How do you factor #4n^2 + 68n + 250 #?

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How do you factor #4n^2 + 68n + 250 #?

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Answer 1 · 2015-10-10T03:13:15

Factor #f(n) = 4n^2 + 68n + 250#

Explanation:

#f(n) = 2y = 2(2n^2 + 34n + 125).#
Factor # y = 2n^2 + 34n + 125 #
#D = b^2 - 4ac = 1156 - 4000 = -2844 < 0#. Since its D < 0, this trinomial y can't be factored.
Finally, #f(n) = 2(2n^2 + 34n + 125)#

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How do you factor #4n^2 + 68n + 250 #?

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