he state $1223|003|01+|10|11.$ We want to write this as the product of two single$-$qubit states, $|$$1|$$0,$ where $|$$1=$ $1|0+$$1|1,|$$0=$ $0|0+$$0|1.$ Then, $1484$ Multiple Quantum Bits $|$$1|$$0=($$1|0+$$1|1)($$0|0+$$0|1)=$ $1$$0|00+$$1$$0|01+$$1$$0|10+$$1$$0|11.$ Matching up the coefficients with our original state, $1$$0=322,$ $1$$0=322,$ $1$$0=122,$ $1$$0=122.$ Using these equations, let us solve for the variables in terms of one of them. Starting with the first equation, we can solve for $1$ in terms of $0$: $1=322$$0.$ Plugging this into the second equation, we can solve for $0$ in terms of $0$: $0=$$0.$ For the third equation, we can solve for $1$ in terms of $0$: $1=122$$0.$ Finally, plugging in $1=1/22$$0$ and $0=$$0$ into the fourth equation, we get $122=122,$ which is a true statement, so it is satisfied, although it does not tell us anything new. So$,$ we have solved for $1,$ $1,$ and $0$ in terms of $0,$ and this is actually sufficient. Plugging into the product state, $|$$1|$$0=($$1|0+$$1|1)($$0|0+$$0|1)=322$$0|0+1221$ $0|1!($$0|0$$0|1).$ We see that $0$ cancels, yielding $|$$1|$$0=322|0+122|1!(|0|1).$ Moving the factor of $1/2$ to the right qubit so that both qubits are normalized, $|$$1|$$0=32|0+12|1!12|012|1$ This is the style of the factorization. Can you try with $1/4(3|003|01+3|10|11)?$