When $3x^2+6x-10$ is divided by x+k, the remainder is 14. How do you determine the value of k?

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Answer 1 · 2017-02-19T07:29:07

The values of $k$ are ${-4,2}$

We apply the remainder theorem

When a polynomial $f(x)$ is divided by $(x-c)$ , we get

$f(x)=(x-c)q(x)+r(x)$

When $x=c$

$f(c)=0+r$

Here,

$f(x)=3x^2+6x-10$

$f(k)=3k^2+6k-10$

which is also equal to $14$

therefore,

$3k^2+6k-10=14$

$3k^2+6k-24=0$

We solve this quadratic equation for $k$

$3(k^2+2k-8)=0$

$3(k+4)(k-2)=0$

So,

$k=-4$

or

$k=2$

When 3x^2+6x-103x^2+6x-10 is divided by x+k, the remainder is 14. How do you determine the value of k?