Try this beautiful problem from Algebra based Roots of the cubic equation.

## Roots of cubic equation – AMC-10A, 2010- Problem 21

The polynomial \(x^3-ax^2+bx-2010\) has three positive integer roots. What is the smallest possible value of \(a\)?

- \(98\)
- \(78\)
- \(83\)
- \(76\)
- \(90\)

**Key Concepts**

Algebra

Vieta’s Relation

roots of the equation

## Check the Answer

But try the problem first…

Answer: \(78\)

AMC-10A (2010) Problem 19

Pre College Mathematics

## Try with Hints

First hint

The given equation is \(x^3-ax^2+bx-2010\).we have to find out the smallest possible value of \(a\).From Vieta’s Relation we know that if \(r_1,r_2,r_3\) are the roots of equation \(ax^3+bx^2+cx+d=0\) then \(r_1 +r_2+r_3= -\frac{b}{a}\) and \(r_1 r_2 r_3=-\frac{d}{a}\)

can you finish the problem……..

Second Hint

Therefore \(a\) is the sum of the three roots of the polynomial \(x^3-ax^2+bx-2010\). and \(2010\) is the product of the three integer roots.Now the factors of \(2010\) =\(2 \times 3 \times 5 \times 67\).there are only three roots to the polynomial so out of four roots we have to choose three roots such that 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,

can you finish the problem……..

Final Step

Therefore the value of \(a\)=\(6+5+67=78\)

## Other useful links

- https://www.cheenta.com/ratio-of-two-triangles-amc-10a-2004-problem-20/
- https://www.youtube.com/watch?v=pYSIvF7jZy4

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