2010 AMC 10A Problems/Problem 21

Problem

The polynomial $x^3-ax^2+bx-2010$ has three positive integer zeros. What is the smallest possible value of $a$ ?

$\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118$

Solution 1

By Vieta's Formulas, we know that $a$ is the sum of the three roots of the polynomial $x^3-ax^2+bx-2010$ . Also, 2010 factors into $2*3*5*67$ . But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with three roots. To minimize $a$ , $2$ and $3$ should be multiplied, which means $a$ will be $6+5+67=78$ and the answer is $\boxed{\textbf{(A)}}$ .

Solution 2

The polynomial $x^{3}-ax^{2}+bx-2010$ as three positive integer zeros. What is the smallest possible value of $a$ ? [/quote] We can expand $(x+a)(x+b)(x+c)$ as $(x^2+ax+bx+ab)(x+c)$ $(x^2+ax+bx+ab)(x+c)=x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc=x^3+x^2(a+b+c)+x(ab+ac+bc)+abc$

We do not care about $+bx$ in this case, because we are only looking for a. We know that the constant term is $-2010=-(2*3*5*67)$ We are trying to minimize a, such that we have $-ax^2$ Since we have three positive solutions, we have $(x-a)(x-b)(x-c)$ as our factors. We have to combine two of the factors of $2*3*5*67$ , and then sum up the $3$ resulting factors. Since we are minimizing, we choose $2$ and $3$ to combine together. We get $(x-6)(x-5)(x-67)$ which gives us a coefficient of $x^2$ of $-6-5-67=-78$ Therefore $-a=-78$ or $a=\boxed{\textbf{(A)78}}$

Page

Toolbox

Search

2010 AMC 10A Problems/Problem 21

Problem

Solution 1

Solution 2