Factor polynomial $$ 4z^{2}-169 $$
Answer
$$ 4z^{2}-169 = ( 2z-13 )( 2z+13 ) $$
Explanation
Rewrite $ 4z^{2}-169 $ as:
$$ 4z^{2}-169 = (2z)^2 - (13)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 2z $ and $ II = 13 $ , we have:
$$ 4z^{2}-169 = (2z)^2 - (13)^2 = ( 2z-13 ) ( 2z+13 ) $$This page was created using
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