How do you find the value of so that the remainder is zero of $x^{2} + k x - 6 \div (x + 1)$ $x^2+kx-6 div (x+1)$ ?

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Answer 1 · 2016-12-09T21:05:15

$k = - 5$ $k=-5$

We need that

$\frac{x^{2} + k x - 6}{x + 1} = q (x)$ $(x^2+kx-6)/(x+1) = q(x)$ with $q (x)$ $q(x)$ a polynomial.

Also equivalently

$x^{2} + k x - 6 = q (x) (x + 1)$ $x^2+kx-6 = q(x)(x+1)$

Now making $x = - 1$ $x = -1$ we have

$1 - k - 6 = 0$ $1-k-6=0$ then $k = - 5$ $k = -5$

How do you find the value of so that the remainder is zero of x2+kx−6÷(x+1)x^2+kx-6 div (x+1)?