Using Vieta's Formulas to find expression involving polynomial roots

Question

I'm having trouble with this problem.

Show that if the roots of $$5x^3-x^2-2x+3=0$$ are $a_1,a_2,a_3$, then $$1/a_1+1/a_2+1/a_3=2/3$$

Answer 1

Use that since $a_1,a_2,a_3$ are the roots of the polynomial then $$5x^3-x^2-2x+3=5(x-a_1)(x-a_2)(x-a_3)$$ Expand the right side and compare the coefficients of equal powers of $x$ to obtain that $$\begin{cases}a_1a_2+a_1a_3+a_2a_3=-\dfrac{2}{5}\\a_1a_2a_3=-\dfrac{3}{5}\end{cases}$$ Now $$\frac{a_1a_2+a_1a_3+a_2a_3}{a_1a_2a_3}=\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}$$ gives you the result.

Answer 2

$$\frac{1}{a_1}+\frac{1}{a_2} + \frac{1}{a_2} = \frac{a_2a_3+a_1a_3+a_1a_2}{a_1a_2a_3} = \frac{(-2)/5}{(-1)^3 \cdot 3/5}=\frac{2}{3}$$

Answer 3

The equation $3y^3-2y^2-y+5=0$ has roots the reciprocals of the roots of $5x^3-x^2-2x+3=0$. To see this, divide the given polynomial through by $x^3$, and reverse the order of summation.

The sum of the roots of $ 3y^3-2y^2-y+5=0$ is $-\frac{-2}{3}$.