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Factor.\newline2q2+9q+92q^2 + 9q + 9

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Q. Factor.\newline2q2+9q+92q^2 + 9q + 9
  1. Identify aa, bb, cc: Identify aa, bb, and cc in the quadratic expression 2q2+9q+92q^2 + 9q + 9. Compare 2q2+9q+92q^2 + 9q + 9 with ax2+bx+cax^2 + bx + c. a=2a = 2 bb00 bb11
  2. Find Factors and Sum: Find two numbers that multiply to aca*c (which is 29=182*9=18) and add up to bb (which is 99).\newlineWe need to find two numbers that satisfy these conditions.\newlineAfter checking possible pairs of factors of 1818, we find that there are no such integers that multiply to 1818 and add up to 99.
  3. Factor by Grouping: Since we cannot find integers that satisfy the conditions, we attempt to factor by grouping or check if the quadratic is a perfect square trinomial.\newlineWe can rewrite the expression as (2q2+6q)+(3q+9)(2q^2 + 6q) + (3q + 9) and try to factor by grouping.
  4. Factor Common Factors: Factor out the greatest common factor from each group.\newlineFrom (2q2+6q)(2q^2 + 6q), we can factor out 2q2q, which gives us 2q(q+3)2q(q + 3).\newlineFrom (3q+9)(3q + 9), we can factor out 33, which gives us 3(q+3)3(q + 3).\newlineNow we have 2q(q+3)+3(q+3)2q(q + 3) + 3(q + 3).
  5. Factor Common Binomial: Factor out the common binomial factor (q+3)(q + 3). We can write the expression as (2q+6)(q+3)(2q + 6)(q + 3).
  6. Verify Factored Form: Verify the factored form by expanding it to ensure it equals the original expression. \newline(2q+3)(q+3)=2q2+3q+6q+9=2q2+9q+9(2q + 3)(q + 3) = 2q^2 + 3q + 6q + 9 = 2q^2 + 9q + 9.\newlineThe expanded form matches the original expression, so the factoring is correct.