Factoring Quadratic Polynomials - 1

Senin, 11 April 2022

This shortcut can help you identify quadratic polynomials that do not factor ‘‘nicely’’ without spending too much time on them. The next three examples are quadratic polynomials that do not factor ‘‘nicely.’’

          x² + x + 1              x² + 14x + 19                  x² – 5x + 10

Quadratic polynomials of the form x² – c² are called the difference of two squares. We can use the shortcut on x² – c² = x² + 0x – c². The factors of c2 must have a difference of 0. This can only happen if they are the same, so the factors of c² we want are c and c.

 

When the sign between x² and c² is plus, the quadratic cannot be factored using real numbers.

 

Practice

 

1.     x² – 4 =

 

2.     x² – 64 =

 

3.     25 – x² =

 

Solutions

 

1.     x² – 4 = (x – 2) (x + 2)

 

2.     x² – 64 = (x – 8) (x + 8)

 

3.     25 – x² = (5 – x) (5 + x)

 

The difference of two squares can come in the form xⁿ – cⁿ where n is any even number. The factorization is xⁿ – cⁿ = (x^n/2 – c^n/2) (x^n/2 + c^n/2). [When n is odd, xⁿ – cⁿ can be factored also but this factorization will not be covered here.]

 

Practice

 

 

Solutions

 

 

When the first term is not x², see if you can factor out the coefficient of x². If you can, then you are left with a quadratic whose first term is x². For example each term in 2x² + 16x – 18 is divisible by 2:

2x² + 16x – 18 = 2(x² + 8x – 9) = 2(x + 9)(x – 1)

 

Practice

 

1.     4x² + 28x + 48 =

 

2.     9x² – 9x – 18 =

 

3.     6x² + 24x + 24 =

 

Solutions

 

1.     4x² + 28x + 48 = 4(x² + 7x + 12) = 4(x + 4)(x + 3)

 

2.     9x² – 9x – 18 = 9(x² – x – 2) = 9(x – 2)(x + 1)

 

3.     6x² + 24x + 24 = 6(x² + 4x + 4)

         = 6(x + 2)(x + 2)

         = 6(x + 2)²

 

The coefficient of the x² term will not always factor away. In order to factor quadratics such as 4x² + 8x + 3 you will need to try all combinations of factors of 4 and of 3: (4x + _)(x + __) and (2x + __) (2x + __). The blanks will be filled in with the factors of 3. You will need to check all of the possibilities: (4x + 1)(x + 3), (4x + 3)(x + 1), and (2x + 1)(2x + 3).

 

Examples

          4x² – 4x – 15 = (2x – 5)(2x + 3)

 

You can see that when the constant term and x²’s coefficient have many factors, this list of factorizations to check can grow rather long. Fortunately there is a way around this problem as we shall see in a later chapter.

 

Practice

 

1.     6x² + 25x – 9 =

 

2.     4x² – 9 =

 

3.     12x² + 32x – 35 =

 

Solutions

 

1.     6x² + 25x – 9 = (2x + 9)(3x – 1)

 

2.     4x² – 9 = (2x – 3)(2x + 3)

 

3.     12x² + 32x – 35 = (6x – 5)(2x + 7)

 

 

Sumber

Labels: Mathematician

Thanks for reading Factoring Quadratic Polynomials - 1. Please share...!