Question: Using the simple multiplication algorithm where we store multiplier in the right half of result register and shift right in each cycle (add multiplicand if

Using the simple multiplication algorithm where we store multiplier in the right half of result register and shift right in each cycle (add multiplicand if the least significant bit of multiplier is 1, or add zero otherwise), perform the following multiplication X*Y

x = -6 = 1111 1010

y = +5 = 0000 0101

Note one of the number is negative, so you will need to takes 2s complement, and multiply only positive numbers. But since the result is negative, you need to complement the result.

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