Question 206660
{{{(y-5)^2-36q^2}}} Start with the given expression.



Let {{{z=y-5}}}



{{{z^2-36q^2}}} Replace {{{y-5}}} with "z"



{{{(z)^2-36q^2}}} Rewrite {{{z^2}}} as {{{(z)^2}}}.



{{{(z)^2-(6q)^2}}} Rewrite {{{36q^2}}} as {{{(6q)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=z}}} and {{{B=6q}}}.



So let's use the difference of squares formula {{{A^2-B^2=(A+B)(A-B)}}} to factor the expression:



{{{A^2-B^2=(A+B)(A-B)}}} Start with the difference of squares formula.



{{{(z)^2-(6q)^2=(z+6q)(z-6q)}}} Plug in {{{A=z}}} and {{{B=6q}}}.



So this shows us that {{{z^2-36q^2}}} factors to {{{(z+6q)(z-6q)}}}.



{{{(y-5+6q)(y-5-6q)}}} Now plug in {{{z=y-5}}}




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Answer:



So {{{(y-5)^2-36q^2}}} completely factors to {{{(y-5+6q)(y-5-6q)}}}



In other words {{{(y-5)^2-36q^2=(y-5+6q)(y-5-6q)}}}