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Given a polynomial $p(x) = x^3 + kx - 2$, where $k \in \mathbb{R}$ is a constant and $p(x)$ has a double root at $x = \alpha$. Prove that $\alpha=-1$ and $k=-3$.

I plugged in $\alpha$ for $p(x)$ and $p'(x)$, but that didn't go anywhere, bc I would just be proving the values for $\alpha$ and $k$ by inspection. So I'm not sure how to tackle this problem.

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  • $\begingroup$ Your method is correct, assuming you equated the two expressions to zero. You have two simultaneous equations to solve. Eliminate $k$. $\endgroup$ Mar 3 at 1:53

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I assume that, by saying "has a double root", you mean $\alpha$ is a root of multiplicity.

Then we can assume

$$f(x) = (x-\alpha)^2(x-\beta)$$

because $\deg(f) = 3$.

So we get

$$f(x) = (x-\alpha)^2(x-\beta)$$

$$= x^3 - (2\alpha+\beta)x^2 + (2\alpha \beta + \alpha^2)x - \alpha^2\beta$$

And this implies

$$x^3 - (2\alpha+\beta)x^2 + (2\alpha \beta + \alpha^2)x - \alpha^2\beta = x^3 + kx -2$$

$$\Rightarrow - (2\alpha+\beta) = 0, 2\alpha \beta + \alpha^2 =k, - \alpha^2\beta = -2$$

After solving this system of equations, we finally get

$$\alpha = -1, \beta = 2, k = -3$$

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  • $\begingroup$ @DavidGao Thank you! This is my bad. I have edited it. $\endgroup$
    – ZYX
    Mar 3 at 7:30

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