+1124 hi friends,
to factorise perfect squares usually is not a problem for me, as long as all the terms are perfect squares. Please help with this.
\(p^2-9s^2+6pq+9q^2\)
Thank you all for helping I do appreciate.
+2 STEP1:Equation at the end of step 1
(((p2)-(9•(s2)))+6pq)+32q2
Equation at the end of step2:
(((p2) - 32s2) + 6pq) + 32q2
STEP3:
Final result :
p2 + 6pq - 9s2 + 9q2
There are many online calculators website who provides many online calculators for you to use. Few of the very useful and important which i've found are:
Prime Factorization Calculator
These online calculators are very useful for the students and it makes calculus and algebra very easy to solve and easy to understand as well.
+33659 Rewrite \(p^2-9s^2+6pq+9q^2\) as \(p^2+6pq+(3q)^2-9s^2\). Rewrite the first three terms: \((p+3q)^2-9s^2\)
This is the difference of two squares, so can be written as \((p+3q+3s)(p+3q-3s)\)
+1124 hi Alan,
I understand that thank you very much, my problem is how to deal with the 6pq?
Ok Alan, I understand...the 6pq is derived at when the \((p+3q)^2\) is calculated. so you convert the problem to a trinomial basically, and leave the 4th term on its own as a square, so when the trinomial portion is factored, you just add the 4th term and then end up with two terms again which are complete squares as well...then carry on from there to get the final answer...
Thank you very much Alan, I do appreciate..
+118667 Thanks Alan,
Hi Jurimagic,
Perhaps you have not looked at what Alan has done with a fresh eye.
This is ultimately a difference of 2 squares factorization.
Incidentally, I know, and expect Alan knows, that you are always grateful for the help we give you.
That is why, if I am on and I see your question, I will always answer quickly. 
\(p^2-9s^2+6pq+9q^2\\ =p^2+6pq+9q^2-9s^2\\ =(p^2+6pq+9q^2) - (9s^2)\\ =(p+3q)(p+3q) - (9s^2)\\ =(p+3q)^2 - (9s^2)\\ \text{This is the difference of 2 squares}\\ =(p+3q\;\textcolor{red}{\textbf{-}}\;3s)(p+3q\;\textcolor{red}{\textbf{+}}3s)\)
LaTex Coding:
p^2-9s^2+6pq+9q^2\\
=p^2+6pq+9q^2-9s^2\\
=(p^2+6pq+9q^2) - (9s^2)\\
=(p+3q)(p+3q) - (9s^2)\\
=(p+3q)^2 - (9s^2)\\
\text{This is the difference of 2 squares}\\
=(p+3q\;\textcolor{red}{\textbf{-}}\;3s)(p+3q\;\textcolor{red}{\textbf{+}}3s)
