How do you find a coefficient for a polynomial whose roots are given?

Question

If the polynomial $x^3 + 2x^2 + kx - 2$ has three real roots: $a, 1/a,$ and $b$, then what is $k$?

I've tried writing the polynomial with the roots filled in, and thus making a system of equations then trying some algebraic manipulations. However, I couldn't find a solution.

So any help or a suggestion to a method that I'm missing is appreciated.

Thanks in advance.

Answer 1

If the roots of a monic polynomial $p(x)$ are $\alpha$, $\beta$, and $\gamma$, then$$p(x)=(x-\alpha)(x-\beta)(x-\gamma)$$and therefore the coefficient of $x$ is $\alpha\beta+\beta\gamma+\gamma\alpha$. Can you take it from here?

Answer 2

Hint: the product of the roots is $a \cdot \dfrac{1}{a}\cdot b = 2$ by Vieta's relations, so one of the roots is $b=2\,$.

Answer 3

$P(x)=(x-a)(x-\frac{1}{a})(x-b)$

$b=2$ since $(-a)\times\frac{1}{a}\times b=-2$

Hence $(x-a)(x-\frac{1}{a})(x-2)=x^3+2x^2+kx-2$

Expand the left side, you get $x^3+(-2-a-\frac{1}{a})x^2+(\frac{2}{a}+2a+1)x-2$

$-2-a-\frac{1}{a}=2$ and $\frac{2}{a}+2a+1=k$

The first equation, $\frac{1}{a}=-a-4$. Substitute $\frac{1}{a}$ with $-a-4$ in equation 2

$2(-a-4)+2a+1=k$

$-2a-8+2a+1=k$

$k=-7$