Where does entanglement fit into quantum computing?
The entanglement of quantum particles is, implicitly or explicitly, the reason why quantum computing is so fast. You will know by now that a qubit (implemented for example as a photon's polarization) can be represented as a unit vector of two complex numbers, and that it may be manipulated by a 2-by-2 matrix of complex numbers. If you have two quantum systems described by their respective state vectors $|a\rangle$ and $|b\rangle$, which describe the two systems in isolation, their composite state is described by the tensor product of these two vectors: $|a\rangle \otimes |b\rangle$. If you operate on $|a\rangle$ with matrix $U$, you get the composite state $(U|a\rangle) \otimes |b\rangle$, and so forth.
Entanglement comes when you operate on this composite state and bring it to a state which cannot emerge as the tensor product of two state vectors. For example, the state $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ is not the tensor product of any two vectors (verify this as an excercise). In such states, measuring qubit $|a\rangle$ has an effect on $|b\rangle$. For example, if you performed a partial measurement (that is, a measurement of ostensibly only one qubit) on the most significant bit of $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$, and you measured it to be $0$, then measuring the least significant bit will yield $0$ with certainty, too, and vice versa.
A good first example for studying entanglement is Deutsch' problem, where the algorithm goes from $(|0\rangle+|1\rangle) \otimes (|0\rangle - |1\rangle)$, a state which can be expressed as the tensor product, to a state which can no longer be expressed as any tensor product, using an operator which cannot be expressed as the tensor product of two operators.