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 $21 xx 13 = 273$

How do you factor 4y^2 + 4yz + z^2 - 14y^2 + 4yz + z^2 - 1?