Question 93455
Start with the given expression


{{{(5y^3*x^2+z)(2y^3*w^2+3z)}}}


When you FOIL, you multiply the terms in this order:

F-First   (i.e. you multiply the first terms in each parenthesis which in this case are {{{5y^3*x^2}}} and {{{2y^3*w^2}}})
O-Outer   (i.e. you multiply the outer terms in each parenthesis which in this case are {{{5y^3*x^2}}} and {{{3z}}})
I-Inner   (i.e. you multiply the inner terms in each parenthesis which in this case are {{{z}}} and {{{2y^3*w^2}}})
L-Last    (i.e. you multiply the last terms in each parenthesis which in this case are {{{z}}} and {{{3z}}})



So lets multiply the first terms: 

{{{5y^3*x^2*2y^3*w^2=10w^2y^6x^2}}}   multiply {{{5y^3*x^2}}} and {{{2y^3*w^2}}} to get {{{10w^2y^6x^2}}}




So lets multiply the outer terms: 

{{{5y^3*x^2*3z=15zy^3x^2}}}   multiply {{{5y^3*x^2}}} and {{{3z}}} to get {{{15zy^3x^2}}}




So lets multiply the inner terms: 

{{{z*2y^3*w^2=2zy^3w^2}}}   multiply {{{z}}} and {{{2y^3*w^2}}} to get {{{2zy^3w^2}}}




So lets multiply the last terms: 

{{{z*3z=3z^2}}}   multiply {{{z}}} and {{{3z}}} to get {{{3z^2}}}


Now lets put all the multiplied terms together

 {{{10w^2y^6x^2+15zy^3x^2+2zy^3w^2+3z^2}}}



 So the expression


 {{{(5y^3*x^2+z)(2y^3*w^2+3z)}}}


 FOILs to:


 {{{10w^2y^6x^2+15zy^3x^2+2zy^3w^2+3z^2}}}