If x is a positive integer such that x3 + 8x2 + 12x-385 = 0, then x is  Choices:
A) less than 6 B) between 7 and 11 C) between 12 and 20 D) between 22 and 34 E) larger than 50
Let f(x) = x3 +8x2 +12x􀀀385. Since   f(6) = 63 +862 + 126-385 = 62(6 +8 +2)-385 = 36 16 - 385 &gt; 0   and f(0) = -385 &lt; 0, there must be some t 2 (0; 6) for which f(t) = 0. Therefore, f(x)   has a positive real root between 0 and 6. Descartes' Rule of Signs implies that there is   only one positive real root, so the answer must be (a). Or one can observe that f(5) = 0   to obtain the answer.   correct answer less than 6
<result>A</result>
If x is a positive integer such that x3 + 8x2 + 12x-385 = 0, then x is