Algebra
UK Secondary (7-11)
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Expand perfect squares
Lesson

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 expand 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

 

Perfect Binomial Squares Fit the Rule:

$\left(x+y\right)^2=x^2+2xy+y^2$(x+y)2=x2+2xy+y2

The square of a binomial appears often, not just in maths 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 colour gene C is dominant, while the white colour gene c is recessive. So a tiger with colour genes of CC or Cc will have a normal skin colour, while a tiger with colour genes of cc will have a white skin colour. 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 colour genes of CC or Cc (i.e. it has a normal skin colour) is $\frac{3}{4}$34, while the probability that it has colour genes of cc (i.e. it has a white skin colour) 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 modelled 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,
 
$\left(\frac{1}{2}C+\frac{1}{2}c\right)^2$(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
 
The final expression shows that the probability that a tiger will have the gene combination CC (i.e. normal skin colour) is $\frac{1}{4}$14, the probability that it will have the gene combination Cc (i.e. normal skin colour) 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.
 

More examples

Question 1

Complete the expansion 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 factorises as $\left(s+4t\right)^2$(s+4t)2.

Question 3

Expand the following perfect square: $\left(4x+7y\right)^2$(4x+7y)2

 

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