# 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 .

 Factoring ax2 + bx + c Example: 6x2 + 7x - 3 Step 1. Using the leading coefficient of a we write the pair of incomplete factors Step 1. The lead coefficient is 6, so we write(6x )(6x ) Step 2. Multiply a and c, the coefficients of x2 and the constant term. Step 2. a × c = (6)(- 3) = -18 Step 3. Find integers whose product is a × c and the constant term Step 3. Two integers whose product is –18 and whose sum is 7 are 9 and –2. Then we write (6x + 9)(6x – 2) Step 4. Discard any common factor within each parenthesis in Step 3. The result is the desired factorization. Step 4. Reducing (6x + 9) to (2x + 3) and (6x – 2) to (3x – 1) we have 6x2 + 7x – 3 = (2x+3)(3x-1)
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