Question

factor completely.
b^2 - 36

Answer 1

@yummydum

Answer 2

this is a perfect square
you set up 2 binomials where \(m^2-m=(m-n)(m+n)\)
so \(b^2-36\) would factor out to: \((b-6)(b+6)\)

Answer 3

i was working out the problem already goformit......

Answer 4

would that be your answer

Answer 5

yep! thats completely factored! :)

Answer 6

thankk you!

Answer 7

anytime! ;D

Answer 8

cpuld you help with this one.... (2r)^4/(2r^4)

Answer 9

yup yup :D

\[(2r)^4\over(2r^4)\]okay so using the rule of PEMDAS (Parenthesis, Exponents, Multiplication/Division, Addition/Subtraction) we do the exponents first:
\[{(2r)^4\over(2r^4)}~~~~~\implies~~~~~{{2^4\times r^4}\over2r^4}\]so that simplified would give you:\[{8r^4}\over{2r^4}\]and using the rules for exponents, when dividing anything that is raised to a power, subtract the exponents:\[{{8r^4}\over{2r^4}}~~~~~\implies~~~~~{8\over2}\]since the r's cancel out because \(4-4\) (the exponents) \(=0\)so we are left with \(\large{8\over2}\)
lastly we just simplify and get:\[{8\over2}~~~~\implies~~~~\huge{4}\]

Answer 10

thats as easy as i could explain it :) @rayleigh i hope this helps! :)

Answer 11

it helped! thankss

Answer 12

no problem! :)