When $3 x^{2} + 6 x - 10$ $3x^2+6x-10$ 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 $k$ $k$ are ${- 4, 2}$ ${-4,2}$

We apply the remainder theorem

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

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

When $x = c$ $x=c$

$f (c) = 0 + r$ $f(c)=0+r$

Here,

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

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

which is also equal to $14$ $14$

therefore,

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

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

We solve this quadratic equation for $k$ $k$

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

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

So,

$k = - 4$ $k=-4$

or

$k = 2$ $k=2$

When 3x^2+6x-103x2+6x−103x^2+6x-10 is divided by x+k, the remainder is 14. How do you determine the value of k?