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

##### 1 Answer

#### Answer:

#### Explanation:

Focusing our attention on just the terms of degree

#(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

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

with

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

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

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

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

**Bonus**

One way of spotting that:

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

is to notice the pattern

You may know that

When the number

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

#= 400+40+1 = 441#

This is like putting

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#