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?
 A: In your last equation, we can divide by $2$, giving us $$10a^3+42a^2+57a+26=0$$
By rational root theorem, it's easy to see that $a=-2$ is a solution, so dividing the polynomial by $a+2$ gives us $10a^2+22a+13=0$, so the remaining roots are $\frac{-11\pm3i}{10}$.
You can substitute these roots into the other polynomials.
A: $f''$ is a cubic, so it has at least one real root.
By trials $\alpha = -2$ is a root of $20 \alpha ^ 3 + 84 \alpha^2 + 114 \alpha + 52 = 0$ and the other two roots are complex. So the only possible triple root is $-2$. It remains to solve:
$$32-2a+b=0,\ -20+a=0$$
Can you end it from here?
A: 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
$\tag {ANS} x^5 + 7 x^4 + 19 x^3 + 26 x^2 + 20 x + 8$
A: $20 \alpha ^ 3 + 84 \alpha^2 + 114 \alpha + 52 = 0$
$10\alpha^3 + 42\alpha^2 + 57\alpha + 26=0$
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 $b=32-112 +152-104 + 40 = 8$.
A: 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. 
