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Factor.\newline2n2+11n+92n^2 + 11n + 9

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Q. Factor.\newline2n2+11n+92n^2 + 11n + 9
  1. Identify aa, bb, cc: Identify aa, bb, and cc in the quadratic expression 2n2+11n+92n^2 + 11n + 9. Compare 2n2+11n+92n^2 + 11n + 9 with ax2+bx+cax^2 + bx + c. a=2a = 2 bb00 bb11
  2. Find numbers multiply and add: Find two numbers that multiply to aca*c (29=182*9 = 18) and add up to bb (1111).\newlineWe need to find two numbers that multiply to 1818 and add up to 1111.\newlineAfter checking possible pairs that multiply to 1818 (11 and 1818, 22 and 29=182*9 = 1800, 29=182*9 = 1811 and 29=182*9 = 1822), we find that 22 and 29=182*9 = 1800 are the numbers we are looking for because 29=182*9 = 1855.
  3. Rewrite middle term: Rewrite the middle term 11n11n using the two numbers found in Step 22.\newlineWe can express 11n11n as the sum of 2n2n and 9n9n.\newline2n2+11n+92n^2 + 11n + 9 becomes 2n2+2n+9n+92n^2 + 2n + 9n + 9.
  4. Factor by grouping: Factor by grouping.\newlineFirst, group the terms: 2n2+2n2n^2 + 2n + 9n+99n + 9.\newlineNow, factor out the common factors from each group.\newlineFrom the first group, we can factor out 2n2n: 2n(n+1)2n(n + 1).\newlineFrom the second group, we can factor out 99: 9(n+1)9(n + 1).
  5. Write factored form: Write the factored form of the expression.\newlineSince both groups contain the common factor (n+1)(n + 1), we can factor this out.\newlineThe factored form is (2n+9)(n+1)(2n + 9)(n + 1).