<p>This question is really bugging me…not really finding anything… so can Bx^4 + Cx^3 + Dx^2 + Ex + K have no real zeroes, if none of the coefficients are zero?</p>
<p>Do you think I could get an 800 if I missed 4, omitted 2?</p>
<p>LoveArtForever: yeah, it can.</p>
<p>Now the problem is whether or not you can make A zero.</p>
<p>@LoveArtForever, yes. Plug 1 in for B, 2 for C, 3 for D, 4 for E, and 5 for K on your graphing calculator. The graph will never cross the x-axis.</p>
<p>And I thought that the leading power (Ax^5) told you how many possible roots there were (5)?</p>
<p>Some of those can be imaginary. And if you set A to 0, that rule doesn’t work. But can you set A to zero? We don’t know.</p>
<p>@PrettyBubbles, your raw score would be a 43, and that typically equates to an 800 or a 790, so it is hard to guarentee your score.</p>
<p>Since the problem doesn’t specifically say that the polynomial expression must be 5th degree, it would be fair to put 0 in for A, simplifying the polynomial expression to 4th degree.</p>
<p>You guys are over thinking the Ax^5 question. Yes, you can set A = 0, nothing prohibited you from doing that. Then it becomes a parabola where it might not intersect the x axis.</p>
<p>The key word was MINIMUM: Yes, you can have one zero assuming A is nonzero. But the thing is, you have to assume you can set EVERYTHING to 0.</p>
<p>The problem here may be the intentions of the college board. They may have intended to test your knowledge that imaginary roots come in pairs, and so therefore put in an odd power…</p>
<p>That was a pretty horrible question.</p>
<p>But there’s still the rule that leading coefficients can’t be zero. Whether or not you can just set A to zero and declare that it isn’t really the leading coefficient seems semantical.</p>
<p>If it specified a 5th degree polynomial, which some say it did, then it’s definitely 1.</p>
<p>Why couldn’t you just use Descarte’s rule then? You can assume that all the coefficients would be positive and realize that it does not change sign. That means there can be 0, 2, 4 zeros.</p>
<p>I am almost certain that it never specified that it was 5th degree polynomial expression.</p>
<p>are double roots considered distinct?</p>
<p>sadly for me it is zero Descarte’s rule only applies to number of postive roots</p>
<p>sillyup20: that doesn’t work. If A isn’t zero, then the endpoints of the fifth polynomial approach ± infinity; it’s impossible for there to be no zeros. The only alternative is to set A to zero.</p>
<p>Here’s a message that I wrote to the College Board just now:</p>
<p>I am extremely concerned that my daughter’s June 2013 SAT subject test scores will be adversely impacted by the mass cheating situations in Korea for May 2013 tests. Since the whole Korea had its SAT scores cancelled in May 2013, they will need to retake the tests in June 2013. The SAT subjects are graded on curves, and since we all know that the Koreans do much better on Math and Science subjects than students in US (hopefully not because of cheating!!), therefore abnormally high participation of SAT in Korea for June 2013 will skew the scores that are graded on curves. In other words, US students that took June 2013 subjects will see lowered scores, especially for Math Level II, Physics, etc.
This is obviously a grossly unfair situation for the hard working kids in US that took the June tests. They should not be the ones that got penalized for the mass cheating in Korea.</p>
<p>^that is…a good point, actually.</p>
<p>@zhuq1688 I don’t think that will impact the scores very much, as I’m sure the College Board has already taken that into account. There’s no harm in double-checking that, though.</p>
<p>is anybody else reporting the ambiguous question?</p>
<p>Yes everyone send in a message about it being ambiguous</p>