How do you factor #49x ^ { 2} - 81#?

2 Answers
Suren Abreu
Aug 6, 2017

#(7x-9)(7x+9)#

Explanation:

#49x^2-81#

To make this more like a quadratic equation, we can add (without changing the value) #(+0x)#, since #0x=0# and its addition does not change the value of the equation.

#49x^2+0x-81#

Factorise.

#49=>7*7#

#81=>9*9#

Hence:

#49x^2+63x-63x-81#

Notice that #(+63x-63x=0)# which is the same as #+0x=0#.

#7x(7x+9)-9(7x+9)#

#(7x-9)(7x+9)#

EZ as pi
Aug 6, 2017

#(7x+9)(7x-9)#

Explanation:

This is an identity known as the difference of two squares:

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

#49x^2-81#

#=(7x+9)(7x-9)#

The clues to look for are:

  • #2# terms with a minus sign
  • perfect squares
  • even indices
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