+349 Suppose the polynomial p(x)=x^3+ax^2+bc+c has the property that the mean of its zeroes, the product of its zeroes, and the sum of its coefficients are all equal. If the y-intercept of the graph of y=p(x) is 0, what is b?
+35 I'm assuming your "bc" is a typo and the polynomial is p(x) = x^3+ax^2+bx+c.
We know that the y intercept occurs at (0, 0), which implies that C = 0. By Vieta's theorem, is C=−r⋅s⋅t, where r, s, t are the roots of p(x). So at the minimum, either r, s, or t is 0. Without loss of generality, let's assume r = 0.
We know rst = (r + s + t)/3 = 1 - (r + s + t) + (rs + st + rt) - (rst), by Vieta's theorem. Plugging in r = 0, we have 0=s+t3=1−(s+t)+(st). Then, we can seperate this big equation into one equation (by ignoring the 3rd equality): s+t=0⇒s=−t . Now, we plug in s = -t to the 3rd expression to get 1−0−t2=0⇒t2=1,t=−1,1. Now, depending on what t is, s is either -1 or 1. So let's just let t = 1 (it won't affect the problem's final answer anyways) and then s = -1. So then b = (rs + st + rt) = (st) = -1.
So, our final answer is -1. This problem is fairly easy, and is a simple application of Vieta's theorem, which is absolutly necessary to know for polynomial Algebra questions, and comes up often in math competitions like the AMCs :)
Other than that, some "tricks" (not really) to note that is if you want the y-intercept, x is 0 and thus the y-intercept will always be C. If you want the sum of the coefficients, this is the equivalent to x = 1 (applies vice versa!).
psss i was too lazy to write out all the latex, writing latex on this website is hard
