How do you factor completely #16x^2-8x+1#?

2 Answers
Mia
Jun 16, 2016

#(4x-1)^2#

Explanation:

This is a polynomial of second degree and since the coefficient of
#" "#
#x<0# , we think of the binomial property that says:
#" "#
#a^2-2ab+b^2=(a-b)^2#
#" "#
In the given polynomial first term #16x^2=(4x)^2#and #1=(1)^2#
#" "#
#16x^2-8x+1#
#" "#
#=(4x)^2-2(4x)(1)+1^2#
#" "#
#=(4x-1)^2#

EZ as pi
Sep 29, 2017

#16x^2 -8x+1#

#=(4x-1)(4x-1)#

Explanation:

The #color(lime)(+)# sign in the third term indicates two things:

  • the factors need to be #color(lime)(ADDED)#
  • the signs in the brackets will #color(lime)("be the same")#

The #color(red)(-)#sign in the second term indicates that the signs will be negative.

#16x^2color(red)(-)8x color(lime)(+) 1#

Find factors of #16 and 1# which add to #8#.

The factors of #1# are just #1#, so we can ignore them.

The factors of #16# which add to #8# are #4 and 4#

#4xx4 = 16 and 4+4=8#

#16x^2 -8x+1#

#=(4x-1)(4x-1)#

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