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For which values $a,b$ $\in$ $\mathbb{R}$ exists a double zero of $f(x) = x^3-ax+b$? For which values $a,b$ exist exactly one, two or three real zeros of $f$, respectively?

I'm not sure how to approach this, I only found out that for $a=0$ and $b=0$ both $f$ and $f^`$ have a zero so I guess I found the double zero. But how do I determine adequate values for $a$ and $b$ to get a specific number of zeros? I thought about using the Newton method somehow but I'm kind of lost at the moment. All help is very much appeciated, so thanks in advance.

Val
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    Hint: it's easy to write down the derivatives of $f(x)$. A double root would have to be a root of both $f(x), f'(x)$. A triple root would also need to be a root of $f''(x)$. – lulu Jan 06 '20 at 20:31

2 Answers2

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If $(x-\alpha)^2(x-\beta)=x^3-ax-b$, then expanding and comparing the coefficients of $x^2$ tells us $-2\alpha-\beta=0 \Rightarrow \beta=-2\alpha$. Now just expand $(x-\alpha)^2(x+2\alpha)=x^3+(\alpha^2-4\alpha)x+2\alpha^2$, so for a double root you need $b \geq 0, a=\frac b2 \pm 4\sqrt{\frac b2}$.

Robert Shore
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The discriminant of your cubic is $4 a^3 - 27 b^2$. The discriminant is $0$ if and only if there is at least one double root. If the discriminant is positive, there are three distinct real roots. If the discriminant is negative, there is one real root and a pair of complex conjugate roots.

Robert Israel
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