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If \(p(x)=x^2-2x-8\), what are the zeroes of \(p(x+3)\)?यदि \(p(x)=x^2-2x-8\), तो \(p(x+3)\) के शून्यक कौन-से हैं?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \((-5,1)\)\((-5,1)\)
Factor the polynomial as \(p(x)=(x-4)(x+2)\), so its zeroes are \(4\) and \(-2\). For \(p(x+3)=0\), we must have \(x+3=4\) or \(x+3=-2\), giving \(x=1\) or \(x=-5\). Therefore, the zeroes are \((-5,1)\). Exam tip: the zeroes of \(p(x+a)\) are obtained by subtracting \(a\) from each zero of \(p(x)\).पहले \(p(x)=(x-4)(x+2)\) लिखें, इसलिए इसके शून्यक \(4\) और \(-2\) हैं। \(p(x+3)=0\) के लिए \(x+3=4\) या \(x+3=-2\) होगा, जिससे \(x=1\) या \(x=-5\) प्राप्त होता है। अतः शून्यकों का युग्म \((-5,1)\) है। परीक्षा-युक्ति: \(p(x+a)\) के शून्यक, \(p(x)\) के प्रत्येक शून्यक में से \(a\) घटाने पर मिलते हैं।
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