How do you factor #6x^2-5x-4#?

1 Answer
Jan 17, 2017

# 6x^2-5x-4 =(3x-4)(2x+1) #

Explanation:

The rule to factorise any quadratic is to find two numbers such that

#"product" = x^2 " coefficient "xx" constant coefficient"#
#"sum" \ \ \ \ \ \ = x " coefficient"#

So for # 6x^2-5x-4 # we seek two numbers such that

#"product" = (6)*(-4) = -24#
#"sum" \ \ \ \ \ \ = -5#

So we look at the factors of #-24#. As the sum is negative and the product is negative then one of the factors must be negative, We can check every combination of the product factors:

# {: ("factor1", "factor2", "sum"), (1,-24,-23),(2,-12,-10), (3,-8,-5) ,(4,-6,-2),(-1,24,23),(-2,12,10),(-3,8,5),(-4,6,2) :} #

So the factors we seek are #color(blue)(3)# and #color(green)(-8)#

Therefore we can factorise the quadratic as follows:

# 6x^2-5x-4= 6x^2 color(blue)(+3)x color(green)(-8)x -4 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= 3x(2x+1) -4(2x+1)#
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= (3x-4)(2x+1) #

Similarly if we grouped the factors the other way around we get the same answer:

# 6x^2-5x-4= 6x^2 color(green)(-8)x color(blue)(+3)x-4 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= 2x(3x-4) + (3x-4)#
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= (2x+1)(3x-4) #

This approach works for all quadratics (assuming it does factorise) , The middle step in the last section can usually be skipped with practice.