Find the value of k so that the remainder is 1 when x^3+5x^2+kx-8 is divided by (x-3) K=

1 answer:

5 0

By the remainder theorem, the remainder upon dividing $x^3+5x^2+kx-8$ by $x-3$ is given by the value of this expression when x = 3 :

$3^3 + 5\cdot3^2 + 3k - 8 = 1$

Solve for k :

$27 + 45 + 3k - 8 = 1 \\\\ 3k = -63 \\\\ \boxed{k = -21}$

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cestrela7 [59]

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Leokris [45]

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