Mastering Algebra - Recognizing Special Products and Their Factors - Part II

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Now that we understand some key algebraic terminology, we are prepared to recognize some special products and to be able to factor them accordingly.  Herein we master how to recognize and factor both differences of perfect squares and perfect square trinomials.

Perfect squares are numbers which have square roots which are integers.  Thus 25, 36, and 49 are all perfect squares because their respective square roots are 5, 6, and 7.  When you have an algebraic binomial which is the difference of two squares, we can always factor this expression in a convenient fashion.  Let us examine a specific example.  Take x^2 - y^2.  Here x^2 and y^2 are the squares.  Because we are taking the difference, this expression is aptly named a "difference of two perfect squares."  We can always factor an expression like this as (x - y)(x + y).  If we insert perfect square coefficients in front of the variables, no matter.  We simply factor the expression by taking the square root of the number and the variable, and placing them in that "-", "+" pattern.  For example 49x^2 - 25y^2 is factored as (7x - 5y)(7x + 5y). 

The difference of two perfect squares has a nice application in short-cut arithmetic.  Namely, we can use this method to perform some lightning multiplications.  For example, take any two numbers that differ from a common "ten" by the same amount.  Specifically, take 36 and 44.  Both these numbers differ from 40, the "common ten," by 4 units.  If we want to multiply 36*44, we can get the answer immediately, which is 1584!  How?  Write 36 as 40 - 4 and 44 as 40 + 4.  Then 36*44 = (40 - 4)(40 + 4) which is 40^2 - 4^2, which is 1600 - 16, which is 1584.  Try some others out on your own to see the beauty of this method.

Perfect square trinomials comprise another set of special products.  These expressions are generated by the product of two identical binomials.  To wit, (x + 1)^2 = (x + 1)(x + 1).  When multiplied out, this yields x^2 + 2x + 1.  All perfect square trinomials are formed this way.  These expressions are such that the middle term coefficient is double the product of the square root of the constant term and the coefficient of first term.  In x^2 + 2x + 1, 2 = 2*1*1, 1 being the square root of the constant term 1, and 1 being the square root of the coefficient of x^2, which is also 1.  Whenever we have a trinomial that meets this condition, we can always factor it using the square of the binomial which meets these conditions. 

Take an example to make this clear.  Look at x^2 + 10x + 25.  The square root of 25 is 5.  The square root of 1, the coefficient of x^2 is 1; 2*5*1 is 10, which is the middle term coefficient.  Therefore x^2 + 10x + 25 can be factored as (x + 5)^2.

Take one more.  Look at x^2 - 12x + 36.  The square root of 36 is 6.  The square root of 1, the coefficient of x^2 is 1; 2*6*1 is 12, which is the middle term coefficient with a "-" sign in front.  No problem.  We adjust our product by writing a "-" instead of a "+" and get  x^2 - 12x + 36 = (x - 6)^2.  That's all there is to it.  You can now handle any perfect square trinomial on the planet!

Perfect square trinomials play a very important role in a process known as "completing the square."  This process is so named because geometrically, this involves making a square from a rectangle by "completing the square on the rectangle."  This method allows us to solve quadratic equations quite easily and also gives us the proof of the famous quadratic formula.  So if you're ever wondering where that strange formula on the cover of your algebra book came from, remember that a perfect square trinomial had something to do with it.  Wow, isn't it great when you learn where things come from?  See you next time...

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Joe Pagano has 1 articles online

Joe is a prolific writer of self-help and educational material and is the creator and author of over a dozen books and ebooks which have been read throughout the world. He is a former teacher of high school and college mathematics and has recently returned as a professor of mathematics at a local community college in New Jersey.

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Mastering Algebra - Recognizing Special Products and Their Factors - Part II

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This article was published on 2010/03/27