I recently bought a bose PS48 powered speaker system but unsure how to connect them to bose 301 series lV speakers. Can anyone

Solution
If you don't have the media center that goes with the subwoofer you can use it at all.
If you do then connect the Bose 301 to the front speaker outputs. In this case they use RCA jacks instead of speaker terminals so you will need an RCA plug at the woofer end of the speaker wire. You can splice the RCA plug off of a pair of interconnect cables to the speaker wire but best not to use the entire interconnect as speaker wire. The wire going to the center pin of the RCA is the plus (usually red).
You may not have enough power to drive the 301s. Many Bose speakers are 2 ohm so they design the amps for that load. The 301s may be 6-8 ohms which will reduce the available power.
It appears the PS48 is specifically designed to use an Bose outboard amp to power the speakers. So if you want to use your 301 instead of the speakers they give you, plug the 301 into the outboard amp, L+R?

I notice the Bose outboard amp speakers connection maybe of a different connector/cable type, if so, you have to purchase some kind of adapters or make your own cables. Is no rocket science, + to +, - to -.

At first glance, this PS48 seems to power tinny cube speakers, are you sure the 301 is matched for this unit? Would be kinda of weird to put truck tires onto a sedan.
 
If you don't have the media center that goes with the subwoofer you can use it at all.
If you do then connect the Bose 301 to the front speaker outputs. In this case they use RCA jacks instead of speaker terminals so you will need an RCA plug at the woofer end of the speaker wire. You can splice the RCA plug off of a pair of interconnect cables to the speaker wire but best not to use the entire interconnect as speaker wire. The wire going to the center pin of the RCA is the plus (usually red).
You may not have enough power to drive the 301s. Many Bose speakers are 2 ohm so they design the amps for that load. The 301s may be 6-8 ohms which will reduce the available power.
 
Solution