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This is one of the pages in the third section about factoring.   It helps you FACTOR TRINOMIALS

In the first page, I presented the perfect square trinomial and the difference of two squares.

In the second page, I presented a method of factoring four-term expressions with pairs of factors.

Pages 3a, 3b, 3c, and 3d give three methods of factoring trinomials:

Page 3a works by listing the factors of the "a" and "c" coefficients.
Page 3b (this page) works by completing the square.
Page 3c is the "AC Method", which has you multiply the "a" and "c" coefficients, and list all the factors of that product.
Page 3d is the "Simplified AC Method"

Here, I will present another method of factoring trinomials.  This method works by "Completing the Square" and the "Difference of Two Squares".

I will start with EXAMPLE 5, but set it equal to y.  That will make it easier to "multiply through" by a constant, and it will let us use the "add to both sides" rule:

(1)  y = 6x² + 19x - 20

Completing a square means looking for a constant term that makes this a perfect square trinomial. Perfect square trinomials are trinomials like

4x� - 24x + 9

The x-coefficient is twice the product of the square roots of the other two coefficients. To begin completing a square, we need to multiply (or divide) every term by a constant so the x� coefficient is a perfect square. Let's start by "multiplying through" by 6, starting with equation (1), above.

(1)  y = 6x² + 19x - 20
(2)  6y = 36x² + 114x - 120

Let a be the x� coefficient, which is a perfect square, and b be the x coefficient.  Then the constant term that completes the square is ((1/2)b/sqrt(a))�.  In other words, divide b in half, then divide it again by the square root of a, and then square the result.  Here, we divide 114 in half getting 57, divide that by the 6 (the square root of 36), getting 19/2, and then square that, getting 361/4.   This is the term that will complete the square.

Now we can see the perfect square that begins with the first two terms 36x² + 114x is

36x² + 114x + 361/4

So we incorporate this perfect square into our equation without changing its meaning -- by adding and then subtracting the same number -- starting from (2):

(2)  6y = 36x² + 114x - 1 20
(3)  6y = 36x² + 114x + 361/4 - 361/4 - 1 20

Now gather together the terms that make the perfect square trinomial, and combine the leftover constants, like this:

(3)  6y = 36x² + 114x + 361/4 - 361/4 - 1 20
(4)  6y = (36x² + 114x + 361/4) - (361/4 + 1 20)
(5)  6y = (6x + 19/2)² - 841/4
(6)  6y = (6x + 19/2)² - (29/2)²

Now we have a perfect square trinomial minus a perfect square.  As you may remember from the first page in this series, the difference of two squares is the product of the sum and the difference of those same numbers:

(6)  6y = (6x + 19/2)² - (29/2)²
(7)  6y = (6x + 19/2 + 29/2) (6x + 19/2 - 29/2)
(8)  6y = (6x + 24) (6x - 5)

Now we have completed the square, and factored the difference of two squares.  Since we multiplied through by 6 in the first place, we should expect to be able to divide through by 6 here.  And we are not disappointed.  The first factor can be divided by 6, like this:

(8)  6y = (6x + 24) (6x - 5)
(9)  y = (x + 4) (6x - 5)

In case you didn't like this method of factoring trinomials, I have another method for you to try: the AC Method, so named because you find the factors of the product of the "a" and "c" coefficients.

Related pages in this website

Next: the AC Method 


The webmaster and author of this Math Help site is Graeme McRae.