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

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We are trying to minimize a, such that we have <math>-ax^2</math>
 
We are trying to minimize a, such that we have <math>-ax^2</math>
 
Since we have three positive solutions, we have <math>(x-a)(x-b)(x-c)</math> as our factors.  We have to combine two of the factors of <math>2*3*5*67</math>, and then sum up the <math>3</math> resulting factors.  Since we are minimizing, we choose <math>2</math> and <math>3</math> to combine together.  We get <math>(x-6)(x-5)(x-67)</math> which gives us a coefficient of <math>x^2</math> of <math>-6-5-67=-78</math>
 
Since we have three positive solutions, we have <math>(x-a)(x-b)(x-c)</math> as our factors.  We have to combine two of the factors of <math>2*3*5*67</math>, and then sum up the <math>3</math> resulting factors.  Since we are minimizing, we choose <math>2</math> and <math>3</math> to combine together.  We get <math>(x-6)(x-5)(x-67)</math> which gives us a coefficient of <math>x^2</math> of <math>-6-5-67=-78</math>
−
Therefore <math>-a=-78</math> or <math>a=\boxed{\textbf{(A)78}}</math>
+
Therefore <math>-a=-78</math> or <math>a=\boxed{\textbf{(A)}78}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2010|ab=A|num-b=20|num-a=22}}
 
{{AMC10 box|year=2010|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 05:08, 19 February 2015

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 tells 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 $\boxed{\textbf{(A)}}$.

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

See Also

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

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