\(P\left(x\right)⋮\left(x+5\right)\\ \Leftrightarrow P\left(-5\right)=625-4\left(-125\right)-19\cdot25+106\left(-5\right)+m=0\\ \Leftrightarrow625+500-475-530+m=0\\ \Leftrightarrow120+m=0\\ \Leftrightarrow m=-120\)
\(\Leftrightarrow P\left(x\right)=x^4-4x^3-19x^2+106x-120\\ P\left(x\right)=x^4-3x^3-x^3+3x^2-22x^2+66x+40x-120\\ P\left(x\right)=\left(x-3\right)\left(x^3-x^2-22x+40\right)⋮\left(x-3\right)\)
Vậy P(x) chia x-3 dư 0