Spehro's answer is correct. Simple and intuitive, with a good use of Occam' s razor for rationale of the equation form.
Since there are good mathematical reasons for its form it seems worth while to show its origin though. First, the thermal coefficient of resistivity is defined by the differential equation
\$\frac{\text{d$\rho $}}{\text{dT}}\$ = \$\alpha \rho (T)\$
When solved with a boundary condition of \$\rho (T_o)\$ = \$\rho _o\$ an exponential form for resitivity as a function of temperature is obtained:
\$\rho(T) \$ = \$\rho _o e^{\alpha \left(T-T_o\right)}\$
This exponential form contains an approximation, namely that the thermal coeffcient \$\alpha\$ is a constant. For the simpler structured materials (mostly metals), and with a restricted temperature range (like between ~250K and 350K) this will be nearly true. \$\alpha\$ is not really a constant for more extreme temperatures, especially low (cryogenic) temperatures.
A second approximation is made by taking a power series expansion of the exponential form, only keeping the first two terms to get a first order (linear) form. No secret, this is the usual way to obtain a first order model of anything.
\$ \rho(T) \$ = \$\rho _o\left[1+\alpha \left(T-T_o\right)\right]\$
So, the linear form expressing resistivity as a function of temperature is really an approximation of an approximation. Usually the linear form is good enough for the region of interest in electronics. For example for Cu \$\alpha\$ is 0.004, and if \$\text{$\Delta $T}\$ is 50K, the difference of resistivity between the exponential and linear equations is less than 2%. Of course, the smaller the exponent, the more accurate the linear form will be.
\Delta T
\$\endgroup\$ – jippie Mar 2 '14 at 21:01