Canonical URI: https://w3id.org/kmath/textbook_solution/visang_solution_body_p029_03
| rdf:type | math:TextbookSolution |
|---|---|
| rdfs:label | 비상 p029 예제 2 풀이 |
| rdfs:comment | 비상 공통수학1 본문 p029 예제 2 풀이. |
| math:answerText | (1) \((x+1)(x-1)(x^2+2)\) (2) \((x+y+2)(x+y-2)\) |
| math:explanationText | (1) \(x^2=X\)로 놓으면 \(x^4+x^2-2=X^2+X-2=(X-1)(X+2)=(x^2-1)(x^2+2)=(x+1)(x-1)(x^2+2)\)이다. (2) \(x+y=X\)로 놓으면 \((x+y+1)(x+y-1)-3=(X+1)(X-1)-3=X^2-4=(X+2)(X-2)=(x+y+2)(x+y-2)\)이다. |
| math:mappingConfidence | 1.0 |
| math:pageStart | 29 |
| math:problem | textbook_problem:visang_vision_p029_03 |
| math:reviewStatus | reviewed |
| math:solutionKind | worked_solution |
| math:usesSolutionPattern | solution_pattern:substitution_factorization |