Find k so that x + 6 is a factor of x^4 + 4x^3 - 21x^2 + kx + 108. Find k so that x + 6 is a factor of x4 + 4x3 - 21x2 + kx + 108. A. -36 C. +32 B. -24 D. +48 Log in or register to post comments Solution Click here to… Jhun Vert Fri, 06/07/2024 - 18:59 Solution Click here to expand or collapse this section $f(x) = x^4 + 4x^3 - 21x^2 + kx + 108$ $(x - r) = x + 6$ $r = -6$ Factor Theorem If $(x - r)$ is a factor of $f(x)$ then $f(r) = 0$ $f(-6) = 0$ $(-6)^4 + 4(-6)^3 - 21(-6)^2 + k(-6) + 108 = 0$ $1296 - 864 - 756 - 6k + 108 = 0$ $6k = -216$ $k = -36$ Log in or register to post comments
Solution Click here to… Jhun Vert Fri, 06/07/2024 - 18:59 Solution Click here to expand or collapse this section $f(x) = x^4 + 4x^3 - 21x^2 + kx + 108$ $(x - r) = x + 6$ $r = -6$ Factor Theorem If $(x - r)$ is a factor of $f(x)$ then $f(r) = 0$ $f(-6) = 0$ $(-6)^4 + 4(-6)^3 - 21(-6)^2 + k(-6) + 108 = 0$ $1296 - 864 - 756 - 6k + 108 = 0$ $6k = -216$ $k = -36$ Log in or register to post comments
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