Question

Can you help me solve this

Answer 1

Keeping in mind that i² = -1,

$\bf \begin{array}{c|ccccc} -1+i&&1&5&1&4\\ &&&-1+i\\ &&---&--&--&--\\ &&1&4+i \end{array}\\\\ -------------------------------\\\\ (-1+i)(4+i)\implies -4-i+4i+\stackrel{-1}{i^2}\implies -4-i+4i-1 \\\\\\ -5+3i\\\\ -------------------------------$

$\bf \begin{array}{c|ccccc} -1+i&&1&5&1&4\\ &&&-1+i&-5+3i\\ &&---&---&----&--\\ &&1&4+i&-4+3i \end{array}\\\\ -------------------------------\\\\ (-1+i)(-4+3i)\implies 4-3i-4i+3\stackrel{-1}{i^2}\implies 4-3i-4i-3 \\\\\\ 1-7i\\\\ -------------------------------\\\\ \begin{array}{c|cccccc} -1+i&&1&5&1&4\\ &&&-1+i&-5+3i&1-7i\\ &&---&---&----&---\\ &&1&4+i&-4+3i&\boxed{5-7i}&\leftarrow remainder \end{array}$


recall that a factor of the expression, gives a remainder of 0, and this one doesn't.