Question

using remainder theorem,find the value of k,so that(4x^2+kx-1) leaves the remainder 2,when divided by (x-3)

Answer 1

[tex]P(x) = 4x^{2} + kx - 1 \\ Q(x) = x - 3 \\ P(x)/Q(x) = 2 \\ By Remainder Theorem, P(3) = 2 \\ Thus, 4*3^{2} + k*3 -1 = 2\\ 3k = -33 k = -11[/tex]

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