Let $f(x)$ be a polynomial. If $(x-a)$ is a factor of $f(x)$, prove that if $f(x)$ is divided by $(x-a)^2$, the remainder will be equal to $n(x-a)$ for some real value $n$.

I have tried setting $f(x) $ to equal $g(x)(x-a)^2+k(x-a)$, but all that really proves that an equation with this form will satisfy it. I have no idea where to go from here, and to say that any equation following this form will work seems really flawed. Any help is appreciated.


Note that $f(x) = (x-a)g(x)$ for some $g(x)$ with degree $1$ less than $f(x)$. Also note that since $\deg(f)\geq 2, g(x)= (x-a)h(x) + r$ for some real $r$ and $h(x)$ with degree $2$ less than $f(x)$. Substituting back, we obtain:

$f(x) = h(x)(x - a)^{2} + r(x - a)$

Thus, $f(x)$'s remainder when divided by $(x-a)^{2}$ is of the form $r(x-a)$. $\blacksquare$


It follows from Euclidean division of polynomials (https://en.m.wikipedia.org/wiki/Polynomial_long_division) that

$f(x) = (x - a)^2g(x) + r(x), \tag 1$

where either

$r(x) = 0 \tag 2$


$\deg r(x) < 2; \tag 3$

in case (2), we take

$n = 0, \tag 4$


$r(x) = 0 = n(x - a) = 0(x - a); \tag 3$

in case (3) we have

$\deg r(x) \le 1, \tag 4$

and thus $r(x)$ may be written

$r(x) = \alpha x + \beta, \; \alpha, \beta \in \Bbb R; \tag 5$

we evaluate (1) at

$x = a, \tag 6$

and obtain

$f(a) = (a - a)^2g(a) + r(a) = r(a); \tag 7$

since $x - a$ is a factor of $f(x)$,

$f(x) = (x - a)h(x), \tag 8$

for some

$h(x) \in \Bbb R[x]; \tag 9$

from (8),

$f(a) = (a - a)h(a) = 0, \tag{10}$

then (5), (7) and (10) together become

$r(a) = \alpha a + \beta = 0, \tag{11}$


$\beta = -\alpha a, \tag{12}$

and then

$r(x) = \alpha x - \alpha a = \alpha (x - a), \tag{13}$

which is of the requisite form with

$n = \alpha; \tag{14}$

that is,

$f(x) = (x - a)^2g(x) + \alpha(x - a). \tag{15}$


Let $f(x)= (x-a)^2g(x)+ r(x)$. Clearly $deg(r)\leq 1$ . Given $(x-a)$ divides $f(x)$. Therefore $(x-a)$ divides $(x-a)^2g(x)+ r(x)$. This implies $(x-a)$ divides $r(x)$. Therefore $r(x)= n(x-a)$ ( where $n$ is a real value ) as $deg(r) \leq 1$.


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