Signed number representation
In C, C++ and many modern languages, signed numbers are represented as two's complement.
So the 1 byte hex number 0x8A is the binary number 10001010. This is a signed number, negative (as it starts with a 1 bit on the most significant position -- left side). To know what number it represents, we have to reverse the two's complement format: subtract 1 bit from from the least significant position -- right side --, then invert the bits. After the subtraction we get 10001001. Inverting this we get 01110110. This equals to the decimal number 118. But we know this was a negative number (remember? most significant bit was 1), so the actual signed byte representation of 0x8A really means -118 decimal.
Okay, now that we cleared this out of the way, it will be probably easier to understand what's going on.
If you look at the specification of the printf() function, you can notice that the format parameter 'x' means unsigned hexadecimal integer. But an integer is not 1 byte, but 4 bytes. So printf() will treat -118 as an unsigned integer. The binary representation of -118 as a signed int (4 bytes) is similar to the -118 as a signed char (1 byte), except everything to the left is all 1s, that is 0xFF (for each byte). Hence the 4 byte signed representation of -118 is 0xFFFFFF8A hex. So this is what printf() puts on the standard output.
A better solution
The two other answers try to fix your input. One of them by changing the actual value (I don't really like this). The other by telling the compiler to interpret the 1 byte 0x8A as an unsigned number, instead of a signed number, so the 4 byte extension would not include all the 0xFFFFFF__ in the front. A bit better, but there's a real simple solution: read the manual. XD
printf() actually supports length modifiers for most integral numbers and their formatting. This includes the hexadecimal display x. You can apply the hh length modifier in front of x format to tell printf() that you will have a char there (1 byte) not an int (4 bytes).
So in my opinion, the proper solution is: