How do you factor the expression #2x^2 - 5x - 3 #?

1 Answer
George C.
Mar 19, 2016

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

Explanation:

Use an AC method. Look for a pair of factors of #AC = 2*3 = 6# with difference #5#.

The pair #6, 1# works, in that #6xx1=6# and #6-1=5#

Use this pair to split the middle term and factor by grouping as follows:

#2x^2-5x-3#

#= 2x^2-6x+x-3#

#= (2x^2-6x)+(x-3)#

#= 2x(x-3)+1(x-3)#

#= (2x+1)(x-3)#

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Alternative method

We will use the difference of squares identity which can be written:

#a^2-b^2 = (a-b)(a+b)#

with #a=(4x-5)# and #b=7# later.

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We will follow the following steps:

  1. Multiply by #2^3 = 8# to simplify the later arithmetic.
  2. Complete the square to make a difference of squares.
  3. Factor the difference of squares.
  4. Simplify.
  5. Divide by #8#.

First multiply by #8#:

#8 xx (2x^2-5x-3) = 16x^2-40x-24#

Then:

#16x^2-40x-24#

#=(4x-5)^2-5^2-24#

#=(4x-5)^2-49#

#=(4x-5)^2 - 7^2#

#=((4x-5)-7)((4x-5)+7)#

#=(4x-12)(4x+2)#

#=(4(x-3))(2(2x+1))#

#=8(x-3)(2x+1)#

Then divide by #8# to find:

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

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