\(x^4+2x^2y^2+y^4-16\) का सही गुणनखंड रूप क्या है?

What is the correct factorised form of \(x^4+2x^2y^2+y^4-16\)?

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

A. (\(x^2+y^2-4\)\(x^2+y^2+4\))

Step 1

Concept

First (x-4+2x-2y-2+y-4=\(x^2+y^2\)2), and then \(16=4^2\). Exam tip: solve mixed identities in two steps.

Step 2

Why this answer is correct

The correct answer is A. (\(x^2+y^2-4\)\(x^2+y^2+4\)). First (x-4+2x-2y-2+y-4=\(x^2+y^2\)2), and then \(16=4^2\). Exam tip: solve mixed identities in two steps.

Step 3

Exam Tip

पहले (x-4+2x-2y-2+y-4=\(x^2+y^2\)2), फिर \(16=4^2\) है। परीक्षा में मिश्रित पहचान को दो चरणों में हल करें।

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

\(x^4+2x^2y^2+y^4-16\) का सही गुणनखंड रूप क्या है? / What is the correct factorised form of \(x^4+2x^2y^2+y^4-16\)?

Correct Answer: A. (\(x^2+y^2-4\)\(x^2+y^2+4\)). Explanation: पहले (x-4+2x-2y-2+y-4=\(x^2+y^2\)2), फिर \(16=4^2\) है। परीक्षा में मिश्रित पहचान को दो चरणों में हल करें। / First (x-4+2x-2y-2+y-4=\(x^2+y^2\)2), and then \(16=4^2\). Exam tip: solve mixed identities in two steps.

Which concept should I revise for this Mathematics MCQ?

First (x-4+2x-2y-2+y-4=\(x^2+y^2\)2), and then \(16=4^2\). Exam tip: solve mixed identities in two steps.

What exam hint can help solve this Mathematics question?

पहले (x-4+2x-2y-2+y-4=\(x^2+y^2\)2), फिर \(16=4^2\) है। परीक्षा में मिश्रित पहचान को दो चरणों में हल करें।