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This section introduces the technique of completing the square.
This section will introduce the technique of completing the square. This is used
heavily in certain topics of calculus and trigonometry and isn’t necessarily a type of
factoring so much as a technique to manipulate something into a more easily factored
form.
_
Completing the Square
The principle idea of completing the square is to rewrite a quadratic form as a
binomial term squared plus/minus a constant. Let’s see an example of what we
mean.
Consider the polynomials and . It may not be initially obvious, but these
polynomials are actually very nearly the same. Rewriting the into its expanded form
we get: So is really just , plus one more. This means that if we were to add zero
cleverly by adding and subtracting 1, we could get the following: So we have
rewritten the polynomial as a binomial squared (ie ) minus . This process of
producing a perfect square out of the and terms, by adding and subtracting some
constant value (in the above case, we added and subtracted ) is the technique of
completing the square.
In our previous example we were given what the binomial square was at
the start but it might not be as obvious what you should choose as your
target perfect square just by looking at your polynomial. Luckily there is a
way to determine what the target perfect square should be based on your
polynomial.
First we will give the general derivation, and then below we will give a concrete
example to help solidify the process with specific numbers. Keep in mind you can
also review the previous example and compare to this general process as
well.
Let’s say we have the polynomial . Our goal is to end up with something of the form:
. As is often the case in mathematics, we will work backwards, expanding our
goal to see how it compares to our original form. Thus expanding we get
Since is the coefficient of on the left, and is the coefficient of on the
right, we can deduce that ; meaning that we want half of the term on the
left to be the value that we will be added to in the completing the square
form. Moreover it also shows us that we need as a constant value on the left
hand side of the equality to mirror the that is on the right hand side of the
equality.
That may seem a bit dense, but the take away is the following: For our
;
(a)
Divide the term in half; getting .
(b)
Square and then add and subtract it from the expression, this yields:
(c)
The left three terms above factor perfectly into which means we can
simplify our polynomial as follows:
(d)
The far right side of the equality above is our Completing the Square end
result.
Ok, the above is a bit of alphabet soup, so it will help to see another concrete
example following these steps.
Complete the square for the expression
(a)
Divide the term in half. So we get .
(b)
Square and then add and subtract it from the expression. So we get:
thus we now get: Notice that we don’t want to simplify the above; that
would defeat the point.
(c)
The left three terms above factor perfectly. So we get:
(d)
The far right side of the equality above is our Completing the Square end
result.
So; completing the square on the expression yields .
