Difference between revisions of "2010 AMC 10A Problems/Problem 21"

Revision as of 19:23, 24 October 2021

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 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

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

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 $2010$ . Since we want the prime factorization of $2010$ is $2 \cdot 3 \cdot 5 \cdot 67$ , we want the smallest difference of the $3$ numbers 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 $5+6+67$ = $\boxed{\textbf{(A)78}}$ .

Video Solution 1

https://youtu.be/LCx0go2BXiY

~IceMatrix

Video Solution 2

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

~ pi_is_3.14

@@ Line 20: / Line 20: @@
 ==Solution 3==
-We want the polynomial <math>x^3-ax^2+bx-2010</math> to have POSITIVE integer roots. That means we want to factor it in to the form <math>(x-a)(x-b)(x-c). We therefore want the prime factorization for </math>2010<math>. Since we want the prime factorization of </math>2010<math> is </math>2 \cdot 3 \cdot 5 \cdot 67<math>, we want the smallest difference of the </math>3<math> numbers since by [[Vieta's formulas]], </math>a<math> is the sum of the </math>3<math> roots.
+We want the polynomial <math>x^3-ax^2+bx-2010</math> to have POSITIVE integer roots. That means we want to factor it in to the form <math>(x-a)(x-b)(x-c).</math> We therefore want the prime factorization for <math>2010</math>. Since we want the prime factorization of <math>2010</math> is <math>2 \cdot 3 \cdot 5 \cdot 67</math>, we want the smallest difference of the <math>3</math> numbers since by [[Vieta's formulas]], <math>a</math> is the sum of the <math>3</math> roots.
-We proceed to factorize it in to </math>(x-5)(x-6)(x-67)<math>. Therefore, our answer is </math>5+6+67<math> = </math>\boxed{\textbf{(A)78}}$.
+We proceed to factorize it in to <math>(x-5)(x-6)(x-67)</math>. Therefore, our answer is <math>5+6+67</math> = <math>\boxed{\textbf{(A)78}}</math>.
 ==Video Solution 1==

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