Using the FOIL distribution method:

$\displaystyle (30+3)(40+4)$

First: $\displaystyle 30*40 = 1200$

Outer: $\displaystyle 30*4= 120$

Inner: $\displaystyle 3*40 = 120$

Last: $\displaystyle 3*4=12$

Resulting in: $\displaystyle 1200 + 120 +120 +12 = 1452$

Using the FOIL distribution method:

$\displaystyle (5x + t)(2x + 3t)$

First: $\displaystyle 5x*2x = 10x^2$

Outer: $\displaystyle 5x*3t= 15x t$

Inner: $\displaystyle t*2x = 2xt$

Last: $\displaystyle t*3t=3t^2$

Resulting in: $\displaystyle 10x^2 + 15xt + 2xt + 3t^2$

Combining like terms, the $\displaystyle xt$'s combine for a final answer of:

$\displaystyle 10x^2 + 17xt + 3t^2$

Using the FOIL distribution method:

$\displaystyle (x + y)(x - y)$

First: $\displaystyle x*x = x^2$

Outer: $\displaystyle x*-y= -xy$

Inner: $\displaystyle y*x = xy$

Last: $\displaystyle y*-y=-y^2$

Resulting in: $\displaystyle x^2 - xy + xy - y^2$

Combining like terms, the $\displaystyle xy$'s cancel for a final answer of:

$\displaystyle x^2 - y^2$

Expressions of this form are commonly referred to as "difference of squares". If you can spot them, they are easy to expand and to factor because the middle terms always cancel.

First: $\displaystyle (x)(x)=x^2$

Outside: $\displaystyle (x)(-2)=-2x$

Inside: $\displaystyle (x)(7)=7x$

Last: $\displaystyle (-2)(7)=-14$

Add the values together and combine like terms:

$\displaystyle x^2-2x+7x-14=x^2+5x-14$

First: $\displaystyle (2n)(-n)=-2n^2$

Outside: $\displaystyle (2n)(3)=6n$

Inside: $\displaystyle (4)(-n)=-4n$

Last: $\displaystyle (4)(3)=12$

Add the values together and combine like terms:

$\displaystyle -2n^2+6n-4n+12=-2n^2+2n+12$

$\displaystyle \left ( 2x+5 \right )\left ( x+3 \right )$

FOIL (first, outside, inside, last) is a device to help students remember to multiply every term by every other term exactly one time. The steps of using FOIL progress as follows:

1) Multiply the first term of each parenthetical expression together:

$\displaystyle (2x)(x)=2x^2+...$

2) Next, multiply the "outside" terms together:

$\displaystyle (2x)(3)=6x$

Add that to the first product, since the terms are all being added together in the parentheses, and the distributive property requires that we maintain the sign:

$\displaystyle 2x^{2}+6x...$

3) Multiply the "inside" terms together:

$\displaystyle (5)(x)=5x$

Add that to the first two products:

$\displaystyle 2x^{2}+6x+5x...$

4) Finally, multiply the last terms from each parenthetical expression together:

$\displaystyle (5)(3)=15$

Add all of the terms together and combine like terms where possible:

$\displaystyle 2x^{2}+5x+6x+15=2x^{2}+11x+15$

$\displaystyle \left ( 5x+7 \right )\left ( 4x+12 \right )$

FOIL (first, outside, inside, last) is a device to help students remember to multiply every term by every other term exactly one time. The steps of using FOIL progress as follows:

1) Multiply the first term of each parenthetical expression together:

$\displaystyle (4x)(5x)=20x^2+...$

2) Next, multiply the "outside" terms together:

$\displaystyle (5x)(12)=60x$

Add that to the first product, since the terms are all being added together in the parentheses, and the distributive property requires that we maintain the sign:

$\displaystyle 20x^{2}+60x...$

3) Multiply the "inside" terms together:

$\displaystyle (7)(4x)=28x$

Add that to the first two products:

$\displaystyle 20x^{2}+60x+28x...$

4) Finally, multiply the last terms from each parenthetical expression together:

$\displaystyle (7)(12)=84$

Add all of the terms together and combine like terms where possible:

$\displaystyle 20x^{2}+60x+28x+84=20x^{2}+88x+84$

Using the FOIL distribution method:

$\displaystyle (50-1)(50+1)$

First: $\displaystyle 50*50 = 2500$

Outer: $\displaystyle 50*1= 50$

Inner: $\displaystyle -1*50 = -50$

Last: $\displaystyle -1*1=-1$

Resulting in: $\displaystyle 2500 + 50 - 50 - 1$

The 50's cancel leaving us with:

$\displaystyle 2500 -1 = 2499$

Using the FOIL distribution method:

$\displaystyle (2x + 3y)(3x + 2y)$

First: $\displaystyle 2x*3x = 6x^2$

Outer: $\displaystyle 2x*2y= 4xy$

Inner: $\displaystyle 3y*3x = 9xy$

Last: $\displaystyle 3y*2y=6y^2$

Resulting in: $\displaystyle 6x^2 + 4xy + 9xy + 6y^2$

Combining like terms, the $\displaystyle xy$'s combine for a final answer of:

$\displaystyle 6x^2 + 13xy + 6y^2$

Begin by considering the first part of this expression, $\displaystyle (5x+3)(2x-1)$. Multiplying the first terms gives 

$\displaystyle (5x)(2x)=10x^{2}$

The product of the outside terms is $\displaystyle -5x$ and that of the inside terms is $\displaystyle 6x$, and the product of the last terms is $\displaystyle -3$. Altogether, this yields 

$\displaystyle 10x^{2}+x-3$ 

as the value of the first part of the expression.

Finally, we just need to add the value of the second part,

$\displaystyle -3(x+4)=-3x-12$

The final, simplified value of the expression is $\displaystyle 10x^{2}-2x-15$.