<p>For the love of God, I can't figure out how to do this...I know how to get "k" when it is the constant term, but not when it's a coefficient.</p>
<p>If x-2 divides evenly into x^3+kx^2+12x-8, what is k?</p>
<p>I can do it using the remainder theorem or whatever (plug in P(2)=0, so k=-6), but I don't know how to do it using long/synthetic division.</p>
<p>Your first step in long division is of course that (x^2)(x - 2) yields an x^3 term so that x^3 - x^3 = 0 and the x^3 term disappears. </p>
<p>Your remainder is:</p>
<p>(k^2) - (-(2x^2)), which is (k + 2)(x^2).</p>
<p>Now here is the key step: after you bring down the 12x term, your goal is to to get rid of the (k + 2)(x^2). To do that, you will need to multiply (x - 2) by (k + 2)x. The result of that multiplication is:</p>
<p>(x - 2)((k + 2)x) = (k + 2)(x^2) - 2(k + 2)x.</p>
<p>Write that below the previous remainder and subtract.</p>
<p>After you subtract, your new remainder is:</p>
<p>12x + 2(k + 2)x,</p>
<p>which simplifies to:</p>
<p>(16 + 2k)x.</p>
<p>You continue by bringing down the -8 term. </p>
<p>Your last remainder will be an expression involving k. That expression must be set equal to zero if (x - 2) is to divide evenly.</p>
<p>Setting that last remainder equal to zero yields the one value for k that satisfies the requirements of the problem. (-6, as you said).</p>
<p>You can also do it synthetically</p>
<p>2 | 1 K 12 -8</p>
<p>bring down the 1</p>
<p>2 + k </p>
<p>multiply by 2</p>
<p>add 12</p>
<p>2k+16</p>
<p>multiply by 2, subtract 8</p>
<p>4k+24 </p>
<p>4k+24 = 0 for x-2 to divide evenly</p>
<p>so k = -6</p>