We've looked at how to apply the distributive law in groups of different ways. This is just a fancy way of saying expanding brackets, which we can do, even if we have algebraic terms.
In this chapter, we are going to look at questions that involves more than one set of brackets. This includes how to expand binomial products, as well as how to simplify algebraic expressions by expanding multiple sets of brackets, then collecting the like terms.
We've already come across binomial expressions when we looked at how to expand brackets. Expressions such as $2\left(x-3\right)$2(x−3) are the product of a term (outside the brackets) and a binomial expression (the sum or difference of two terms).
So a binomial is a mathematical expression in which two terms are added or subtracted. They are usually surrounded by brackets or parentheses, such as ($x+2$x+2).
The most common way to multiply binomials of the form $\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d) is to use the FOIL method, which stands for First, Outer, Inner, Last. The picture below shows how the terms are multiplied when we use the FOIL method.
Let's 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)
Finally, If we add up the individual parts of this rectangle, we get $x^2+5x+2x+10$x2+5x+2x+10, which simplifies down to $x^2+7x+10$x2+7x+10 - the same answer we got when expanded with the FOIL method.
There are two special cases of binomial expansion that we can use to simplify our calculations. The first case occurs when the expanded form results in a difference of two squares. To see this, let's expand the expression $\left(A-B\right)\left(A+B\right)$(A−B)(A+B) and collect the like terms.
$\left(A-B\right)\left(A+B\right)$(A−B)(A+B) | $=$= | $A^2+AB-BA-B^2$A2+AB−BA−B2 |
$=$= | $A^2-B^2$A2−B2 |
The second case occurs when the product of two binomial expressions are the same, which we call a perfect square. Let's start by expanding the expression $\left(A+B\right)^2$(A+B)2.
$\left(A+B\right)^2$(A+B)2 | $=$= | $\left(A+B\right)\left(A+B\right)$(A+B)(A+B) |
$=$= | $A^2+AB+BA+B^2$A2+AB+BA+B2 | |
$=$= | $A^2+2AB+B^2$A2+2AB+B2 |
So to summarise, when we identify a product in the form of $\left(A-B\right)\left(A+B\right)$(A−B)(A+B), we can expand this expression and get $A^2-B^2$A2−B2. When we see a expression in the form of $\left(A+B\right)^2$(A+B)2, we can expand and get $A^2+2AB+B^2$A2+2AB+B2.
Expand and simplify the following:
$\left(m-6\right)\left(m+3\right)$(m−6)(m+3)
Expand the following perfect square: $\left(6y+\frac{1}{2}\right)^2$(6y+12)2
Expand and simplify the following:
$5x\left(2x-9y\right)\left(2x+9y\right)$5x(2x−9y)(2x+9y)
Expand the following:
$\left(2x+1\right)\left(5x+7\right)\left(2x-1\right)$(2x+1)(5x+7)(2x−1)
Manipulate rational, exponential, and logarithmic algebraic expressions
Apply algebraic methods in solving problems