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OpenStudy (mertsj):
25a^1 ????
OpenStudy (anonymous):
maybe \(25a^2\)?
OpenStudy (anonymous):
whoops yes i meant 25 squared!
OpenStudy (anonymous):
so the problem is 25a squared - (3a +2b) squared
OpenStudy (anonymous):
\[25a^2-(3a+2b)^2\] use \(x^2-y^2)=(x+y)(x-y)\) with \(x=5a,y=(3a+2b)\)
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OpenStudy (anonymous):
That's the difference of two squares: so
\[a^2-b^2=(a-b)(a+b) \]
set \[a=\sqrt{25a^1} \] or \[a=\sqrt{25a^2} \] whichever one it is. Do the same for the second term, set b to equal the square root of the second term.
OpenStudy (anonymous):
YES. my book says factoring when either square is a square of a polynomial
OpenStudy (anonymous):
so satellite73, would the answer be (5a+3a+2b)(5a-3a+2b)??
OpenStudy (anonymous):
It works whether they are squares or not.
OpenStudy (anonymous):
i see, so would that be the answer?? or would you factor both of those trinomials and multiply?
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OpenStudy (anonymous):
What you have is almost right:
\[ [5a+(3a+2b)]*[5a-(3a+2b))\]
\[ [5a+(3a+2b)]*[5a-3a-2b))\]
check that -2b, you had "+2b" instead.
OpenStudy (anonymous):
i see, so when i multiply that, i should get my original question?