Question 77977
Since these expressions represent a difference of squares, we can use that identity. For instance, {{{25(x+3)^2-9(x-1)^2}}} is equivalent to

{{{5^2(x+3)^2-3^2(x-1)^2}}} 

which can be written as

{{{(5(x+3))^2-(3(x-1))^2}}} 

So now we use the difference of squares formula to get

{{{(5(x+3))^2-(3(x-1))^2=(5(x+3)+3(x-1))(5(x+3)-3(x-1))}}}


So {{{5^2(x+3)^2-3^2(x-1)^2}}} factors to

{{{(5(x+3)+3(x-1))(5(x+3)-3(x-1))}}}

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The same idea can be applied to  {{{16(x+1)^2-9(x+2)^2}}}


{{{16(x+1)^2-9(x+2)^2}}}

{{{4^2(x+1)^2-3^2(x+2)^2}}}

{{{(4(x+1))^2-(3(x+2))^2=(4(x+1)+3(x+2))(4(x+1)-3(x+2))}}}



So {{{16(x+1)^2-9(x+2)^2}}} factors to

{{{(4(x+1)+3(x+2))(4(x+1)-3(x+2))}}}