# Problem with reversing Vieta's formulas

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?

• 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 May 10, 2020 at 10:00
• It's easy to see numbers are 2,3,4 so expression in question is equal to 234 May 10, 2020 at 10:01
• How did you see this?
– user745970
May 10, 2020 at 10:04

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