2010 AMC 10A Problems/Problem 21

Problem

The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. 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

Solution 1

By Vieta's Formulas, we know that $a$ is the sum of the three roots of the polynomial $x^3-ax^2+bx-2010$ . Again Vieta's Formulas tell us that $2010$ is the product of the three integer roots. Also, $2010$ factors into $2\cdot3\cdot5\cdot67$ . 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 $\boxed{\textbf{(A)78}}$ . ~JimPickens

Solution 2

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 $-2010=-(2\cdot 3\cdot 5\cdot 67)$ 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 $2\cdot 3\cdot 5\cdot 67$ , 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 $a=\boxed{\textbf{(A)}78}$

Video Solution 1

https://youtu.be/LCx0go2BXiY

~IceMatrix

Video Solution 2

https://youtu.be/3dfbWzOfJAI?t=2352

~ pi_is_3.14

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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