How do you factor #49x^2-144#?

2 Answers
ScienceGuy2002
May 14, 2018

(7x+12)(7x-12)

Explanation:

This problem is known as a 'perfect square.' The square root of 49 is 7 and the square root of #x^2# is #x#. The next number, 144, is also a perfect square and the answer to that is 12.
Firstly, multiply the first numbers/variables: 7#x*#7#x#=49#x^2#
Secondly, multiply the outer numbers/vars.: 7#x*#-12=-84#x#
Thirdly, multiply the inner numbers/vars.: 7#x*#12=84#x#
Lastly, multiply the last numbers/vars.:12#*#12=144
The two number, 84#x# and 84#x#, cancel themselves out and leaves one with 49#x#-144
One CAN'T use this perfect square method IF the equation has a plus instead of a minus.

Meave60
May 14, 2018

#(7x)^2-12^2=(7x+12)(7-12)#

Explanation:

Factor:

#49x^2-144#

This is a difference of squares. Use the formula:

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

where:

#a=7x# and #b=12#

Plug in the known values for #a# and #b#.

#(7x)^2-12^2=(7x+12)(7-12)#

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