Factor `1-16y+64y^2` :

There are special forms that you should strive to memorize. In this case we have the square of a binomial. Any time you have a trinomial whose first and last terms (assuming it is in standard form) are squares, you should suspect a square of a binomial form.

`(a-b)^2=a^2-2ab+b^2`

Here a=1 and b=8y

`(1-8y)^2=1^2-2(1)(8y)+(8y)^2=1-16y+64y^2`

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So `1-16y+64y^2=(1-8y)^2=(1-8y)(1-8y)`

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If you don't recognize the special form, you can try to find two numbers p and q whose product is 64 and whose sum is -16. It is clear that both numbers will be -8 as (-8)(-8)=64 and (-8)+(-8)=-16

Then `1-16y+64y^2=1-8y-8y+64y^2`

`=1(1-8y)-8y(1-8y)`

`=(1-8y)(1-8y)` as before.

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