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Factoring Trinomials with Leading Coefficient of 1
Factor: x 2  4 x  3
Solution :
a  1; b  4; c  3
We are looking for two numbers whose product is c  3 and
whose sum is b  4.
(1)(3) = 3
(-1)(-3) = 3
(1) + (3) = 4
(-1) + (-3) = -4
Thus, factors of x 2  4 x  3 are ( x  1)( x  3).
Factor: x 2  5 x  6
Solution :
a  1; b  5; c  6
We are looking for two numbers whose product is c  6 and
whose sum is b  5.
(1)(6) = 6
(-1)(-6) = 6
(2)(3) = 6
(-2)(-3) = 6
(1) + (6) = 7
(-1) + (-6) = -7
(2) + (3) = 5
(-2) + (-3) = -5
Thus, factors of x 2  5 x  6 are ( x  2)( x  3).
Factoring Trinomials with Leading
Coefficient Not 1
Factor: 2x 2  7 x  3
Solution :
a  2; b  7; c  3
a  c  (2)(3)  6
We are looking for two numbers whose product is 6 and
whose sum is  7.
Note: (-6)(-1) = 6 and (-6) + (-1) = -7
Rewrite 2x 2  7 x  3 as
2x 2  6 x  1x  3
2x
2
 6 x    1x  3
2x  x  3  ( 1)  x  3
 x  3 (2 x  1)
Thus, factors of 2x 2  7 x  3 are  x  3  (2 x  1).
Factor: 8x 2  14 x  3
Solution :
a  8; b  14; c  3
a  c  (8)(3)  24
We are looking for two numbers whose product is 24 and
whose sum is  14.
Note: (-12)(-2) = 24 and (-12) + (-2) = -14
Rewrite 8x 2  14 x  3 as
8x 2  12 x  2 x  3
8 x
2
 12 x    2 x  3
4x  2 x  3  ( 1)  2 x  3 
 2 x  3 (4 x  1)
Thus, factors of 8x 2  14 x  3 are  2 x  3  (4 x  1).