**Official June 2013 SAT II Math 2 Thread**

<p>There was also a question about where does some graph and y = 6 intersect? Answer was 9.7 I think. There was also log (base a) of 25 = .7. Answer was 99.32 I think.</p>

<p>Not that I doubt you zach…but can anyone corroborate him that the polynomial question specified a fifth degree?</p>

<p>For the polynomial question, I think it had something to do with Descartes’ rule of signs.</p>

<p>[Descartes</a>’ rule of signs - Wikipedia, the free encyclopedia](<a href="http://en.wikipedia.org/wiki/Descartes’_rule_of_signs]Descartes"&gt;http://en.wikipedia.org/wiki/Descartes’_rule_of_signs)</p>

<p>There were no sign changes in the polynomial, therefore there were no real roots.</p>

<p>there was a question that asked what the remainder was when you divided a polynomial by (x+1), I think the answer might’ve been 3?</p>

<p>ATPUT73: no, because you can make the coefficients negative if you want.</p>

<p>@crushroller, I remember those two questions, and I got the same answers.</p>

<p>OK, for the polynomial, I’m going into semantics here.</p>

<p>Even if the “rules” state that a leading coefficient can’t be zero, the question specifies that the coefficients can be real numbers. And much like the “golden rule” of most trading card games…the question’s directions win, right?</p>

<p>Until I realized that it might have only said “are real numbers”, not “can be any real numbers”.</p>

<p>NOOOOO!!!</p>

<p>Ok, so for the minimum number of real roots of Ax^5+Bx^4+Cx^3+Dx^2+Ex, where they are all rational and distinct coefficients, why can’t it be zero? If A=0 and there are 2 pairs of complex roots?</p>

<p>I think the College Board was testing our ability to apply Descartes’ rule; however, the question is unfair since there could be one root in the case that A=0. This is still a polynomial expression because it simplifies to x^4, and the problem never explicitly states that the polynomial expression is a 5th degree one.</p>

<p>LoveArtForever: the leading coefficient can’t be zero.</p>

<p>johnstucky: people need to report this question. IDK if college board has ever thrown a question out for ambiguity, or if its curves are pre-made or real…?</p>

<p>Yeah… so what is the answer then??</p>

<p>@LoveArtForever you are forgetting the +K, which would make zero real roots possible.</p>

<p>The leading coefficient in a polynomial can’t be zero, but that doesn’t necessarily mean that A can’t be zero as 0x^5 + Bx^4 + Cx^3 + Dx^2 + Ex + K is equal to Bx^4 + Cx^3 + Dx^2 + Ex + K, where the leading coefficient would be a nonzero. Hence, there is still ambiguity.</p>

<p>Ohhh… ok</p>

<p>I got 1 real root for that problem</p>

<p>johnstucky: also important is whether or not the question specified that it was a 5th degree polynomial, which some say it did.</p>

<p>Y1 = sin(x)
Y2 = cos(x)</p>

<p>2ND>CALC>intersect>spam enter</p>

<p>x = 0.7854 = pi/4</p>

<p>Ok, yeah I’m googling it haha</p>

<p>@RMIBstudent the College Board has indeed thrown out ambiguous questions in the past, although they rarely do so. The only instance that I know of was on the 2005 October SAT for a writing question where there was a subject verb agreement error and both the subject and verb were underlined, thus both were acceptable answers, so the question did not count toward the score.</p>

<p>well…physics went well, but I’m probably retaking this come October.</p>

<p>And I thought Barron’s was supposed to overprep you.</p>