How do you factor #2x^2-3x-9#?

1 Answer
Steve M
Dec 6, 2016

# 2x^2-3x-9 = (2x+3)(x-3) #

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 #2x^2-3x-9# we seek two numbers such that

#"product" = (2)*(-9) = -18#
#"sum" \ \ \ \ \ \ = -3#

So we look at the factors of #-18#. As the product is negative one of the factors must also be negative and the other positive, We compute their sum we get

# {: ("factor1", "factor2", "sum"), (18,-1,17), (9,-2,7), (6,-3,3), (-18,1,-17),(-9,2,-7), (-6,3,-3) :} #

So the factors we seek are #color(blue)(-6)# and #color(green)(3)#

Therefore we can factorise the quadratic as follows:

# \ \ \ \ \ 2x^2-3x-9 = 2x^2 color(blue)(-6)x + color(green)(3)x -9 #
# :. 2x^2-3x-9 = 2x(x-3) + 3(x -3) #
# :. 2x^2-3x-9 = (2x+3)(x-3) #

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

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