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(sqrt3+2sqrt2)(sqrt3-2)=

(3+22)(32)= (\sqrt{3}+2 \sqrt{2})(\sqrt{3}-2)=

Full solution

Q. (3+22)(32)= (\sqrt{3}+2 \sqrt{2})(\sqrt{3}-2)=
  1. Apply FOIL method: Use the FOIL method to multiply the two binomials.\newline(3+22)(32)=(3×3)+(3×2)+(22×3)+(22×2)(\sqrt{3}+2\sqrt{2})(\sqrt{3}-2) = (\sqrt{3} \times \sqrt{3}) + (\sqrt{3} \times -2) + (2\sqrt{2} \times \sqrt{3}) + (2\sqrt{2} \times -2)
  2. Calculate each term: Calculate each term separately.\newline(3×3)=3(\sqrt{3} \times \sqrt{3}) = 3\newline(3×2)=23(\sqrt{3} \times -2) = -2\sqrt{3}\newline(22×3)=26(2\sqrt{2} \times \sqrt{3}) = 2\sqrt{6}\newline(22×2)=42(2\sqrt{2} \times -2) = -4\sqrt{2}
  3. Combine the terms: Combine the terms. 323+26423 - 2\sqrt{3} + 2\sqrt{6} - 4\sqrt{2}
  4. Final answer: There's no like terms to combine, so this is the final answer. 323+26423 - 2\sqrt{3} + 2\sqrt{6} - 4\sqrt{2}