Question

Can someone help me pls?

Answer 1

Expanding this: \[-3(y+2)^2 : -3y^2-12y-12\]
\[-3(y+2)^2\]
Expanding this as well:\[(y+2)^2 :y^2+4y+4\]
\[(y+2)^2\]
Apply perfect square formula, which is like so (example here):\[(a+b)^2=a^2+2\]
So therefore:\[(y+2)^2=y^2+2y \times2+2^2\]
Simplify: \[y^2+2y \times2+2^2 \]
\[=y^2+4y+4\]
So now it should look like this:\[=-3(y^2+4y+4y)\]
Apply distributive law, which is like so (example here): \[m(a+b+c)=ma+mb+mc\]
So:\[-3(y^2+4y+4)=-3y^2-3\times4y-3\times4\]
\[=-3y^2-3\times4y-3\times4\]
Simplify:
\[=-3y^2-12y-12\]
Should look like this now:\[=-3y^2-12y-12-5+6y\]
Group like terms:\[-3y^2-12y+6y-12-5\]
Add similar elements, like your y numbers:
\[=-3y^2-6y-17\]
So, your final answer is:
\[=-3y^2-6y-17\]

Answer 2

You're welcome.