Hello how would I factor 81x2 −9 y2 ? the x and y are squared
what u need help on
81x2 −9 y2 9xy(9x-y)
9xy(9x-y)
well\[81x^{2} -9y^{2} i need help with factoring 50 cent\]
well how did you guys get that answer
They factored it.
Take the common.
first they divided by nine
9xy was the common
for me to understand i need you to write it out for me please or take it step by step with me because i still do not understand
9xy....? \[\large 9xy(9x-y) \qquad=\qquad 81x^2y-9xy^2\] That doesn't look like the equation he started with :\ hmm
\[\large 81x^2-9y^2\]81 = 9*9, so let's rewrite it that way, \[\large 9\cdot9x^2-9y^2\] It appears they only thing these terms have in common is a 9. Divide a 9 out of each term,\[\large 9x^2-y^2\]Then we have to multiply each term by 9 to keep it balanced,\[\large 9(9x^2-y^2)\] Do you understand how factoring works Sunlight?
Hello Zepdrix I factored both 81 AND 9
Oh I made a boo boo... we have the `difference of squares`.. so we can simplify this down a bit further.
and i got 9*9 xx - 3*3yy
Ah yes very good :) So you can rewrite each term as a square.\[\large (9x)^2-(3y)^2\]
And from here, we can apply a special rule.
\[\large a^2-b^2=(a-b)(a+b)\]
what is the a and the b suppose to stand for?
I used different letters to show you a rule. We can apply that rule to our problem. So in our case, our a=9x, b=3y and we can break it down using the rule I posted above.
\[\large \color{royalblue}{a}^2-\color{red}{b}^2=(a-b)(a+b)\]\[\large \color{royalblue}{(9x)}^2-\color{red}{(3y)}^2=?\]
(9x-3y)(9x+3y)
is that okay?
Yay good job \c:/
Subtracting squared terms, try to remember that rule. It comes up a lot as you advance in math classes.
lol uhm okay but can i use the foiling method afterwards to work backwards?
Yes, that's a great way to check your work.
Hey I just did the foil and I got it yay thank you thank you :o))
cool c:
well okay good bye for now
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