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Factor.\newline2u2+9u+92u^2 + 9u + 9

Full solution

Q. Factor.\newline2u2+9u+92u^2 + 9u + 9
  1. Identify coefficients: Identify the coefficients aa, bb, and cc in the quadratic expression 2u2+9u+92u^2 + 9u + 9 by comparing it to the standard form ax2+bx+cax^2 + bx + c.a=2a = 2, b=9b = 9, c=9c = 9
  2. Find factorable numbers: Find two numbers that multiply to acac (a×ca \times c) and add up to bb. In this case, ac=2×9=18ac = 2 \times 9 = 18, and we need two numbers that multiply to 1818 and add up to 99.
  3. Rewrite middle term: The two numbers that satisfy the conditions are 33 and 66 because 3×6=183 \times 6 = 18 and 3+6=93 + 6 = 9.
  4. Group terms for factoring: Rewrite the middle term 9u9u using the two numbers found in the previous step. This will allow us to split the middle term for factoring by grouping.\newline2u2+9u+92u^2 + 9u + 9 becomes 2u2+3u+6u+92u^2 + 3u + 6u + 9.
  5. Factor out common factors: Group the terms into two pairs and factor out the common factor from each pair.\newlineFrom 2u2+3u2u^2 + 3u, factor out uu to get u(2u+3)u(2u + 3).\newlineFrom 6u+96u + 9, factor out 33 to get 3(2u+3)3(2u + 3).
  6. Final factored form: Notice that both groups now have a common binomial factor 2u+32u + 3. Factor out the common binomial factor.\newlineThe factored form is u+3u + 32u+32u + 3.