 # 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.)

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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