We've already come across binomial expressions when we looked at how to expand brackets using the distributive law. Expressions such as $2\left(x3\right)$2(x−3) are the product of a term (outside the brackets) and a binomial expression (the sum or difference of two terms). So a binomial is a mathematical expression in which two terms are added or subtracted. They are usually surrounded by brackets or parentheses, such as $\left(x+7\right)$(x+7).
Recall that to expand $2\left(x3\right)$2(x−3) we use the distributive law: $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC
Now we want to look at how to multiply two binomials together, such as $\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d).
When we multiply binomials of the form $\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d) we can treat the second binomial $\left(cx+d\right)$(cx+d) as a constant term and apply the distributive property in the form $\left(B+C\right)\left(A\right)=BA+CA$(B+C)(A)=BA+CA. The picture below shows this in action:
As you can see in the picture, we end up with two expressions of the form $A\left(B+C\right)$A(B+C). We can expand these using the distributive property again to arrive at the final answer:
$ax\left(cx+d\right)+b\left(cx+d\right)$ax(cx+d)+b(cx+d)  $=$=  $acx^2+adx+bcx+bd$acx2+adx+bcx+bd 
$=$=  $acx^2+\left(ad+bc\right)x+bd$acx2+(ad+bc)x+bd 
Let's try an example.
Expand and simplify $\left(x+5\right)\left(x+2\right)$(x+5)(x+2).
Think: We need to multiply both terms inside $\left(x+5\right)$(x+5) by both terms inside $\left(x+2\right)$(x+2).
Do:
$\left(x+5\right)\left(x+2\right)$(x+5)(x+2)  $=$=  $x\left(x+2\right)+5\left(x+2\right)$x(x+2)+5(x+2) 

$=$=  $x^2+2x+5x+10$x2+2x+5x+10 
Distribute both sets of brackets 

$=$=  $x^2+7x+10$x2+7x+10 
Simplify by collecting like terms 
Reflect: We chose to split the binomial $\left(x+5\right)$(x+5) up multiply $\left(x+2\right)$(x+2) by both $x$x and $5$5. We could just have easily chosen to split up $\left(x+2\right)$(x+2) and multiply each term by $\left(x+5\right)$(x+5), as follows:
$\left(x+5\right)\left(x+2\right)$(x+5)(x+2)  $=$=  $x\left(x+5\right)+2\left(x+5\right)$x(x+5)+2(x+5) 

$=$=  $x^2+5x+2x+10$x2+5x+2x+10 
Distribute both sets of brackets 

$=$=  $x^2+7x+10$x2+7x+10 
Simplify by collecting like terms 
Let's take the same example as above, $\left(x+5\right)\left(x+2\right)$(x+5)(x+2) and see how this expansion works diagramatically by finding the area of a rectangle.
Notice that the length of the rectangle is $x+5$x+5 and the width is $x+2$x+2. So one expression for the area would be $\left(x+5\right)\left(x+2\right)$(x+5)(x+2).
Another way to express the area would be to split the large rectangle into two smaller rectangles. This way, the area would be $x\left(x+2\right)+5\left(x+2\right)$x(x+2)+5(x+2). Notice that this is the same expression we get after using the distributive property as shown above.
Finally, if we add up the individual parts of this rectangle, we get $x^2+5x+2x+10$x2+5x+2x+10, which simplifies to $x^2+7x+10$x2+7x+10  the same final answer we found above.
The area of the figure below is $\left(y+3\right)\left(y+6\right)$(y+3)(y+6). We want to find another expression for this area by finding the sum of the areas of the four smaller rectangles.
What is the area of rectangle $A$A?
What is the area of rectangle $B$B?
What is the area of rectangle $C$C?
What is the area of rectangle $D$D?
Using the areas of each rectangle, write an equivalent expression for the area of the figure.
Expand and simplify the following:
$\left(x+6\right)\left(x12\right)$(x+6)(x−12)
Expand and simplify the following:
$3\left(y+3\right)\left(y+5\right)$3(y+3)(y+5)
Fill in the blanks to make the expression true.
Simplify algebraic expressions by applying properties of operations of numbers, using various representations and tools, in different contexts.