2.04 Special products
We have been looking at distributing and simplifying products of polynomials. While we can always use the distributive property to distribute, sometimes there are patterns we can use to be more efficient.
Squaring a binomial
Just like perfect square numbers $9$9$\left(3^2\right)$(32) and $144$144$\left(12^2\right)$(122), algebraic expressions such as $a^2$a2, $x^2$x2 and $a^2b^2$a2b2 are also called perfect squares. Squares of binomial expressions, such as $\left(a+b\right)^2$(a+b)2, are also perfect squares, and we can distribute these binomial products in the following way:
| $\left(a+b\right)^2$(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 | Since $ab=ba$ab=ba |
$\left(x+y\right)^2=x^2+2xy+y^2$(x+y)2=x2+2xy+y2
An application of squaring binomials
The square of a binomial appears often, not just in math but in the real world as well.
Squares like these can also be used to show the possible ways that genes can combine in offspring. For example, among tigers, the normal color gene C is dominant, while the white color gene c is recessive. So a tiger with color genes of CC, Cc, or cC will have a normal skin color, while a tiger with color genes of cc will have a white skin color. The following square shows all four possible combinations of these genes.
Since the square for each gene combination represents $\frac{1}{4}$14 of the area of the larger square, the probability that a tiger has color genes of CC or Cc (i.e. it has a normal skin color) is $\frac{3}{4}$34, while the probability that it has color genes of cc (i.e. it has a white skin color) is $\frac{1}{4}$14.
| $\left(\frac{1}{2}C+\frac{1}{2}c\right)^2$(12C+12c)2 | $=$= | $\frac{1}{4}C^2+2\left(\frac{1}{4}C\right)\left(\frac{1}{4}c\right)+\frac{1}{4}c^2$14C2+2(14C)(14c)+14c2 |
| $=$= | $\frac{1}{4}C^2+\frac{1}{2}Cc+\frac{1}{4}c^2$14C2+12Cc+14c2 |
Practice questions
Question 1
Complete the distribution of the perfect square: $\left(x-3\right)^2$(x−3)2
$\left(x-3\right)^2=x^2-\editable{}x+\editable{}$(x−3)2=x2−x+
Question 2
Write the perfect square trinomial that factors as $\left(s+4t\right)^2$(s+4t)2.
Question 3
Distribute the following perfect square: $\left(4x+7y\right)^2$(4x+7y)2
Sum and difference
We've already looked at how to distribute parentheses when we were multiplying a binomial by a single number, as well as how to distribute binomial products. Now we are going to look at a special case of distributing binomial products, called the sum and difference or the difference of two squares.
When we have a sum and difference in a factored form, it looks something like this:
$\left(a+b\right)\left(a-b\right)$(a+b)(a−b)
To distribute this binomial product, will still use the distributive law, making sure we multiply both terms in the first set of parentheses by both terms in the second set of parentheses, as shown in the picture below.
By doing this we get the distributed expression:
$a^2-ab+ab-b^2$a2−ab+ab−b2
If we then collect any like terms and simplify the expression we are left with the difference of two squares:
$a^2-b^2$a2−b2
$\left(x+y\right)\left(x-y\right)=x^2-y^2$(x+y)(x−y)=x2−y2
So in general, if we see something of the form $\left(x+y\right)\left(x-y\right)$(x+y)(x−y) (or equivalently $\left(x-y\right)\left(x+y\right)$(x−y)(x+y)) we know it's distribution will be $x^2-y^2$x2−y2. Let's confirm this result by distributing using our rectangle method.
Set up the rectangle with the binomial expressions on each side
Multiply each term and write the answers in the relevant boxes.
When collecting like terms we can see the $ab$ab will cancel out with the $ba$ba. leaving us just with $a^2-b^2$a2−b2. A difference of two squares.
Practice questions
Question 4
Distribute the following:
$\left(m+3\right)\left(m-3\right)$(m+3)(m−3)
Question 5
Distribute the following:
$\left(3x-8\right)\left(3x+8\right)$(3x−8)(3x+8)
Question 6
Distribute the following:
$6\left(8x-9y\right)\left(8x+9y\right)$6(8x−9y)(8x+9y)