<p>thanks a lot mystic :)</p>
<p>SATQ how do you solve that?</p>
<p>It’s “c”, 2… right?</p>
<p>yup, (C), or k=2. I used synthetic division but there are probably hundreds of ways of solving this problem. since you know a factor, you can use synthetic division and then set the “remainder” of the synthetic division (which, in this case, is 16-8k) equal to 0.</p>
<p>So:
16-8k = 0
16 = 8k
k = 2</p>
<p>I imagine most people wouldn’t use this method because it might be time-consuming, but I did it in about 45 seconds with synthetic division.</p>
<p>@efeens</p>
<p>I actually think that synthetic division is a really nice quick and efficient way to do this problem. </p>
<p>But for those that aren’t comfortable with synthetic division, plugging in 8 for t and setting the result equal to 0 serves the same purpose.</p>
<p>In fact, by the Remainder Theorem (from precalculus) these two methods are equivalent.</p>
<p>Yes, t-8 is a factor of the equation if and only if 8 is a root. So plug in t = 8 and set to 0:</p>
<p>64 - 8k - 48 = 0
k = 2</p>
<p>The answer is indeed k = 2(Choice C)</p>
<p>If (t-8) is one of the factors of t² - kt - 48, then </p>
<p>t² - kt - 48 = (t-8)(t-a), where (t-a) is the other factor(we don’t need to worry about the other factor). </p>
<p>Now, if we set t to be equal to 8 on both sides(why is that a good choice?), then</p>
<p>8² - k(8) - 48 = (8-8)(8-a)</p>
<p>64 - 8k - 48 = 0 </p>
<p>16 = 8k </p>
<p>or k = 2</p>
<p>The approach of using synthetic division suggested by @efeens44 and @DrSteve also works well for this problem.</p>
<p>Question:
If (t-8) is a factor of t² - kt - 48, then k=</p>
<p>A) -6
B) -2
C) 2
D) 6
E) 14</p>