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Factor \(x^2+6x+9-y^2\) completely.\(x^2+6x+9-y^2\) को पूर्ण रूप से गुणनखंडों में कीजिए।
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \((x+3-y)(x+3+y)\)\((x+3-y)(x+3+y)\)
Group the expression as \(x^2+6x+9-y^2=(x^2+6x+9)-y^2=(x+3)^2-y^2\). This is a difference of squares: \((x+3)^2-y^2=[(x+3)-y][(x+3)+y]=(x+3-y)(x+3+y)\).हम अभिव्यक्ति को ऐसे समूह बनाते हैं: \(x^2+6x+9-y^2=(x^2+6x+9)-y^2=(x+3)^2-y^2\). यह वर्गों का अंतर है: \((x+3)^2-y^2=[(x+3)-y][(x+3)+y]=(x+3-y)(x+3+y)\).
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