Factoring Trinomials of the Form x^{2} + bx + c and x^{2}  bx + c
Just as the product of two binomials can often be rewritten as a
trinomial, trinomials of the form ax^{2} + bx + c can often be
rewritten as the product of two binomials. For example, x^{2} + 3x + 2 = (x + 1)(x + 2).
We now know that the product of two binomials of the form (x + d )
and (x + e) is given by:
(x + d )(x + e) = x^{2} + xe + dx + de = x^{2} + (d + e)x + de 

Thus, in order to rewrite a binomial
x^{2} + bx + c as the product of
two binomials (
b positive or negative), we must find numbers
d and
e such that
d + e = b and
de = c. Since
c is positive,
d
and
e must have the same sign.
Here are the steps to factoring a trinomial of the form x^{2} + bx + c, with c > 0. We assume that the coefficients are integers, and
that we want to factor into binomials with integer coefficients.
 Write out all the pairs of numbers which can be multiplied to produce c.
 Add each pair of numbers to find a pair that produce b when added. Call the
numbers in this pair d and e.
 If b > 0, then the factored form of the trinomial is (x + d )(x + e). If b < 0,
then the factored form of the trinomial is (x  d )(x  e).
 Check: The binomials, when multiplied, should equal the original trinomial.
Note: Some trinomials cannot be factored. If none of the pairs total
b, then
the trinomial cannot be factored.
Example 1: Factor x^{2} + 5x + 6.
 Pairs of numbers which make 6 when multiplied: (1, 6) and (2, 3).
 1 + 6≠5. 2 + 3 = 5. Thus, d = 2 and e = 3 (or vice versa).
 (x + 2)(x + 3)
 Check: (x + 2)(x + 3) = x^{2} +3x + 2x + 6 = x^{2} + 5x + 6
Thus,
x^{2} + 5x + 6 = (x + 2)(x + 3).
Example 2: Factor x^{2}  7x + 12.
 Pairs of numbers which make 12 when multiplied: (1, 12), (2, 6), and (3, 4).
 1 + 12≠7. 2 + 6≠7. 3 + 4 = 7. Thus, d = 3 and e = 4.
 (x  3)(x  4)
 Check: (x  3)(x  4) = x^{2} 4x  3x + 12 = x^{2}  7x + 12
Thus,
x^{2}  7x + 12 = (x  3)(x  4).
Example 3: Factor 2x^{3} +4x^{2} + 2x.
First, remove common factors: 2x^{3} +4x^{2} +2x = 2x(x^{2} + 2x + 1)
 Pairs of numbers which make 1 when multiplied: (1, 1).
 1 + 1 = 2. Thus, d = 1 and e = 1.
 2x(x + 1)(x + 1) (don't forget the common factor!)
 Check: 2x(x + 1)(x + 1) = 2x(x^{2} +2x + 1) = 2x^{3} +4x^{2} + 2x
Thus,
2x^{3} +4x^{2} +2x = 2x(x + 1)(x + 1) = 2x(x + 1)^{2}.
x^{2} + 2x + 1 is a perfect square trinomial.