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 (Alcumus)

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 the polynomial has only three roots, 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

Note:

If you are feeling unconfident about $78$ , you can try to expand the expression $(x-5)(x-6)(x-67)$ which has roots $5$ , $6$ , and $67$ . $(x-5)(x-6)(x-67)=(x^{2}-11x+30)(x-67)$ $=x\left(x^{2}-11x+30\right)-67\left(x^{2}-11x+30\right)=x^{3}-11x^{2}+30x-67x^{2}+737x-2010=x^{3}-78x^{2}+767x-2010$

As we can see, $a=78$ , and since this is the least answer choice, we can be confident that the right option is $\boxed{\textbf{(A) } 78}$ .

~JH. L

Solution 2

We can expand $(x+a)(x+b)(x+c)$ as \begin{align*}(x+a)(x+b)(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 \end{align*}

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 $a=\boxed{\textbf{(A)}78}$

Solution 3

We want the polynomial $x^3-ax^2+bx-2010$ to have POSITIVE integer roots. That means we want to factor it in to the form $(x-a)(x-b)(x-c).$ We therefore want the prime factorization for $2010$ . The prime factorization of $2010$ is $2 \cdot 3 \cdot 5 \cdot 67$ . We want the smallest difference of the $3$ roots since by Vieta's formulas, $a$ is the sum of the $3$ roots.

We proceed to factorize it in to $(x-5)(x-6)(x-67)$ . Therefore, our answer is $5+6+67$ = $\boxed{\textbf{(A)78}}$ .

~Arcticturn

Suggestion for the author: The variables $a$ and $b$ are already defined as coefficients of the cubic polynomial. So, consider using different variables when describing how we want to factor this polynomial in the second sentence. Thanks!

~Jwarner

Notes

We can check $5 \cdot 6 \cdot 67$ has the smallest possible sum because of the following reasons:

$1)$ : We don't want to multiply $67$ by anything since that would make the sum of the roots too big.

$2)$ : The smallest number should multiply since that would make the numbers optimally small. Therefore, we want $2$ times $3$ .

Vieta's Formulas

~Arcticturn

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search