Factoring For Second-Degree Polynomials

Factoring involves a certain amount of trial and error, which can become frustrating, especially
when the leading coefficient is not 1. You might want to try a rather neat scheme that will
greatly reduce the number of candidates.

We’ll demonstrate the method for the polynomial
4x² +11x + 6 (1)

Using the leading coefficient of 4 we write the pair of incomplete factors
(4x )(4x ) (2)

Next, multiply the coefficient of x² and the constant term in (1) to produce 4 × 6 = 24 . Now find
two integers whose product is 24 and whose sum is 11, the coefficient of the middle term of (1).
It’s clear that 8 and 3 will do nicely, so we write
(4x+8)(4x+3) (3)

Finally, within each parenthesis in (3) discard any common divisor . Thus (4x+8) reduces to
(x+2) and we write
(x + 2)(4x + 3) (4)

which is the factorization of 4x² +11x + 6.
Will the method always work? Yes—if you first remove all common factors in the original
polynomial. That is, you must first write
6x² + 15x + 6 = 3(2x² + 5x + 2)
and apply the method to the polynomial 2x² + 5x + 2.

(For a proof that the method works, see M. A. Autrie and J. D. Austin, “A Novel Way to
Factor Quadratic Polynomials .”, The Mathematics Teacher 72 no. 2[1979].)

We’ ll use the polynomial 2x² – x – 6 of Example 7 to demonstrate the method when
some of the coefficients are negative .