How do you determine the value of k if the remainder is 3 given $(x^{3} + x^{2} + k x - 15) \div (x - 2)$ $(x^3+x^2+kx-15) div(x-2)$ ?

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Answer 1 · 2016-12-24T01:33:20

Use the remainder theorem to find $k = 3$ $k = 3$

Let $f (x) = x^{3} + x^{2} + k x - 15$ $f(x) = x^3+x^2+kx-15$

By the remainder theorem, the remainder when $f (x)$ $f(x)$ is divided by $(x - 2)$ $(x-2)$ will be $f (2)$ $f(2)$

So we have:

$3 = f (2) = 8 + 4 + 2 k - 15 = 2 k - 3$ $3 = f(2) = 8+4+2k-15 = 2k-3$

Add $3$ $3$ to both ends to get:

$6 = 2 k$ $6 = 2k$

Divide both sides by $2$ $2$ and transpose to get:

$k = 3$ $k = 3$

How do you determine the value of k if the remainder is 3 given (x^3+x^2+kx-15) div(x-2)(x3+x2+kx−15)÷(x−2)(x^3+x^2+kx-15) div(x-2)?