Question 88875
1.

Start with the given expression


{{{(3y-1)(2y+7)}}}


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 {{{3y}}} and {{{3y}}})
O-Outer   (i.e. you multiply the outer terms in each parenthesis which in this case are {{{3y}}} and {{{-1}}})
I-Inner   (i.e. you multiply the inner terms in each parenthesis which in this case are {{{-1}}} and {{{3y}}})
L-Last    (i.e. you multiply the last terms in each parenthesis which in this case are {{{-1}}} and {{{-1}}})



So lets multiply the first terms: 

{{{3y*2y=6y^2}}}   multiply {{{3y}}} and {{{2y}}} to get {{{6y^2}}}




So lets multiply the outer terms: 

{{{3y*7=21y}}}   multiply {{{3y}}} and {{{7}}} to get {{{21y}}}




So lets multiply the inner terms: 

{{{-1*2y=-2y}}}   multiply {{{-1}}} and {{{2y}}} to get {{{-2y}}}




So lets multiply the last terms: 

{{{-1*7=-7}}}   multiply {{{-1}}} and {{{7}}} to get {{{-7}}}


Now lets put everything together

 {{{(3y-1)(2y+7)=6y^2+21y-2y-7=6y^2+19y-7}}}


 So the expression


 {{{(3y-1)(2y+7)}}}


 FOILs to:


 {{{6y^2+19y-7}}}


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2.

Start with the given expression


{{{(4x-5)(2x+9)}}}


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 {{{4x}}} and {{{4x}}})
O-Outer   (i.e. you multiply the outer terms in each parenthesis which in this case are {{{4x}}} and {{{-5}}})
I-Inner   (i.e. you multiply the inner terms in each parenthesis which in this case are {{{-5}}} and {{{4x}}})
L-Last    (i.e. you multiply the last terms in each parenthesis which in this case are {{{-5}}} and {{{-5}}})



So lets multiply the first terms: 

{{{4x*2x=8x^2}}}   multiply {{{4x}}} and {{{2x}}} to get {{{8x^2}}}




So lets multiply the outer terms: 

{{{4x*9=36x}}}   multiply {{{4x}}} and {{{9}}} to get {{{36x}}}




So lets multiply the inner terms: 

{{{-5*2x=-10x}}}   multiply {{{-5}}} and {{{2x}}} to get {{{-10x}}}




So lets multiply the last terms: 

{{{-5*9=-45}}}   multiply {{{-5}}} and {{{9}}} to get {{{-45}}}


Now lets put everything together

 {{{(4x-5)(2x+9)=8x^2+36x-10x-45=8x^2+26x-45}}}


 So the expression


 {{{(4x-5)(2x+9)}}}


 FOILs to:


 {{{8x^2+26x-45}}}


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3.

Start with the given expression


{{{(y-3)(y-3)}}}


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 {{{y}}} and {{{y}}})
O-Outer   (i.e. you multiply the outer terms in each parenthesis which in this case are {{{y}}} and {{{-3}}})
I-Inner   (i.e. you multiply the inner terms in each parenthesis which in this case are {{{-3}}} and {{{y}}})
L-Last    (i.e. you multiply the last terms in each parenthesis which in this case are {{{-3}}} and {{{-3}}})



So lets multiply the first terms: 

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




So lets multiply the outer terms: 

{{{y*-3=-3y}}}   multiply {{{y}}} and {{{-3}}} to get {{{-3y}}}




So lets multiply the inner terms: 

{{{-3*y=-3y}}}   multiply {{{-3}}} and {{{y}}} to get {{{-3y}}}




So lets multiply the last terms: 

{{{-3*-3=9}}}   multiply {{{-3}}} and {{{-3}}} to get {{{9}}}


Now lets put everything together

 {{{(y-3)(y-3)=y^2-3y-3y+9=y^2-6y+9}}}


 So the expression


 {{{(y-3)(y-3)}}}


 FOILs to:


 {{{y^2-6y+9}}}


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