Factor polynomial $$ y^{2}-289 $$
Answer
$$ y^{2}-289 = ( y-17 )( y+17 ) $$
Explanation
Rewrite $ y^{2}-289 $ as:
$$ y^{2}-289 = (y)^2 - (17)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = y $ and $ II = 17 $ , we have:
$$ y^{2}-289 = (y)^2 - (17)^2 = ( y-17 ) ( y+17 ) $$This page was created using
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