How do you factor #2x^2-x-21#?

1 Answer
Aug 10, 2015

Answer:

#2x^2-x-21 = (2x-7)(x+3)#

Explanation:

Assuming that #2x^2-x-21#
can be written as #(ax+-c)(bx+-d)#
#color(white)("XXXX")##color(white)("XXXX")#with #a, b, c, d in ZZ#

Since the sign of the last term (#-21#) is negative the interior signs of the two binomials must be opposite
#color(white)("XXXX")##rarr# we want #(ax-c)(bx+d)#;
and
since the sign of the middle term (#-x#) is negative
#cb > ad#

Looking at factors of 2: #(ab) in {(2,1)}#
and factors of 21: #(c,d) in { (3,7), (7,3), (1,21), (21,1)}#

we find
#color(white)("XXXX")##(a,b) = (2,1)# and #(c,d) = (7,3)#
give us
#color(white)("XXXX")##a*d - b*c = (-1)#

So our factors are
#color(white)("XXXX")##(2x-7)(x+3)#