<p>Find the number of second-degree polynomials f(x) with integer coefficients and integer zeros for which f(0) = 2010.</p>
<p>A.) 2010
B.) 163
C.) 123
D.) 82
E.) 2</p>
<p>Please Explain :)</p>
<p>B.) 163</p>
<p>Just out of curiosity, where did this question come from? I’m in the process of typing up an explanation, but I gave you that number, maybe you can see where it comes from yourself.</p>
<p>wow it took me like 5 minutes just to figure out what that question meant and I got a 790 on math II :P.</p>
<p>Let f(x) = a (x - b) (x - c). This is because it is a second degree polynomial with integer coefficients(represented by when a is multiplied into the product of x - b and x - c ) and integer zeroes(represented by b and c). f (0) = abc = 2010. The prime factors of 2010 are 2 x 3 x 5 x 67. First consider the case where r and s (and thus a) are positive. There are 3^4 = 81 ways to split up the prime factors between a, r, and s. However, r and s are indistinguishable. In one case, (a,b,c) = (2010, 1 , 1). Here we have r = s.
The other 80 cases are double counting, so there are 40. We must now consider the various cases of signs. For the 40 cases where |r|=/= |s|,there are a total of four possibilities, e.g. (r , s) = (2,3); (-2,3); (2,-3); (-2,-3). For the case of |r|=|s|=1 , there are only three possibilities, (r,s) = (1,1); (1,-1); (-1,-1) , as (-1, 1) is not distinguishable from the second of those three.
Thus the grand total is 4 x 40 + 3 = 163</p>
<p>@rainbowrose: haha me too xD except I haven’t taken it yet, but I have some high expectations :D</p>
<p>163 is correct. My tutor gave me the question. Here is what I did so far:</p>
<p>P(x) = a(x-p)(x-q), where p and q are the roots
P(x) = ax^2 + … + apq FOIL
P(0) = apq = 2010</p>
<p>Factor 2010: {3,5,67}</p>
<p>Need 3 factors because apq = 2010</p>
<p>Find all possible sets? {3,5,67}, {-3,-5, 67}, etc. {2010,1,1} etc.</p>
<p>Am I on the right track?</p>
<p>I think I understand the problem now.</p>
<p>I think finding all of the possible sets may have taken too long. Just remember that when it comes to math, there’s always a shorter way! …well most of the time!</p>
<p>Is it wrong that I feel extremely excited for being able to solve that problem?</p>
<p>schoolisfun, can you solve my other problem?</p>
<p>Sure, I’ll see if I can do it!</p>
<p>lol just understanding what the question asks is hard enough</p>
<p>I got 800 Math II too haha</p>
<p>^haha actually these problems are AIME problems. Lucky for me too, I still have to take the math II and these problems would have scared me xD</p>
<p>I got an 800 and I was about to guess A lol.</p>
<p>This sounds more like an AMC problem? I’m surprised this appeared on the test, though I did realize how to do it in a minute and I haven’t take, the SAT II Math 2 yet (I will in November)</p>