\(16x^4-81y^4\) का पूर्ण गुणनखंड रूप कौन-सा है?

Which is the complete factorised form of \(16x^4-81y^4\)?

Author: Muft Shiksha Editorial Team Published:
Explanation opens after your attempt
Correct Answer

B. ((2x-3y)(2x+3y)\(4x^2+9y^2\))सही पूर्ण रूप

Step 1

Concept

It is (\(4x^2\)2-\(9y^2\)2), and \(4x^2-9y^2\) factors again. In exams check difference of squares repeatedly.

Step 2

Why this answer is correct

The correct answer is B. ((2x-3y)(2x+3y)\(4x^2+9y^2\)) / सही पूर्ण रूप. It is (\(4x^2\)2-\(9y^2\)2), and \(4x^2-9y^2\) factors again. In exams check difference of squares repeatedly.

Step 3

Exam Tip

यह (\(4x^2\)2-\(9y^2\)2) है और \(4x^2-9y^2\) फिर टूटता है। परीक्षा में बार-बार वर्गों का अंतर जांचें।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

\(16x^4-81y^4\) का पूर्ण गुणनखंड रूप कौन-सा है? / Which is the complete factorised form of \(16x^4-81y^4\)?

Correct Answer: B. ((2x-3y)(2x+3y)\(4x^2+9y^2\)) / सही पूर्ण रूप. Explanation: यह (\(4x^2\)2-\(9y^2\)2) है और \(4x^2-9y^2\) फिर टूटता है। परीक्षा में बार-बार वर्गों का अंतर जांचें। / It is (\(4x^2\)2-\(9y^2\)2), and \(4x^2-9y^2\) factors again. In exams check difference of squares repeatedly.

Which concept should I revise for this Mathematics MCQ?

It is (\(4x^2\)2-\(9y^2\)2), and \(4x^2-9y^2\) factors again. In exams check difference of squares repeatedly.

What exam hint can help solve this Mathematics question?

यह (\(4x^2\)2-\(9y^2\)2) है और \(4x^2-9y^2\) फिर टूटता है। परीक्षा में बार-बार वर्गों का अंतर जांचें।