6.2,6.3 Factoring Trinomials

							6.2,6.3 Factoring Trinomials

ü Trinomials of the Form x 2 + b x + c

If a polynomial of the form x2 + b x + c is factorable then it can be written as Hx + uL Hx + vL.
Note: Hx + uL Hx + vL = x2 + v x + u x + u ÿ v = x2 + Hu + vL x + u ÿ v
Looking at this carefully we see that if a trinomial of the form x2 + b x + c is factorable then b = u + v and c = u ÿ v, for some
numbers u and v. So to factor this form of a polynomial we need to find two numbers that have product c and that have sum b.


    ü Examples

 Example 1: Factor x2 + 7 x + 12


Solution: We need to find two numbers that multiply to be 12, but add to be 7. These two numbers are 3 and 4. This then gives

                                               x2 + 7 x + 12 = Hx + 3L Hx + 4L
the factorization



 Example 2: Factor x2 - 12 x + 35


Solution: We need to find two numbers that multiply to be 35, but add to be -12. These two numbers are -7 and - 5. This then

                                                      x2 - 12 x + 35 = Hx - 7L Hx - 5L
gives the factorization



 Example 3: Factor x2 + 5 x - 24


Solution: We need to find two numbers that multiply to be -24, but add to be 5. These two numbers are -3 and 8. This then

                                                       x2 + 5 x - 24 = Hx - 3L Hx + 8L
gives the factorization




ü Trinomials of the Form a x 2 + b x + c

If a polynomial of the form a x2 + b x + c is factorable then it can be written as Hm x + uL Hn x + vL.

                                           Hm x + uL Hn x + vL = m ÿ n x2 + m ÿ v x + n ÿ u x + u ÿ v
Note:


                                                               = m ÿ n x2 + Hm ÿ v + n ÿ uL x + u ÿ v


a ÿ c = m ÿ n ÿ u ÿ v = Hm ÿ vL Hn ÿ uL, for some numbers m, n, u and v. So to factor this form of a polynomial we need to find two
Looking at this carefully we see that if a trinomial of the form a x2 + b x + c is factorable then b = m ÿ v + n ÿ u and


numbers that have product a ÿ c and that have sum b. We can then rewrite the polynomial as four terms (similar to the first step
above) and then factor by grouping.
   ü Examples

 Example 1: Factor 6 x2 + 7 x + 2


Solution: First we find the product of a and c. In this case it's 6 ÿ 2 = 12. So we need two numbers that multiply to be 12, but add
to be 7. These numbers are 3 and 4. We use these numbers to rewrite the middle term as 3 x + 4 x and then factor by grouping as
follows:
                                               6 x2 + 7 x + 2 = 6 x2 + 3 x + 4 x + 2
                                                     = I6 x2 + 3 xM + H4 x + 2L
                                                    = 3 xH2 x + 1L + 2 H2 x + 1L
                                                       = H2 x + 1L H3 x + 2L


 Example 2: Factor 10 x2 - 23 x + 12


Solution: First we find the product 10 ÿ 12 = 120. So we need two numbers that multiply to be 120 and add to be -23. These two
numbers are -15 and - 8. So the middle term can be rewritten as -15 x - 8 x. Now factor by grouping.
                                         10 x2 - 23 x + 12 = 10 x2 - 15 x - 8 x + 12
                                                           = I10 x2 - 15 xM + H-8 x + 12L
                                                           = 5 xH2 x - 3L - 4 H2 x - 3L
                                                           = H2 x - 3L H5 x - 4L


 Example 3: Factor 24 x2 - 14 x - 3


Solution: First we find the product 24 ÿ -3 = -72. So we need two numbers that multiply to be -72 and add to be -14. These
two numbers are -18 and 4. So the middle term can be rewritten as -18 x + 4 x. Now factor by grouping.
                                             24 x2 - 14 x - 3 = 24 x2 - 18 x + 4 x - 3
                                                    = I24 x2 - 18 xM + H4 x - 3L
                                                    = 6 xH4 x - 3L + 1 H4 x - 3L
                                                       = H4 x - 3L H6 x + 1L