\(a^4-16b^4\) का पूर्ण गुणनखंड रूप कौन सा है?
Which is the complete factorised form of \(a^4-16b^4\)?
#higher powers
#complete factorisation
#difference of squares
A (\(a^2-4b^2\)\(a^2+4b^2\))
B ((a-2b)(a+2b)\(a^2+4b^2\))
C ((a-4b)(a+4b))
D ((a-2b)2 (a+2b)2 )
Explanation opens after your attempt
Correct Answer
B. ((a-2b)(a+2b)\(a^2+4b^2\))
Step 1
Concept
First (a-4 -16b-4 =\(a^2-4b^2\)\(a^2+4b^2\)), then \(a^2-4b^2\) factors further. Exam tip: stop only after complete factorisation.
Step 2
Why this answer is correct
The correct answer is B. ((a-2b)(a+2b)\(a^2+4b^2\)). First (a-4 -16b-4 =\(a^2-4b^2\)\(a^2+4b^2\)), then \(a^2-4b^2\) factors further. Exam tip: stop only after complete factorisation.
Step 3
Exam Tip
पहले (a-4 -16b-4 =\(a^2-4b^2\)\(a^2+4b^2\)), फिर \(a^2-4b^2\) टूटता है। परीक्षा में पूर्ण गुणनखंड तक रुकें।
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\(x^4-81\) का पूर्ण गुणनखंड रूप क्या है?
What is the complete factorised form of \(x^4-81\)?
#complete factorisation
#difference of squares
#higher powers
A (\(x^2-9\)\(x^2+9\))
B ((x-3)(x+3)\(x^2+9\))
C ((x-9)(x+9))
D ((x-3)2 (x+3)2 )
Explanation opens after your attempt
Correct Answer
B. ((x-3)(x+3)\(x^2+9\))
Step 1
Concept
First (x-4 -81=\(x^2-9\)\(x^2+9\)), then (x-2 -9=(x-3)(x+3)). Exam tip: factor further when possible.
Step 2
Why this answer is correct
The correct answer is B. ((x-3)(x+3)\(x^2+9\)). First (x-4 -81=\(x^2-9\)\(x^2+9\)), then (x-2 -9=(x-3)(x+3)). Exam tip: factor further when possible.
Step 3
Exam Tip
पहले (x-4 -81=\(x^2-9\)\(x^2+9\)), फिर (x-2 -9=(x-3)(x+3)) होता है। परीक्षा में आगे टूट सकने वाले भाग को फिर तोड़ें।
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\(x^4-81\) को पहले किस पहचान से तोड़ना सबसे उचित है?
Which identity is most suitable first for factorising \(x^4-81\)?
#difference of squares
#higher powers
#factorisation
A दो वर्गों का अंतर / Difference of two squares
B पूर्ण वर्ग त्रिपद / Perfect square trinomial
C सामान्य गुणनखंड / Common factor
D घनों का योग / Sum of cubes
Explanation opens after your attempt
Correct Answer
A. दो वर्गों का अंतर / Difference of two squares
Step 1
Concept
(x-4 -81=\(x^2\)2 -92 ). Exam tip: view higher powers as squares when possible.
Step 2
Why this answer is correct
The correct answer is A. दो वर्गों का अंतर / Difference of two squares. (x-4 -81=\(x^2\)2 -92 ). Exam tip: view higher powers as squares when possible.
Step 3
Exam Tip
(x-4 -81=\(x^2\)2 -92 ) है। परीक्षा में बड़ी घात को भी वर्ग के रूप में देखें।
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\(16x^4-1\) का पूर्ण गुणनखंडन कौन सा है?
Which is the complete factorisation of \(16x^4-1\)?
#higher powers
#difference of squares
#complete factorisation
A (\(4x^2-1\)\(4x^2+1\))
B ((2x-1)(2x+1)\(4x^2+1\))
C ((4x-1)(4x+1))
D ((2x-1)2 (2x+1)2 )
Explanation opens after your attempt
Correct Answer
B. ((2x-1)(2x+1)\(4x^2+1\))
Step 1
Concept
(16x-4 -1=\(4x^2-1\)\(4x^2+1\)). Then (4x-2 -1=(2x-1)(2x+1)) also factors.
Step 2
Why this answer is correct
The correct answer is B. ((2x-1)(2x+1)\(4x^2+1\)). (16x-4 -1=\(4x^2-1\)\(4x^2+1\)). Then (4x-2 -1=(2x-1)(2x+1)) also factors.
Step 3
Exam Tip
(16x-4 -1=\(4x^2-1\)\(4x^2+1\)) है। फिर (4x-2 -1=(2x-1)(2x+1)) भी टूटता है।
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\(x^4-81\) का पूर्ण गुणनखंडन क्या होगा?
What will be the complete factorisation of \(x^4-81\)?
#higher powers
#complete factorisation
#difference of squares
A (\(x^2-9\)\(x^2+9\))
B ((x-9)(x+9)\(x^2+9\))
C ((x-3)(x+3)\(x^2+9\))
D ((x-3)2 (x+3)2 )
Explanation opens after your attempt
Correct Answer
C. ((x-3)(x+3)\(x^2+9\))
Step 1
Concept
(x-4 -81=\(x^2-9\)\(x^2+9\)) and (x-2 -9=(x-3)(x+3)). Exam tip: check difference of squares repeatedly.
Step 2
Why this answer is correct
The correct answer is C. ((x-3)(x+3)\(x^2+9\)). (x-4 -81=\(x^2-9\)\(x^2+9\)) and (x-2 -9=(x-3)(x+3)). Exam tip: check difference of squares repeatedly.
Step 3
Exam Tip
(x-4 -81=\(x^2-9\)\(x^2+9\)) और (x-2 -9=(x-3)(x+3)) है। परीक्षा में बार बार वर्गों के अंतर को जांचें।
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\(a^4-b^4\) का पूर्ण गुणनखंडन कौन सा है?
Which is the complete factorisation of \(a^4-b^4\)?
#higher powers
#difference of squares
#complete factorisation
A (\(a^2-b^2\)\(a^2+b^2\))
B ((a-b)(a+b)\(a^2+b^2\))
C ((a-b)2 (a+b)2 )
D (\(a^2-b^2\)2 )
Explanation opens after your attempt
Correct Answer
B. ((a-b)(a+b)\(a^2+b^2\))
Step 1
Concept
First write (a-4 -b-4 =\(a^2-b^2\)\(a^2+b^2\)). Then split (a-2 -b-2 =(a-b)(a+b)) too.
Step 2
Why this answer is correct
The correct answer is B. ((a-b)(a+b)\(a^2+b^2\)). First write (a-4 -b-4 =\(a^2-b^2\)\(a^2+b^2\)). Then split (a-2 -b-2 =(a-b)(a+b)) too.
Step 3
Exam Tip
पहले (a-4 -b-4 =\(a^2-b^2\)\(a^2+b^2\)) करें। फिर (a-2 -b-2 =(a-b)(a+b)) भी तोड़ें।
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\( 16x^4-81y^4 \) का गुणनखंडों में सही पहला चरण क्या है?
What is the correct first factorization step for \( 16x^4-81y^4 \)?
#higher powers
#difference of squares
#factorization
A ( \(4x^2-9y^2\)\(4x^2+9y^2\) )
B ( (4x-9y)(4x+9y) )
C ( \(2x^2-3y^2\)\(2x^2+3y^2\) )
D ( \(16x^2-81y^2\)\(x^2+y^2\) )
Explanation opens after your attempt
Correct Answer
A. ( \(4x^2-9y^2\)\(4x^2+9y^2\) )
Step 1
Concept
(16x-4 =\(4x^2\)2 ) and (81y-4 =\(9y^2\)2 ). Exam tip: identify larger squares first.
Step 2
Why this answer is correct
The correct answer is A. ( \(4x^2-9y^2\)\(4x^2+9y^2\) ). (16x-4 =\(4x^2\)2 ) and (81y-4 =\(9y^2\)2 ). Exam tip: identify larger squares first.
Step 3
Exam Tip
(16x-4 =\(4x^2\)2 ) और (81y-4 =\(9y^2\)2 ) है। परीक्षा में पहले बड़े वर्ग पहचानें।
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