<s>[INST] 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?
 [/INST] We'll equate
$\tag 1 (x-u)^3 \,(x^2+vx+w)$
to 
$\tag 2  x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b$
The coefficient of $x^4$ in the expansion of $\text{(1)}$ is $v - 3u$. Equating coefficients, $v = 7 + 3u$. So now we want to equate $\text{(2)}$ with
$\tag 3  (x-u)^3 \,(x^2+(7+3u)x+w)$
Equating the coefficients of $x^3$ from $\text{(3)}$ with  $\text{(2)}$ now gives
$\quad -6u^2 - 21u + w = 19 \; \text{ iff } \; w = 19  +6u^2 + 21u$
Eliminating $w$, 
$\tag 4  (x-u)^3 \,(x^2+(7+3u)x+ 19  +6u^2 + 21u)$
Equating the coefficients of $x^2$ from $\text{(4)}$ with  $\text{(2)}$ now gives
$\quad - 10 u^3 - 42 u^2 - 57 u = 26$
So we need to solve
$\tag 5  10 u^3 + 42 u^2 + 57 u - 26 = 0$
Using the rational root theorem you get that $u = -2$.
So $v = 7 + 3u = 1$ and $w = 19  +6u^2 + 21u = 1$ and expanding
$\quad (x+2)^3 \,(x^2+x+1)$
gives
$\tag {ANS} x^5 + 7 x^4 + 19 x^3 + 26 x^2 + 20 x + 8$
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