Let $\alpha \leqslant \beta \leqslant \gamma$ and $$\alpha+\beta+\gamma = 9$$ $$\alpha\beta+\beta\gamma+\gamma\alpha=26$$ $$\alpha\beta\gamma=24.$$ What is $100\alpha+10\beta+\gamma$?

From Vieta's formulas we get

$$\frac{-b}{a}=9$$ $$\frac{c}{a}=26$$ $$\frac{-d}{a}=24$$

so from here we can form the cubic $f(x)=x^3-9x^2+26x-24$. My question is how can I find the roots from this cubic in order to compute the expression they asked or do I even have to find the actual roots?

  • $\begingroup$ If you suspect that there are integer solutions, you might try some small divisors of $24$ such as $\pm1$ and $\pm2$. Once you have found one, you can factor this out and solve the resulting quadratic which also factors $\endgroup$
    – Henry
    May 10, 2020 at 10:00
  • $\begingroup$ It's easy to see numbers are 2,3,4 so expression in question is equal to 234 $\endgroup$
    – Hrishabh
    May 10, 2020 at 10:01
  • 1
    $\begingroup$ How did you see this? $\endgroup$
    – user745970
    May 10, 2020 at 10:04

1 Answer 1


Note that $f(2)=0$, so by the factor theorem, $x-2$ is a factor. You can then factorise $f$ as $(x-2)(x^2-7x+12)=(x-2)(x-3)(x-4)$.


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