + 5x^2 + dx + 6$ is divided by $x - 2$ the remainder is 16. When it is divided by $x + 1$ the remainder is 10. Find the value of constant $d$. A. 7 C. -5 B. -7 D. 5 Answer Key Click here to show or hide the answer key [ A ] Solution Click here to expand or collapse this section $f(x) = x^4 + bx^3 + 5x^2 + dx + 6$ By Remainder Theorem When f(x) is divided by x - 2, the remainder is f(2) $\color{#FC6255}{f(2) =} 2^4 + b(2^3) + 5(2^2) + d(2) + 6 = 16$ $8b + 2d = -26$ ← Equation (1) When f(x) is divided by x + 1, the remainder is f(-1) $\color{#FC6255}{f(-1) =} (-1)^4 + b(-1)^3 + 5(-1)^2 + d(-1) + 6 = 10$ $-b - d = -2$ ← Equation (2) From Equation (1) and Equation (2) $b = -5$ $d = 7$ ← answer Category Algebra Remainder Theorem Polynomial in x Log in or register to post comments