Find $a, b \in \mathbb{R}$ such that the polynomial $f(x)=x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b = 0$ has a triple root. I am given the following polynomial:
$$f(x) = x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b = 0$$
with $a, b \in \mathbb{R}$.
I have to find $a$ and $b$ such that the given polynomial has a triple root.
I know that if a polynomial has a triple root $\alpha$ then we have:
$$f(\alpha) = 0$$
$$f'(\alpha) = 0$$
$$f''(\alpha) = 0$$
And in previous exercises (where the degree of the polynomial was $4$, not $5$) I could use $f''(\alpha) = 0$ to find $2$ $\alpha$'s and find $a$ and $b$ for each $\alpha$. I could do that because the second derivative of a $5$th degree polynomial is a quadratic so I could find the $\alpha$'s. Here, it's different. We have:
$$f(x) = x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b$$
$$f'(x) = 5x^4+28x^3+57x^2+52x+a$$
$$f''(x) = 20x^3+84x^2+114x+52$$
And when I try to solve $f''(\alpha) = 0$ I get:
$$20 \alpha ^ 3 + 84 \alpha^2 + 114 \alpha + 52 = 0$$
And I don't know how to find the $\alpha$'s. So, what approach should I use to solve this exercise?

Simplify the equation as
$$10 a^ 3 + 42 a^2 + 57a + 26= 0$$
which may be factorized as follows,
$$(10 a ^ 3 + 20a^2)+ (22 a^2 + 57a + 26 ) $$
$$= 10(a + 2) a^2+ (a + 2)(22a + 13 ) = (a + 2)(10^2+22a + 13 ) =0$$
which yields $a = -2$.
Alternatively, the cubic equation can always be solved with the Cardano's formula. First, make the variable change $x = t - \frac{42}{3\cdot 10} = t - \frac{7}{5}$ to write the equation in the depressed form $t^3+pt+q=0$, i.e.
$$t^3-\frac9{50}t+\frac{27}{250}=0$$
Then, apply the Cardano's formula to get the analytical solution $$t = \sqrt[3]{-\frac q2+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}=-\frac35$$
which gives the same result $a= -2$. The 'standard' approach may appear brute force, but it is guaranteed to yield the solution. 
