Question

factor: 5x+10-2z-zx+5y-zy
=5(x+2)-z(2-x)+y(5-z)
now what can i do??

Answer 1

factor the stuff on the right of the equal sign
distribute the 5 -z and y

Answer 2

\[10+5 x+5 y-2 z-x z-y z\text{= }5 (2+x)+y (5-z)-(2-x) z\]\[10+5 x+5 y-2 z-x z-y z\text{=}10+5 x+5 y-2 z+x z-y z \]\[10+5 x+5 y-2 z-x z-y z-(10+5 x+5 y-2 z+x z-y z)=0 \]\[-2 x z=0\]

Answer 3

if you factor out -2z-zx it is -z(2+x) therefore your factorization should be 5(x+2)-z(x+2)+y(5-z) and upon grouping in to fator (5-z)(x+2)+y(5-z) but 5-z is a common factor on the equation hence factoring it out you'll have (x+2+y)(5-z)