How do you use the perfect square trinomial formula to factor #16y^2-8y+1#?

1 Answer
Aug 4, 2018

Answer:

#16y^2-8y+1=(4y-1)(4y-1)#

Explanation:

Here,
#f(y)=16y^2-8y+1#
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To complete perfect square let #M# be the third term,
such that #16y^2-8y +M # is a perfect square.

So,

#color(blue)((i)1^(st) term=16y^2#
#color(blue)((ii)2^(nd)term=-8y#
#color(blue)((iii)3^(rd)term=M#

Formula :
#color(blue)(3^(rd)term=(2^(nd)term)^2/(4xx 1^(st) term)#

#M=(-8y)^2/(4xx16y^2)=(64y^2)/(64y^2)=1#

#i.e. 16y^2-8y+1 # is a perfect square.
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So ,

#f(y)=16y^2-8y+1#

#:.f(y)=(4y)^2-2(4y)(1)+(1)^2#

#:.f(y)=(4y-1)^2#