Find d so that x + 6 is a factor of x^4 + 4x^3 - 21x^2 + dx + 108. Find d so that x + 6 is a factor of x4 + 4x3 - 21x2 + dx + 108. A. +48 C. -24 B. +32 D. -36 Log in or register to post comments Solution Click here to… Jhun Vert Wed, 07/10/2024 - 15:44 Solution Click here to expand or collapse this section Remainder Theorem (x – r) is a factor of f(x) if and only if f(r) = 0. Try this video discussion of this topic. https://mathalino.com/crs/node/18121 Given: $f(x) = x^4 + 4x^3 - 21x^2 + dx + 108$ $x - r = x + 6$, hence, $r = -6$ Apply the factor theorem $f(-6) = 0$ $(-6)^4 + 4(-6)^3 - 21(-6)^2 + d(-6) + 108 = 0$ $d = -36$ Log in or register to post comments
Solution Click here to… Jhun Vert Wed, 07/10/2024 - 15:44 Solution Click here to expand or collapse this section Remainder Theorem (x – r) is a factor of f(x) if and only if f(r) = 0. Try this video discussion of this topic. https://mathalino.com/crs/node/18121 Given: $f(x) = x^4 + 4x^3 - 21x^2 + dx + 108$ $x - r = x + 6$, hence, $r = -6$ Apply the factor theorem $f(-6) = 0$ $(-6)^4 + 4(-6)^3 - 21(-6)^2 + d(-6) + 108 = 0$ $d = -36$ Log in or register to post comments
Solution Click here to…