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

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<math>\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118</math>
 
<math>\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118</math>
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==Solution==
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==Solution 1==
 
By Vieta's Formulas, we know that <math>a</math> is the sum of the three roots of the polynomial <math>x^3-ax^2+bx-2010</math>.  
 
By Vieta's Formulas, we know that <math>a</math> is the sum of the three roots of the polynomial <math>x^3-ax^2+bx-2010</math>.  
 
Also, 2010 factors into <math>2*3*5*67</math>. 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 <math>a</math>, <math>2</math> and <math>3</math> should be multiplied, which means <math>a</math> will be <math>6+5+67=78</math> and the answer is <math>\boxed{\textbf{(A)}}</math>.
 
Also, 2010 factors into <math>2*3*5*67</math>. 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 <math>a</math>, <math>2</math> and <math>3</math> should be multiplied, which means <math>a</math> will be <math>6+5+67=78</math> and the answer is <math>\boxed{\textbf{(A)}}</math>.
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==Solution 2==
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The polynomial <math> x^{3}-ax^{2}+bx-2010 </math> as three positive integer zeros. What is the smallest possible value of <math>a</math>? [/quote]
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We can expand <math>(x+a)(x+b)(x+c)</math> as <math>(x^2+ax+bx+ab)(x+c)</math>
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<math>(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</math>
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We do not care about <math>+bx</math> in this case, because we are only looking for a.  We know that the constant term is <math>-2010=-(2*3*5*67)</math>
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We are trying to minimize a, such that we have <math>-ax^2</math>
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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>
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Therefore <math>-a=-78</math> or <math>a=\boxed{\textbf{(A)78}}</math>

Revision as of 13:27, 6 June 2011

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

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