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?

$20 \alpha ^ 3 + 84 \alpha^2 + 114 \alpha + 52 = 0$
$10\alpha^3 + 42\alpha^2 + 57\alpha + 26=0$
Rational roots theorem says if there is are rational root it is $\pm\frac {1,2,13}{1,2,5}$ and it's clearly negative.
And we can try by brute force $\alpha  =-2$.
If we don' want to try them all we can note such large coefficients can be reduced by letting $\alpha = b-1$ and 
$10(b-1)^3 + 42(b-1)^2 + 57(b-1) + 26=0$
$10b^3 +(-30+42)b^2 + (30-2*42+57)b +(-10+42-57+26) =0$
$10b^3 +12b^2 +3b +1=0$ and as $10+3 = 12+1$ we can see $b=-1$ is a solution.
Or we could let $b=c-1$ and get
$10(c-1)^3 + 12(c-1)^2 + 3(c-1) + 1 =0$
$10c^3 + (-30 + 12)c^2 + (30-24 + 3)c + (-10+12-3+) = 0$
$10c^3 -18c^2 +9c =0$ which has $c=0$ as solutions as well as $\frac {18\pm\sqrt{18^2-4*9*10}}{20}$ which are not real solutions.
but at any rate $\alpha = -2$ is a solution.
So $f'(-2) = 5*16-28*8+57*4-52*2+a=0$
$a = -80 + 224 - 228 + 104= 20$
And $f(-2) = -32 + 7*16 - 19*8 + 26*4 -2*20 + b$
so $b=32-112 +152-104 + 40 = 8$.
