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
(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
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.
|Factoring ax2 + bx + c||Example: 6x2 + 7x - 3|
| Step 1. Using the leading coefficient
we write the pair of incomplete factors
|Step 1. The lead coefficient is 6, so we
(6x )(6x )
|Step 2. Multiply a and c, the
of x2 and the constant term.
|Step 2. a × c = (6)(- 3) = -18|
|Step 3. Find integers whose product
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
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)