How do you factor #4y^2 + 4yz + z^2 - 1#?

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Answer 1 · 2016-05-15T08:24:51

#4y^2+4yz+z^2-1 = (2y+z-1)(2y+z+1)#

Focusing our attention on just the terms of degree #2#, notice that both #4y^2 = (2y)^2# and #z^2# are perfect squares, and we find:

#(2y+z)^2 = (2y)^2+2(2y)z+z^2 = 4y^2+4yz+z^2#

So we have:

#4y^2+4yz+z^2-1 = (2y+z)^2 - 1#

Then since #(2y+z)^2# and #1=1^2# are both perfect squares, we can use the difference of squares identity:

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

with #a=(2y+z)# and #b=1# as follows:

#(2y+z)^2 - 1#

#= (2y+z)^2-1^2#

#= ((2y+z)-1)((2y+z)+1)#

#= (2y+z-1)(2y+z+1)#

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Bonus

One way of spotting that:

#4y^2+4yz+z^2 = (2y+z)^2#

is to notice the pattern #4, 4, 1# of coefficients on the left hand side.

You may know that #441=21^2# so notice that the pattern of coefficients on the right hand side is #2, 1#.

When the number #21# is squared to give #441# there are no carries from one column to another, so we have:

#21xx21 = (20+1)xx(20+1) = (20xx20)+2(20xx1)+(1xx1)#

#= 400+40+1 = 441#

This is like putting #y=10# and #z=1# in #4y^2+4yz+z^2# and #(2y+z)#.

This works for some other quadratics with small coefficients. For example:

#(x+3)^2 = x^2+6x+9# like #13^2 = 169#

#(2x+1)(x+3) = 2x^2+7x+3# like #21 xx 13 = 273#