# Thread: Proving that (x-a)^2 is always a factor of \Delta(x)?

1. ## Proving that (x-a)^2 is always a factor of \Delta(x)?

Okay heres the question:

Let $f$ be any polynomial function. Let:

$\tau_a(x) = f'(a)(x-a)+f(a)$

Where:

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Also, let the difference between the tangent and $f$ be denoted by $\Delta$, which is defined:

$\Delta_a(x) = f(x) - \tau_a(x) = f(x) - f'(a)(x-a)-f(a)$

Lets look at one example:

For

$f(x) = x^2$

we know that:

$\frac{d}{dx} [f(x)] = f'(x) = 2x$

and that:

$\tau_a(x) = (2a)(x-a)+a^2$

and also that:

$\Delta_a(x) = x^2 - (2ax - 2a^2 +a^2) = x^2 - 2ax +a^2$

Its obvious that since:

$(p-q)^2 = p^2 -2pq + q^2$

We know that $(x-a)^2$ is a factor of $\Delta_a(x)$, specifically that:

$\Delta_a(x) = (1)(x-a)^2$

Now, prove generally, for all polynomial functions $f$, that $\Delta_a(x)$ can always be written in the form:

$\Delta_a(x) = g(x)(x-a)^2$

For some function $g(x)$.

In other words, prove that $(x-a)^2$ is always a factor of $\Delta_a(x)$ for all polynomial $f$.
Okay, so I've been trying to do this with polynomial long division and other such attempts, and I just can't figure it out. Any help would be much appreciated. Thanks in advance

2. Since $\Delta_a(x) = f(x) - f(a) - (x-a)f'(a)$, it is obvious that $\Delta_a(x)$ is a polynomial. Also, $\Delta_a(a) = 0$, so by the factor theorem $x-a$ is a factor of $\Delta_a(x)$.

Now differentiate: $\Delta_a'(x) = f'(x) - f'(a)$. Thus $\Delta_a'(a) = 0$, and again by the factor theorem $x-a$ is a factor of $\Delta_a'(x)$. But that means that $x-a$ is a repeated factor of $\Delta_a(x)$. In other words, $(x-a)^2$ divides $\Delta_a(x)$, and so $\Delta_a(x) = g(x)(x-a)^2$ for some polynomial g(x).