Hey, everybody! I really need help figuring out how to factor this.
If someone can please provide step by step solution/instruction on how to factor 4y^2+4yz+z^2-1 ...
it would really help me out. Thanks!
When you have four terms and there's nothing that factors out of all of them,
then look at pairs of terms. Can you see anything that might factor out of 4y^2 + 4yz?
Can you think of any factoring formula which might apply to z^2 - 1? Then see where this leads.
Normally I would look at pairs of terms to see if it would factor by grouping,
but in this case, it does not.
No matter how you associate the terms in pairs, that is,
\(\displaystyle 1) \ \ 4y^2 + 4yz \ \ + \ \ z^2 - 1 \ =\)
\(\displaystyle 4y(y + z) \ \ + \ \ (z - 1)(z + 1) \ \ \ or\)
\(\displaystyle 2) \ \ 4y^2 + z^2 \ \ + \ \ 4yz - 1 \ =\)
\(\displaystyle <Each \ \ pair \ \ is \ \ already \ \ prime.> \ \ \ or\)
\(\displaystyle 3) \ \ 4y^2 - 1 \ \ + \ \ 4yz - 1 \ = \)
\(\displaystyle (2y - 1)(2y + 1) \ \ + \ \ 4yz - 1, \)
it won't work.
But the first three terms form a perfect square trinomial, and 1 is a perfect square:
\(\displaystyle (4y^2 + 4yz + z^2) \ - \ 1\)
This is a difference of two squares, and now you should be able to get to the next step
and factor it as a binomial in y and z, which is then squared, and later subtract 1 from that:
\(\displaystyle ( \ \ \ \ \ \ \ \ \ \ \ \ \ \ )^2 \ - \ 1 \ \ \ \ \ \ \ \ \)<----------
Get it to this point at least.
