# A question about derivative and polynomial

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.

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

## 1 Answer

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

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

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