When does polynomial equation have a double root?

polynomials

Solution 1:

Given any polynomial $P(x)$, we have the general principle that

$(x - c)^2 \mid P(x) \tag 0$

if and only if

$(x - c) \mid P(x), \; (x - c) \mid P'(x), \tag 1$

for (0) implies

$P(x) = (x - c)^2Q(x) \tag 2$

for some polynomial $Q(x)$; then

$P'(x) = 2(x - c)Q(x) + (x - c)^2Q'(x), \tag 3$

and thus clearly

$(x - c) \mid P'(x), \tag 4$

and (1) binds. Going the other way, from (1) we may write

$P(x) = (x - c)R(x) \tag 5$

whence

$P'(x) = R(x) + (x - c)R'(x), \tag 6$

$R(x) = P'(x) - (x - c)R'(x); \tag 7$

since $(x - c) \mid P'(x)$ it follows that

$(x - c) \mid R(x), \tag 8$

and so we have

$R(x) = (x - c)S(x) \tag 9$

for some polynomial $S(x)$; now from (5),

$P(x) = (x - c)R(x)$ $= (x - c)(x - c)S(x) = (x - c)^2 S(x), \tag{10}$

and we see that (0) is in fact the case.

We apply this concept to

$P(x) = x^n - ax^{n - 1} + bx - 1, \tag{11}$

observing that

$(x - 1)^2 \mid P(x) \Longrightarrow P(1) = 0$ $\Longrightarrow -a + b = 1 - a + b - 1 = 0 \Longrightarrow a = b; \tag{12}$

thus (11) becomes

$P(x) = x^n - ax^{n - 1} + ax - 1, \tag{13}$

whence

$P'(x) = nx^{n - 1} - a(n - 1)x^{n - 2} + a; \tag{14}$

thus,

$P'(1) = 0 \Longrightarrow n - a(n - 1) + a = 0 \Longrightarrow n - an + a + a = 0$ $\Longrightarrow n + (2 - n)a = 0 \Longrightarrow a = \dfrac{n}{n - 2}; \tag{15}$

then

$P(x) = x^n - \dfrac{n}{n - 2}x^{n - 1} + \dfrac{n}{n - 2}x - 1, \tag{16}$

$P'(x) = nx^{n - 1} - \dfrac{n(n - 1)}{n - 2}x^{ - 2} + \dfrac{n}{n - 2}. \tag{17}$

We Check:

From (6),

$P(1) = 1 - \dfrac{n}{n - 2} + \dfrac{n}{n - 2} - 1 = 0; \tag{18}$

and from (7),

$P'(1) = n - \dfrac{n(n - 1)}{n - 2} + \dfrac{n}{n - 2} = \dfrac{n(n - 2)}{n - 2} - \dfrac{n(n - 1)}{n - 2} + \dfrac{n}{n - 2} = \dfrac{n^2 - 2n -n^2 + n + n}{n - 2} = 0. \tag{19}$