1. ## Polynomials

Question:
If $ax^3 + bx^2 + d=0$ has a double root, show that $27a^2d+4b^3 = 0$

I do not know how to start this question. However, I know that it uses this theorem:

If a polynomial has a root $\alpha$ of multiplicity m, then P'(x) has the root $\alpha$ with multiplicity (m-1).

Any help is appreciated.

2. It's pretty straight forward:

suppose r is the double root of the given polynomial, thus:

$
P(x = r) = ar^3 + br^2 + d = 0
$

now according to the theorem you've already stated:

If a polynomial has a root \alpha of multiplicity m, then P'(x) has the root \alpha with multiplicity (m-1)
$P'(x) = 3ax^2 + 2bx$

so r must be one of the roots of P'(x), thus:

$P'(r) = 3ar^2 + 2br = 0$

$r\left( {3ar + 2b} \right) = 0$

r=0 is a double root of P(x) only if d = 0, which is not necessarily true, so we'll choose: $r = - \frac{{2b}}{{3a}}$

now substitute this value of r in the first equation:

$
a\left( { - \frac{{2b}}
{{3a}}} \right)^3 + b\left( { - \frac{{2b}}
{{3a}}} \right)^2 + d = 0
$

... simplify and obtain the answer.

3. Thank you very much . I was stuck on the r (3ar+2b) part. I never thought to find that root and substitute it.