 7.04 Special products

Lesson

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

Squaring a binomial pattern:

$\left(x+y\right)^2=x^2+2xy+y^2$(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.

It is possible to model the probabilities of the gene combinations as the square of a binomial. Since any parent tiger has a $50%$50% chance of having a C gene and a $50%$50% chance of having a c gene, the genetic makeup of a parent tiger can be modeled as $\frac{1}{2}C+\frac{1}{2}c$12C+12c and that of its offspring as $\left(\frac{1}{2}C+\frac{1}{2}c\right)^2$(12C+12c)2.

Squaring a binomial the usual way,

 $\left(\frac{1}{2}C+\frac{1}{2}c\right)^2$(12​C+12​c)2 $=$= $\frac{1}{4}C^2+2\left(\frac{1}{4}C\right)\left(\frac{1}{4}c\right)+\frac{1}{4}c^2$14​C2+2(14​C)(14​c)+14​c2 $=$= $\frac{1}{4}C^2+\frac{1}{2}Cc+\frac{1}{4}c^2$14​C2+12​Cc+14​c2

The final expression shows that the probability that a tiger will have the gene combination CC (i.e. normal skin color) is $\frac{1}{4}$14, the probability that it will have the gene combination Cc (i.e. normal skin color) is $\frac{1}{2}$12 and the probability that it will have the gene combination cc (i.e. white skin) is $\frac{1}{4}$14, which are exactly the same results as those shown in the gene table above.

Practice questions

Question 1

Complete the distribution of the perfect square: $\left(x-3\right)^2$(x3)2

1. $\left(x-3\right)^2=x^2-\editable{}x+\editable{}$(x3)2=x2x+

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)(ab)

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:

$a^2-ab+ab-b^2$a2ab+abb2

If we then collect any like terms and simplify the expression we are left with the difference of two squares:

$a^2-b^2$a2b2

Sum and difference or difference of two squares

$\left(x+y\right)\left(x-y\right)=x^2-y^2$(x+y)(xy)=x2y2

So in general, if we see something of the form $\left(x+y\right)\left(x-y\right)$(x+y)(xy) (or equivalently $\left(x-y\right)\left(x+y\right)$(xy)(x+y)) we know it's distribution will be $x^2-y^2$x2y2. 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 $a^2-b^2$a2b2.  A difference of two squares.

Practice questions

Question 4

Distribute the following:

$\left(m+3\right)\left(m-3\right)$(m+3)(m3)

Question 5

Distribute the following:

$\left(3x-8\right)\left(3x+8\right)$(3x8)(3x+8)

Question 6

Distribute the following:

$6\left(8x-9y\right)\left(8x+9y\right)$6(8x9y)(8x+9y)

Outcomes

A1.10.B

Multiply polynomials of degree one and degree two

A1.10.D

Rewrite polynomial expressions of degree one and degree two in equivalent forms using the distributive property

A1.10.E

Factor, if possible, trinomials with real factors in the form ax^2 + b^x + c, including perfect square trinomials of degree two