Okay heres the question:

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

Let $\displaystyle f$ be any polynomial function. Let:

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

Where:

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

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

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

Lets look at one example:

For

$\displaystyle f(x) = x^2$

we know that:

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

and that:

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

and also that:

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

Its obvious that since:

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

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

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

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

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

For some function $\displaystyle g(x)$.

In other words, prove that $\displaystyle (x-a)^2$ is always a factor of $\displaystyle \Delta_a(x)$ for all polynomial $\displaystyle f$.