Factoring Quadratic Expressions

arithmetic examples - working with numbers

The quadratic formula for finding roots of expressions  comes follow from (i) completing the square and then (ii) factoring if (i) results in the difference of two squares. Numerical examples follow to illustrates the step. The quadratic formula itself is derived in the next lesson.

In other parts of this site (see chapters 14 an 15 of the site volume 2, Three Skills for Algebra, arithmetic and algebraic solutions to questions have been developed. This lesson provides the arithmetic solution for solving quadratics equations or factoring them. The next lesson gives the algebraic solution or route. In the next lesson, the general case in which the letters a, b and c are not immediately replaced by numbers follows the same route and leads to the quadratic formula.

1. Quadratic with two roots

x²-8x + 12 =

=  x²+2(-4)x + 12  take Q = -4
= (x+-4)² - 4² + 12
= (x-4)² - 4
= (x-4)² -  2²  = a difference of two squares
= (x -4 +2) (x -4 -2)
= (x -2)(x-6)

Now x²-8x + 12 = (x -2)(x - 6) = 0 when and only when x = 2 or x = 6 by the zero product law.

Exercise: Identify the intervals where the factors (x--2) and (x-6) are postive and negative. Then do a sign analysis of the product

y = (x -2)(x-6) =  x²-8x + 12

The format for this sign analysis appears in Chapter 3, Slope Sign Analysis, of Volume 3, Why Slopes and More Math.

Remark: Factoring by inspection would be quicker here as -8 = -2 + -6 and 12 =( 6)(1).  But the aim here was to illustrate the route: Complete the square and then factor, if possible, using the difference of two squares.

2. Quadratic with two roots

x²+6x + 5 =

=  x²+2(3)x + 5  take Q = 3
= (x+3)² - 3² + 5
= (x+3)² - 4
= (x+3)² - 2² 

= a difference of two squares

Factoring the difference of two squares (x+3)² - 2² will tell us when x²+6x + 5 may have the value zero,

0 = x²+6x + 5
   = (x+3)² -  2²
    =  (x+3+2)(x+3-2)
   =  (x+5)(x+1)

The zero product law says the latter product is zero when and only when

x + 5 = 0 or x+ 1 = 0.

That is, when and only when

x = -5 or x = -1

respectively. 

Exercise: Identify the intervals where the factors (x+5) and (x+1) are postive and negative. Then do a sign analysis of the product. 

y = x²+6x + 5 = (x+5)(x+1)

The format for this sign analysis appears in Chapter 3, Slope Sign Analysis, of Volume 3, Why Slopes and More Math.

Remark: Factoring by inspection would be quicker here as 6 = 5 + 1 and 5 = (5)(1).  But the aim here was to illustrate the route: Complete the square and then factor, if possible, using the difference of two squares.

3.  Quadratic with no  roots

x²-8x + 25 =

=  x²+2(-4)x + 25  take Q = -4
= (x+-4)² - 4² + 25
= (x+3)² + 9
= (x+3)² +  3²  = a sum of two squares
> 9
> 0 

So x²-8x + 25 > 9 is never zero

Exercise: Identify the interval where (x+3)² + 9 is positive. 

Optional Exercise: Read about complex numbers and then write  9 = - (3 i)² to obtain a difference of two squares. Here i = sqrt(-1)

4. Quadratics with coefficient a of x² not unity (not 1)

3x²- 24x + 10 =

= 3 [x²-8x + 12]

=  3[x²+2(4)x - 4² +12]  take Q = 4
= 3[(x+4)² - 4]

5. Using q = [sqrt(q)]² in factoring a quadratic

Example:  Factor x²-10x + 8

x²-10x + 8 =

=  x²+2(-5)x + 8  take Q = -5
= (x+-5)² - 5² + 8
= (x-5)² - 17
= (x-5)² - [(17)^½]²

 = a difference of two squares, namely the square of x -5 and the square 17 of the square root (17)^½

Factoring the difference of two squares will tell us when

 (x-5)² -  [(17)^½]²

may have the value zero:

Here: (x-5)² -  [(17)^½]² 

    = [(x-5) +   (17)^½] [(x-5) -  (17)^½]
   = [(x-5) +   (17)^½] [(x-5) -   (17)^½]  
   = [ x  - {5 -   (17)^½}]  [ x  - {5 +  (17)^½]} ]

The Zero Product law implies the latter can  be zero when and only when

  x  =  5 -   (17)^½    or  x  =  5 +   (17)^½

Use a calculator to find the approximate decimal values to locate these roots on the x-axis.

Exercise: Identify the intervals where the factors (x--2) and (x-6) are postive and negative. Then do a sign analysis of the product

y =   x²-10x + 8 = [ x  - {5 -   (17)^½}]  [ x  - {5 +  (17)^½]} ]

The format for this sign analysis appears in Chapter 3, Slope Sign Analysis, of Volume 3, Why Slopes and More Math.

Remark: Factoring by inspection would be quicker here as -8 = -2 + -6 and 12 =( 6)(1).  But the aim here was to illustrate the route: Complete the square and then factor, if possible, using the difference of two squares.

Remark: Any number  q  > 0 equals to the square of its square root.  That is, = [sqrt(q)]² allows to write 17 as as [(17)^½]²  in the foregoing.

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