Question 90298
Start with the given expression


{{{(x-10)(x+9)}}}


In order to multiply this, we need to FOIL the two parenthesis


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 {{{x}}} and {{{x}}})
O-Outer   (i.e. you multiply the outer terms in each parenthesis which in this case are {{{x}}} and {{{-10}}})
I-Inner   (i.e. you multiply the inner terms in each parenthesis which in this case are {{{-10}}} and {{{x}}})
L-Last    (i.e. you multiply the last terms in each parenthesis which in this case are {{{-10}}} and {{{-10}}})



So lets multiply the first terms: 

{{{x*x=x^2}}}   multiply {{{x}}} and {{{x}}} to get {{{x^2}}}




So lets multiply the outer terms: 

{{{x*9=9x}}}   multiply {{{x}}} and {{{9}}} to get {{{9x}}}




So lets multiply the inner terms: 

{{{-10*x=-10x}}}   multiply {{{-10}}} and {{{x}}} to get {{{-10x}}}




So lets multiply the last terms: 

{{{-10*9=-90}}}   multiply {{{-10}}} and {{{9}}} to get {{{-90}}}


Now lets put all the multiplied terms together

 {{{x^2+9x-10x-90}}}


 Now combine like terms


 {{{x^2-x-90}}}


 So the expression


 {{{(x-10)(x+9)}}}


 FOILs to:


 {{{x^2-x-90}}}