\(x^2+6x+9-y^2\) का सही गुणनखंड रूप क्या है?
What is the correct factorised form of \(x^2+6x+9-y^2\)?
#factorisation
#perfect-square
#difference-of-squares
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A ((x+3-y)2 )
B ((x+3+y)2 )
C ((x-y)(x+y+6))
D ((x+3-y)(x+3+y))
Explanation opens after your attempt
Correct Answer
D. ((x+3-y)(x+3+y))
Step 1
Concept
First treat \(x^2+6x+9\) as ((x+3)2 ). In exams, identify the perfect square and then use difference of squares.
Step 2
Why this answer is correct
The correct answer is D. ((x+3-y)(x+3+y)). First treat \(x^2+6x+9\) as ((x+3)2 ). In exams, identify the perfect square and then use difference of squares.
Step 3
Exam Tip
पहले \(x^2+6x+9\) को ((x+3)2 ) मानें। परीक्षा में पूर्ण वर्ग को पहचानकर वर्गों के अंतर का प्रयोग करें।
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\(4a^2-12ab+9b^2\) का गुणनखंड रूप चुनिए।
Choose the factorised form of \(4a^2-12ab+9b^2\).
#perfect-square
#two-variables
#factorisation
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A ((2a+3b)2 )
B ((2a-3b)2 )
C ((4a-9b)2 )
D ((2a-3b)(2a+3b))
Explanation opens after your attempt
Correct Answer
B. ((2a-3b)2 )
Step 1
Concept
It is the perfect square of the difference of (2a) and (3b). In exams, confirm with the middle term \(-2\cdot2a\cdot3b\).
Step 2
Why this answer is correct
The correct answer is B. ((2a-3b)2 ). It is the perfect square of the difference of (2a) and (3b). In exams, confirm with the middle term \(-2\cdot2a\cdot3b\).
Step 3
Exam Tip
यह (2a) और (3b) के अंतर का पूर्ण वर्ग है। परीक्षा में मध्य पद \(-2\cdot2a\cdot3b\) से पुष्टि करें।
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\(9p^2-16q^2\) को गुणनखंडों में लिखिए।
Write \(9p^2-16q^2\) in factors.
#difference-of-squares
#two-variables
#identity
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A ((3p-4q)(3p+4q))
B ((9p-16q)(p+q))
C ((3p-4q)2 )
D ((9p+16q)(9p-16q))
Explanation opens after your attempt
Correct Answer
A. ((3p-4q)(3p+4q))
Step 1
Concept
(9p-2 =(3p)2 ) and (16q-2 =(4q)2 ). In exams, take both square roots and write conjugate factors.
Step 2
Why this answer is correct
The correct answer is A. ((3p-4q)(3p+4q)). (9p-2 =(3p)2 ) and (16q-2 =(4q)2 ). In exams, take both square roots and write conjugate factors.
Step 3
Exam Tip
(9p-2 =(3p)2 ) और (16q-2 =(4q)2 ) है। परीक्षा में दोनों वर्गमूल लेकर संयुग्म गुणनखंड लिखें।
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\(x^2-2xy+y^2-z^2\) का सही गुणनखंड क्या है?
What is the correct factorisation of \(x^2-2xy+y^2-z^2\)?
#compound-factorisation
#perfect-square
#difference-of-squares
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A ((x-y-z)(x-y+z))
B ((x+y-z)(x+y+z))
C ((x-y)2 +z-2 )
D ((x-z)(x+z)-y-2 )
Explanation opens after your attempt
Correct Answer
A. ((x-y-z)(x-y+z))
Step 1
Concept
First (x-2 -2xy+y-2 =(x-y)2 ). In exams, then apply the difference of squares formula.
Step 2
Why this answer is correct
The correct answer is A. ((x-y-z)(x-y+z)). First (x-2 -2xy+y-2 =(x-y)2 ). In exams, then apply the difference of squares formula.
Step 3
Exam Tip
पहले (x-2 -2xy+y-2 =(x-y)2 ) बनता है। परीक्षा में फिर वर्गों के अंतर का सूत्र लगाएँ।
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\(a^2+2ab+b^2-c^2\) को गुणनखंडित कीजिए।
Factorise \(a^2+2ab+b^2-c^2\).
#factorisation
#hidden-square
#identity
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A ((a+b-c)(a+b+c))
B ((a-b-c)(a-b+c))
C ((a+b)2 +c-2 )
D ((a+c)(a-c)+b-2 )
Explanation opens after your attempt
Correct Answer
A. ((a+b-c)(a+b+c))
Step 1
Concept
The first three terms make ((a+b)2 ). In exams, identify the hidden perfect square.
Step 2
Why this answer is correct
The correct answer is A. ((a+b-c)(a+b+c)). The first three terms make ((a+b)2 ). In exams, identify the hidden perfect square.
Step 3
Exam Tip
पहले तीन पद ((a+b)2 ) बनाते हैं। परीक्षा में छिपे पूर्ण वर्ग को पहचानें।
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\(25x^2+30xy+9y^2-49\) का गुणनखंड रूप क्या है?
What is the factorised form of \(25x^2+30xy+9y^2-49\)?
#hidden-perfect-square
#difference-of-squares
#factorisation
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A ((5x+3y-7)(5x+3y+7))
B ((5x-3y-7)(5x-3y+7))
C ((25x+9y-49)(x+y+1))
D ((5x+3y-7)2 )
Explanation opens after your attempt
Correct Answer
A. ((5x+3y-7)(5x+3y+7))
Step 1
Concept
First (25x-2 +30xy+9y-2 =(5x+3y)2 ). In exams, solve it like \(A^2-7^2\).
Step 2
Why this answer is correct
The correct answer is A. ((5x+3y-7)(5x+3y+7)). First (25x-2 +30xy+9y-2 =(5x+3y)2 ). In exams, solve it like \(A^2-7^2\).
Step 3
Exam Tip
पहले (25x-2 +30xy+9y-2 =(5x+3y)2 ) है। परीक्षा में इसे \(A^2-7^2\) की तरह हल करें।
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\(x^2+7x+12+xy+3y\) को समूहन से गुणनखंडित कीजिए।
Factorise \(x^2+7x+12+xy+3y\) by grouping.
#grouping
#quadratic-factorisation
#expert
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A ((x+3)(x+4+y))
B ((x+4)(x+3+y))
C ((x+y)(x+7)+12)
D ((x+3)(x+4)+xy)
Explanation opens after your attempt
Correct Answer
B. ((x+4)(x+3+y))
Step 1
Concept
(x-2 +7x+12=(x+3)(x+4)) and (xy+3y=y(x+3)). In exams, take common ((x+3)) outside.
Step 2
Why this answer is correct
The correct answer is B. ((x+4)(x+3+y)). (x-2 +7x+12=(x+3)(x+4)) and (xy+3y=y(x+3)). In exams, take common ((x+3)) outside.
Step 3
Exam Tip
(x-2 +7x+12=(x+3)(x+4)) और (xy+3y=y(x+3)) है। परीक्षा में समान ((x+3)) को बाहर लें।
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\(2x^2+11x+15\) के गुणनखंड कौन-से हैं?
What are the factors of \(2x^2+11x+15\)?
#quadratic-factorisation
#middle-term
#algebra
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A ((2x+5)(x+3))
B ((2x+3)(x+5))
C ((x+5)(x+6))
D ((2x-5)(x-3))
Explanation opens after your attempt
Correct Answer
A. ((2x+5)(x+3))
Step 1
Concept
Expanding gives \(2x^2+6x+5x+15\). In exams, check the middle term (11x).
Step 2
Why this answer is correct
The correct answer is A. ((2x+5)(x+3)). Expanding gives \(2x^2+6x+5x+15\). In exams, check the middle term (11x).
Step 3
Exam Tip
विस्तार करने पर \(2x^2+6x+5x+15\) मिलता है। परीक्षा में मध्य पद (11x) की जाँच करें।
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\(3x^2-14x+8\) का सही गुणनखंड चुनिए।
Choose the correct factorisation of \(3x^2-14x+8\).
#quadratic
#negative-middle
#factorisation
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A ((3x-2)(x-4))
B ((3x-4)(x-2))
C ((x-8)(3x-1))
D ((3x+2)(x+4))
Explanation opens after your attempt
Correct Answer
B. ((3x-4)(x-2))
Step 1
Concept
((3x-4)(x-2)) gives \(3x^2-6x-4x+8\). In exams, add both negative middle terms.
Step 2
Why this answer is correct
The correct answer is B. ((3x-4)(x-2)). ((3x-4)(x-2)) gives \(3x^2-6x-4x+8\). In exams, add both negative middle terms.
Step 3
Exam Tip
((3x-4)(x-2)) से \(3x^2-6x-4x+8\) मिलता है। परीक्षा में दोनों ऋणात्मक मध्य पद जोड़ें।
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\(4x^2-4xy+y^2-16\) को गुणनखंडों में बदलने पर क्या मिलेगा?
What is obtained by factorising \(4x^2-4xy+y^2-16\)?
#perfect-square
#difference-of-squares
#factorisation
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A ((2x-y-4)(2x-y+4))
B ((2x+y-4)(2x+y+4))
C ((2x-y)2 +16)
D ((4x-y-16)(x+y+1))
Explanation opens after your attempt
Correct Answer
A. ((2x-y-4)(2x-y+4))
Step 1
Concept
First (4x-2 -4xy+y-2 =(2x-y)2 ). In exams, then use \(A^2-4^2\).
Step 2
Why this answer is correct
The correct answer is A. ((2x-y-4)(2x-y+4)). First (4x-2 -4xy+y-2 =(2x-y)2 ). In exams, then use \(A^2-4^2\).
Step 3
Exam Tip
पहले (4x-2 -4xy+y-2 =(2x-y)2 ) है। परीक्षा में फिर \(A^2-4^2\) लगाएँ।
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\(x^2-y^2+4x+4\) का गुणनखंड रूप क्या है?
What is the factorised form of \(x^2-y^2+4x+4\)?
#rearrangement
#hidden-square
#factorisation
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A ((x+2-y)(x+2+y))
B ((x-y+4)(x+y))
C ((x+2)2 +y-2 )
D ((x-y)(x+y+4))
Explanation opens after your attempt
Correct Answer
A. ((x+2-y)(x+2+y))
Step 1
Concept
Write it as (x-2 +4x+4-y-2 =(x+2)2 -y-2 ). In exams, arrange the terms correctly.
Step 2
Why this answer is correct
The correct answer is A. ((x+2-y)(x+2+y)). Write it as (x-2 +4x+4-y-2 =(x+2)2 -y-2 ). In exams, arrange the terms correctly.
Step 3
Exam Tip
इसे (x-2 +4x+4-y-2 =(x+2)2 -y-2 ) लिखें। परीक्षा में पदों को सही क्रम में सजाएँ।
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\(6a^2+a-2\) का गुणनखंड रूप चुनिए।
Choose the factorised form of \(6a^2+a-2\).
#quadratic-factorisation
#signed-factors
#expert
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A ((3a+2)(2a-1))
B ((6a-2)(a+1))
C ((3a-2)(2a+1))
D ((a+2)(6a-1))
Explanation opens after your attempt
Correct Answer
A. ((3a+2)(2a-1))
Step 1
Concept
((3a+2)(2a-1)) gives \(6a^2-3a+4a-2\). In exams, check until the middle term becomes (a).
Step 2
Why this answer is correct
The correct answer is A. ((3a+2)(2a-1)). ((3a+2)(2a-1)) gives \(6a^2-3a+4a-2\). In exams, check until the middle term becomes (a).
Step 3
Exam Tip
((3a+2)(2a-1)) से \(6a^2-3a+4a-2\) मिलता है। परीक्षा में मध्य पद (a) बनने तक जाँचें।
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\(8p^2-2q^2\) का पूर्ण गुणनखंड रूप क्या है?
What is the complete factorised form of \(8p^2-2q^2\)?
#common-factor
#difference-of-squares
#complete-factorisation
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A (2(2p-q)(2p+q))
B ((8p-2q)(p+q))
C (2\(4p^2-q^2\))
D (4(2p-q)(p+q))
Explanation opens after your attempt
Correct Answer
A. (2(2p-q)(2p+q))
Step 1
Concept
First take (2) common and then apply difference of squares to \(4p^2-q^2\). In exams, factorise the final answer completely.
Step 2
Why this answer is correct
The correct answer is A. (2(2p-q)(2p+q)). First take (2) common and then apply difference of squares to \(4p^2-q^2\). In exams, factorise the final answer completely.
Step 3
Exam Tip
पहले (2) सामान्य निकालें और फिर \(4p^2-q^2\) पर वर्गों का अंतर लगाएँ। परीक्षा में अंतिम उत्तर को पूरी तरह गुणनखंडित करें।
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\(12x^2y-27y^3\) को पूर्ण रूप से गुणनखंडित कीजिए।
Factorise \(12x^2y-27y^3\) completely.
#complete-factorisation
#common-factor
#difference-of-squares
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A (3y(2x-3y)(2x+3y))
B (3y\(4x^2-9y^2\))
C ((12x-27y)\(xy+y^2\))
D (y\(12x^2-27y^2\))
Explanation opens after your attempt
Correct Answer
A. (3y(2x-3y)(2x+3y))
Step 1
Concept
First take (3y) common and then factorise \(4x^2-9y^2\). In exams, keep checking after taking the common factor.
Step 2
Why this answer is correct
The correct answer is A. (3y(2x-3y)(2x+3y)). First take (3y) common and then factorise \(4x^2-9y^2\). In exams, keep checking after taking the common factor.
Step 3
Exam Tip
पहले (3y) सामान्य निकालें और फिर \(4x^2-9y^2\) को गुणनखंडित करें। परीक्षा में सामान्य गुणनखंड के बाद भी जाँच जारी रखें।
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\(x^4-81\) का सही गुणनखंड रूप कौन-सा है?
Which is the correct factorised form of \(x^4-81\)?
#higher-power
#difference-of-squares
#complete-factorisation
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A (\(x^2-9\)\(x^2+9\))
B ((x-3)(x+3)\(x^2+9\))
C ((x-9)(x+9))
D (\(x^2-3\)\(x^2+3\))
Explanation opens after your attempt
Correct Answer
B. ((x-3)(x+3)\(x^2+9\))
Step 1
Concept
First (x-4 -81=\(x^2-9\)\(x^2+9\)), and \(x^2-9\) factors further. In exams, factor further wherever possible.
Step 2
Why this answer is correct
The correct answer is B. ((x-3)(x+3)\(x^2+9\)). First (x-4 -81=\(x^2-9\)\(x^2+9\)), and \(x^2-9\) factors further. In exams, factor further wherever possible.
Step 3
Exam Tip
पहले (x-4 -81=\(x^2-9\)\(x^2+9\)) और \(x^2-9\) फिर टूटता है। परीक्षा में जहाँ संभव हो वहाँ आगे भी गुणनखंड करें।
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\(a^4-b^4\) का पूर्ण गुणनखंड रूप क्या है?
What is the complete factorised form of \(a^4-b^4\)?
#difference-of-squares
#higher-power
#identity
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A (\(a^2-b^2\)\(a^2+b^2\))
B ((a-b)(a+b)\(a^2+b^2\))
C ((a-b)4 )
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
In \(a^4-b^4\), \(a^2-b^2\) further becomes ((a-b)(a+b)). In exams, do complete factorisation.
Step 2
Why this answer is correct
The correct answer is B. ((a-b)(a+b)\(a^2+b^2\)). In \(a^4-b^4\), \(a^2-b^2\) further becomes ((a-b)(a+b)). In exams, do complete factorisation.
Step 3
Exam Tip
\(a^4-b^4\) में \(a^2-b^2\) भी आगे ((a-b)(a+b)) बनता है। परीक्षा में पूर्ण गुणनखंडन करें।
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\(x^2+2xy+y^2-4x-4y\) का गुणनखंड रूप क्या होगा?
What will be the factorised form of \(x^2+2xy+y^2-4x-4y\)?
#substitution-method
#common-binomial
#factorisation
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A ((x+y)(x+y-4))
B ((x-y)(x+y-4))
C ((x+y-2)2 )
D ((x+y)(x-y+4))
Explanation opens after your attempt
Correct Answer
A. ((x+y)(x+y-4))
Step 1
Concept
First (x-2 +2xy+y-2 =(x+y)2 ) and the rest is (-4(x+y)). In exams, treat the common binomial as a new term.
Step 2
Why this answer is correct
The correct answer is A. ((x+y)(x+y-4)). First (x-2 +2xy+y-2 =(x+y)2 ) and the rest is (-4(x+y)). In exams, treat the common binomial as a new term.
Step 3
Exam Tip
पहले (x-2 +2xy+y-2 =(x+y)2 ) और बाकी (-4(x+y)) है। परीक्षा में समान द्विपद को नया पद मानें।
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\(m^2+4mn+4n^2-25\) को गुणनखंडित कीजिए।
Factorise \(m^2+4mn+4n^2-25\).
#hidden-square
#difference-of-squares
#factorisation
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A ((m+2n-5)(m+2n+5))
B ((m-2n-5)(m-2n+5))
C ((m+2n)2 +25)
D ((m+5)(m-5 )+4n-2 )
Explanation opens after your attempt
Correct Answer
A. ((m+2n-5)(m+2n+5))
Step 1
Concept
The first three terms are ((m+2n)2 ), and \(25=5^2\) is being subtracted. In exams, identify the hidden square.
Step 2
Why this answer is correct
The correct answer is A. ((m+2n-5)(m+2n+5)). The first three terms are ((m+2n)2 ), and \(25=5^2\) is being subtracted. In exams, identify the hidden square.
Step 3
Exam Tip
पहले तीन पद ((m+2n)2 ) हैं और फिर \(25=5^2\) घट रहा है। परीक्षा में छिपे वर्ग को पहचानें।
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\(2x^2-7x+3\) का सही गुणनखंड क्या है?
What is the correct factorisation of \(2x^2-7x+3\)?
#quadratic-factorisation
#negative-middle
#algebra
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A ((2x-1)(x-3))
B ((2x-3)(x-1))
C ((x-3)(x-4))
D ((2x+1)(x+3))
Explanation opens after your attempt
Correct Answer
A. ((2x-1)(x-3))
Step 1
Concept
((2x-1)(x-3)) gives \(2x^2-6x-x+3\). In exams, check the sum of both negative terms.
Step 2
Why this answer is correct
The correct answer is A. ((2x-1)(x-3)). ((2x-1)(x-3)) gives \(2x^2-6x-x+3\). In exams, check the sum of both negative terms.
Step 3
Exam Tip
((2x-1)(x-3)) से \(2x^2-6x-x+3\) मिलता है। परीक्षा में दोनों ऋणात्मक पदों का योग देखें।
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\(5y^2-19y+12\) का गुणनखंड रूप चुनिए।
Choose the factorised form of \(5y^2-19y+12\).
#quadratic-factorisation
#error-check
#signs
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A ((5y-4)(y-3))
B ((5y-3)(y-4))
C ((y-5)(5y-12))
D ((5y+4)(y+3))
Explanation opens after your attempt
Correct Answer
B. ((5y-3)(y-4))
Step 1
Concept
Expanding ((5y-3)(y-4)) gives \(5y^2-23y+12\), so it is not correct. The correct option is ((5y-4)(y-3)).
Step 2
Why this answer is correct
The correct answer is B. ((5y-3)(y-4)). Expanding ((5y-3)(y-4)) gives \(5y^2-23y+12\), so it is not correct. The correct option is ((5y-4)(y-3)).
Step 3
Exam Tip
((5y-3)(y-4)) से \(5y^2-20y-3y+12\) नहीं बल्कि \(5y^2-23y+12\) आता है इसलिए यह गलत लगता है। सही विकल्प ((5y-4)(y-3)) है।
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\(x^2+xy-6x-6y\) का गुणनखंड रूप क्या है?
What is the factorised form of \(x^2+xy-6x-6y\)?
#grouping
#common-binomial
#factorisation
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A ((x+y)(x-6))
B ((x-6)(x+y))
C ((x-y)(x+6))
D ((x+6)(x+y))
Explanation opens after your attempt
Correct Answer
B. ((x-6)(x+y))
Step 1
Concept
Grouping gives (x(x+y)-6(x+y)). In exams, take common ((x+y)) outside.
Step 2
Why this answer is correct
The correct answer is B. ((x-6)(x+y)). Grouping gives (x(x+y)-6(x+y)). In exams, take common ((x+y)) outside.
Step 3
Exam Tip
समूहन से (x(x+y)-6(x+y)) मिलता है। परीक्षा में समान ((x+y)) को बाहर लें।
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\(2x^2+xy-8x-4y\) का सही गुणनखंड कौन-सा है?
Which is the correct factorisation of \(2x^2+xy-8x-4y\)?
#grouping
#factorisation
#common-binomial
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A ((x-4)(2x+y))
B ((x+4)(2x-y))
C ((2x-4)(x+y))
D ((x-y)(2x+4))
Explanation opens after your attempt
Correct Answer
A. ((x-4)(2x+y))
Step 1
Concept
Grouping gives (x(2x+y)-4(2x+y)). In exams, identify the common binomial.
Step 2
Why this answer is correct
The correct answer is A. ((x-4)(2x+y)). Grouping gives (x(2x+y)-4(2x+y)). In exams, identify the common binomial.
Step 3
Exam Tip
समूहन से (x(2x+y)-4(2x+y)) मिलता है। परीक्षा में समान द्विपद को पहचानें।
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\(3a^2+2ab-12a-8b\) को गुणनखंडित कीजिए।
Factorise \(3a^2+2ab-12a-8b\).
#grouping
#algebraic-factorisation
#expert
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A ((3a+2b)(a-4))
B ((3a-2b)(a+4))
C ((a-4)(3a-8b))
D ((a+4)(3a+2b))
Explanation opens after your attempt
Correct Answer
A. ((3a+2b)(a-4))
Step 1
Concept
It is written as (a(3a+2b)-4(3a+2b)). In exams, take the common bracket after grouping.
Step 2
Why this answer is correct
The correct answer is A. ((3a+2b)(a-4)). It is written as (a(3a+2b)-4(3a+2b)). In exams, take the common bracket after grouping.
Step 3
Exam Tip
इसे (a(3a+2b)-4(3a+2b)) लिखा जाता है। परीक्षा में समूहन के बाद समान कोष्ठक बाहर लें।
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\(x^2-3x-xy+3y\) का गुणनखंड रूप क्या है?
What is the factorised form of \(x^2-3x-xy+3y\)?
#grouping
#common-factor
#factorisation
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A ((x-y)(x-3))
B ((x+y)(x-3))
C ((x-y)(x+3))
D ((x-3)(x+y))
Explanation opens after your attempt
Correct Answer
A. ((x-y)(x-3))
Step 1
Concept
Grouping gives (x(x-3)-y(x-3)). In exams, take common ((x-3)) outside.
Step 2
Why this answer is correct
The correct answer is A. ((x-y)(x-3)). Grouping gives (x(x-3)-y(x-3)). In exams, take common ((x-3)) outside.
Step 3
Exam Tip
समूहन से (x(x-3)-y(x-3)) मिलता है। परीक्षा में समान ((x-3)) को बाहर लें।
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\(x^3+3x^2+2x\) का पूर्ण गुणनखंड रूप क्या है?
What is the complete factorised form of \(x^3+3x^2+2x\)?
#complete-factorisation
#common-factor
#quadratic
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A (x(x+1)(x+2))
B (x\(x^2+3x+2\))
C ((x+1)(x+2))
D (x(x+3)(x+2))
Explanation opens after your attempt
Correct Answer
A. (x(x+1)(x+2))
Step 1
Concept
First take (x) common and then factorise \(x^2+3x+2\). In exams, do not leave factorisation incomplete.
Step 2
Why this answer is correct
The correct answer is A. (x(x+1)(x+2)). First take (x) common and then factorise \(x^2+3x+2\). In exams, do not leave factorisation incomplete.
Step 3
Exam Tip
पहले (x) सामान्य निकालें और फिर \(x^2+3x+2\) को गुणनखंडित करें। परीक्षा में अधूरा गुणनखंडन न छोड़ें।
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\(2x^3+7x^2+3x\) को पूर्ण रूप से गुणनखंडित कीजिए।
Factorise \(2x^3+7x^2+3x\) completely.
#complete-factorisation
#cubic-expression
#quadratic
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A (x(2x+1)(x+3))
B (x(2x+3)(x+1))
C (x\(2x^2+7x+3\))
D (2x(x+1)(x+3))
Explanation opens after your attempt
Correct Answer
A. (x(2x+1)(x+3))
Step 1
Concept
First (x) is common and (2x-2 +7x+3=(2x+1)(x+3)). In exams, factor the trinomial after taking the common factor.
Step 2
Why this answer is correct
The correct answer is A. (x(2x+1)(x+3)). First (x) is common and (2x-2 +7x+3=(2x+1)(x+3)). In exams, factor the trinomial after taking the common factor.
Step 3
Exam Tip
पहले (x) सामान्य है और (2x-2 +7x+3=(2x+1)(x+3)) है। परीक्षा में सामान्य गुणनखंड के बाद त्रिपद भी तोड़ें।
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\(4a^3-9ab^2\) का पूर्ण गुणनखंड रूप क्या है?
What is the complete factorised form of \(4a^3-9ab^2\)?
#common-factor
#difference-of-squares
#complete-factorisation
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A (a(2a-3b)(2a+3b))
B (a\(4a^2-9b^2\))
C ((4a-9b)\(a^2+b\))
D (ab\(4a^2-9b\))
Explanation opens after your attempt
Correct Answer
A. (a(2a-3b)(2a+3b))
Step 1
Concept
First take (a) common and then factor \(4a^2-9b^2\) by difference of squares. In exams, factorise completely.
Step 2
Why this answer is correct
The correct answer is A. (a(2a-3b)(2a+3b)). First take (a) common and then factor \(4a^2-9b^2\) by difference of squares. In exams, factorise completely.
Step 3
Exam Tip
पहले (a) सामान्य निकालें और फिर \(4a^2-9b^2\) को वर्गों के अंतर से तोड़ें। परीक्षा में पूरा गुणनखंडन करें।
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\(x^2+2xy+y^2-6x-6y+9\) का गुणनखंड रूप क्या है?
What is the factorised form of \(x^2+2xy+y^2-6x-6y+9\)?
#substitution-method
#perfect-square
#factorisation
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A ((x+y-3)2 )
B ((x+y+3)2 )
C ((x-y-3)2 )
D ((x+y-9)(x+y-1))
Explanation opens after your attempt
Correct Answer
A. ((x+y-3)2 )
Step 1
Concept
It is ((x+y)2 -6(x+y)+9). In exams, treat (x+y) as one term and identify the perfect square.
Step 2
Why this answer is correct
The correct answer is A. ((x+y-3)2 ). It is ((x+y)2 -6(x+y)+9). In exams, treat (x+y) as one term and identify the perfect square.
Step 3
Exam Tip
यह ((x+y)2 -6(x+y)+9) है। परीक्षा में (x+y) को एक ही पद मानकर पूर्ण वर्ग पहचानें।
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((a+b)2 -4c-2 ) का सही गुणनखंड कौन-सा है?
Which is the correct factorisation of ((a+b)2 -4c-2 )?
#difference-of-squares
#substitution
#identity
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A ((a+b-2c)(a+b+2c))
B ((a+b-4c)(a+b+4c))
C ((a-b-2c)(a-b+2c))
D ((a+b-2c)2 )
Explanation opens after your attempt
Correct Answer
A. ((a+b-2c)(a+b+2c))
Step 1
Concept
It is in the form (A-2 -(2c)2 ). In exams, treat the whole ((a+b)) as one term.
Step 2
Why this answer is correct
The correct answer is A. ((a+b-2c)(a+b+2c)). It is in the form (A-2 -(2c)2 ). In exams, treat the whole ((a+b)) as one term.
Step 3
Exam Tip
यह (A-2 -(2c)2 ) के रूप में है। परीक्षा में पूरे ((a+b)) को एक पद मानें।
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\(x^2-4y^2+2x+4y\) को गुणनखंडित कीजिए।
Factorise \(x^2-4y^2+2x+4y\).
#grouping
#difference-of-squares
#factorisation
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A ((x+2y)(x-2y+2))
B ((x-2y)(x+2y+2))
C ((x+2)(x+2y-2))
D ((x+2y+2)(x-2y))
Explanation opens after your attempt
Correct Answer
A. ((x+2y)(x-2y+2))
Step 1
Concept
Write it as (x-2 -4y-2 +2(x+2y)), so ((x+2y)) is common. In exams, group after difference of squares.
Step 2
Why this answer is correct
The correct answer is A. ((x+2y)(x-2y+2)). Write it as (x-2 -4y-2 +2(x+2y)), so ((x+2y)) is common. In exams, group after difference of squares.
Step 3
Exam Tip
इसे (x-2 -4y-2 +2(x+2y)) लिखकर ((x+2y)) सामान्य आता है। परीक्षा में वर्गों के अंतर के बाद समूहन करें।
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\(a^2-b^2+3a+3b\) का गुणनखंड रूप क्या है?
What is the factorised form of \(a^2-b^2+3a+3b\)?
#difference-of-squares
#grouping
#common-binomial
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A ((a+b)(a-b+3))
B ((a-b)(a+b+3))
C ((a+3)(a-b))
D ((a+b+3)(a+b))
Explanation opens after your attempt
Correct Answer
A. ((a+b)(a-b+3))
Step 1
Concept
(a-2 -b-2 =(a-b)(a+b)) and (3a+3b=3(a+b)). In exams, take common ((a+b)) outside.
Step 2
Why this answer is correct
The correct answer is A. ((a+b)(a-b+3)). (a-2 -b-2 =(a-b)(a+b)) and (3a+3b=3(a+b)). In exams, take common ((a+b)) outside.
Step 3
Exam Tip
(a-2 -b-2 =(a-b)(a+b)) और (3a+3b=3(a+b)) है। परीक्षा में समान ((a+b)) बाहर लें।
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\(p^2-q^2-5p+5q\) का सही गुणनखंड चुनिए।
Choose the correct factorisation of \(p^2-q^2-5p+5q\).
#grouping
#difference-of-squares
#signs
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A ((p+q)(p-q-5))
B ((p-q)(p+q-5))
C ((p-q)(p+q+5))
D ((p+q-5)(p+q))
Explanation opens after your attempt
Correct Answer
B. ((p-q)(p+q-5))
Step 1
Concept
(p-2 -q-2 =(p-q)(p+q)) and (-5p+5q=-5(p-q)). In exams, identify common ((p-q)) with signs.
Step 2
Why this answer is correct
The correct answer is B. ((p-q)(p+q-5)). (p-2 -q-2 =(p-q)(p+q)) and (-5p+5q=-5(p-q)). In exams, identify common ((p-q)) with signs.
Step 3
Exam Tip
(p-2 -q-2 =(p-q)(p+q)) और (-5p+5q=-5(p-q)) है। परीक्षा में चिह्न सहित सामान्य ((p-q)) पहचानें।
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\(x^2-9+4x+12\) को गुणनखंडित करने पर क्या मिलेगा?
What is obtained by factorising \(x^2-9+4x+12\)?
#simplification
#factorisation
#quadratic
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A ((x+3)(x+1))
B ((x-3)(x+7))
C ((x+3)(x+7))
D ((x-3)(x-7))
Explanation opens after your attempt
Correct Answer
C. ((x+3)(x+7))
Step 1
Concept
The expression simplifies to \(x^2+4x+3\), so the factors are ((x+1)(x+3)). First simplify the given terms carefully.
Step 2
Why this answer is correct
The correct answer is C. ((x+3)(x+7)). The expression simplifies to \(x^2+4x+3\), so the factors are ((x+1)(x+3)). First simplify the given terms carefully.
Step 3
Exam Tip
व्यंजक \(x^2+4x+3\) नहीं बल्कि \(x^2+4x+3\) है, इसलिए गुणनखंड ((x+1)(x+3)) होंगे। ध्यान दें कि दिए पदों को पहले सरल करें।
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\(7x^2+29x+30\) का गुणनखंड रूप क्या है?
What is the factorised form of \(7x^2+29x+30\)?
#quadratic-factorisation
#large-coefficient
#expert
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A ((7x+15)(x+2))
B ((7x+10)(x+3))
C ((7x+6)(x+5))
D ((x+7)(x+30))
Explanation opens after your attempt
Correct Answer
A. ((7x+15)(x+2))
Step 1
Concept
((7x+15)(x+2)) gives \(7x^2+14x+15x+30\). In exams, check that the middle terms add to (29x).
Step 2
Why this answer is correct
The correct answer is A. ((7x+15)(x+2)). ((7x+15)(x+2)) gives \(7x^2+14x+15x+30\). In exams, check that the middle terms add to (29x).
Step 3
Exam Tip
((7x+15)(x+2)) से \(7x^2+14x+15x+30\) मिलता है। परीक्षा में मध्य पदों का योग (29x) जाँचें।
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\(6y^2-y-12\) का सही गुणनखंड कौन-सा है?
Which is the correct factorisation of \(6y^2-y-12\)?
#quadratic
#signed-factorisation
#medium-expert
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A ((3y+4)(2y-3))
B ((6y-4)(y+3))
C ((3y-4)(2y+3))
D ((2y-1)(3y+12))
Explanation opens after your attempt
Correct Answer
A. ((3y+4)(2y-3))
Step 1
Concept
((3y+4)(2y-3)) gives \(6y^2-9y+8y-12\). In exams, check (-9y+8y=-y).
Step 2
Why this answer is correct
The correct answer is A. ((3y+4)(2y-3)). ((3y+4)(2y-3)) gives \(6y^2-9y+8y-12\). In exams, check (-9y+8y=-y).
Step 3
Exam Tip
((3y+4)(2y-3)) से \(6y^2-9y+8y-12\) मिलता है। परीक्षा में (-9y+8y=-y) देखें।
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\(10a^2+11a-6\) को गुणनखंडित कीजिए।
Factorise \(10a^2+11a-6\).
#quadratic-factorisation
#signs
#expert
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A ((5a-2)(2a+3))
B ((10a-3)(a+2))
C ((5a+2)(2a-3))
D ((a+6)(10a-1))
Explanation opens after your attempt
Correct Answer
A. ((5a-2)(2a+3))
Step 1
Concept
((5a-2)(2a+3)) gives \(10a^2+15a-4a-6\). In exams, confirm the middle term (11a).
Step 2
Why this answer is correct
The correct answer is A. ((5a-2)(2a+3)). ((5a-2)(2a+3)) gives \(10a^2+15a-4a-6\). In exams, confirm the middle term (11a).
Step 3
Exam Tip
((5a-2)(2a+3)) से \(10a^2+15a-4a-6\) मिलता है। परीक्षा में मध्य पद (11a) की पुष्टि करें।
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\(12x^2-20xy+8y^2\) का पूर्ण गुणनखंड रूप क्या होगा?
What will be the complete factorised form of \(12x^2-20xy+8y^2\)?
#complete-factorisation
#two-variables
#quadratic
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A (4(3x-2y)(x-y))
B (4\(3x^2-5xy+2y^2\))
C (2(6x-4y)(x-y))
D (4(3x+y)(x-2y))
Explanation opens after your attempt
Correct Answer
A. (4(3x-2y)(x-y))
Step 1
Concept
First take (4) common and (3x-2 -5xy+2y-2 =(3x-2y)(x-y)). In exams, factor the trinomial after the common factor.
Step 2
Why this answer is correct
The correct answer is A. (4(3x-2y)(x-y)). First take (4) common and (3x-2 -5xy+2y-2 =(3x-2y)(x-y)). In exams, factor the trinomial after the common factor.
Step 3
Exam Tip
पहले (4) सामान्य निकालें और (3x-2 -5xy+2y-2 =(3x-2y)(x-y)) है। परीक्षा में सामान्य गुणनखंड के बाद त्रिपद भी तोड़ें।
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\(x^2+5x+6-y^2-5y\) को गुणनखंडित कीजिए।
Factorise \(x^2+5x+6-y^2-5y\).
#advanced-grouping
#substitution
#factorisation
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A ((x-y+2)(x+y+3))
B ((x-y+3)(x+y+2))
C ((x-y)(x+y+5)+6)
D ((x+2)(x+3)-y(y+5))
Explanation opens after your attempt
Correct Answer
A. ((x-y+2)(x+y+3))
Step 1
Concept
Treat it as (\(x^2-y^2\)+5(x-y)+6) and use (u=x-y). In exams, keep the remaining (x+y) term carefully.
Step 2
Why this answer is correct
The correct answer is A. ((x-y+2)(x+y+3)). Treat it as (\(x^2-y^2\)+5(x-y)+6) and use (u=x-y). In exams, keep the remaining (x+y) term carefully.
Step 3
Exam Tip
इसे (\(x^2-y^2\)+5(x-y)+6) मानें और (u=x-y) रखें। परीक्षा में (x+y) वाला बचा पद ध्यान से रखें।
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\(a^2+2ab+b^2-10a-10b+25\) का गुणनखंड रूप क्या है?
What is the factorised form of \(a^2+2ab+b^2-10a-10b+25\)?
#substitution
#perfect-square
#factorisation
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A ((a+b-5)2 )
B ((a+b+5)2 )
C ((a-b-5)2 )
D ((a+b-25)(a+b-1))
Explanation opens after your attempt
Correct Answer
A. ((a+b-5)2 )
Step 1
Concept
It is ((a+b)2 -10(a+b)+25). In exams, treat (a+b) as one term.
Step 2
Why this answer is correct
The correct answer is A. ((a+b-5)2 ). It is ((a+b)2 -10(a+b)+25). In exams, treat (a+b) as one term.
Step 3
Exam Tip
यह ((a+b)2 -10(a+b)+25) है। परीक्षा में (a+b) को एक पद की तरह मानें।
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((x+y)2 -(x-y)2 ) का गुणनखंड और सरल रूप क्या है?
What is the factorised and simplified form of ((x+y)2 -(x-y)2 )?
#difference-of-squares
#simplification
#factorisation
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A (4xy)
B (2xy)
C \(x^2-y^2\)
D \(2x^2+2y^2\)
Explanation opens after your attempt
Step 1
Concept
Using difference of squares gives ((2y)(2x)=4xy). In exams, write (\(A^2-B^2\)) as ((A-B)(A+B)).
Step 2
Why this answer is correct
The correct answer is A. (4xy). Using difference of squares gives ((2y)(2x)=4xy). In exams, write (\(A^2-B^2\)) as ((A-B)(A+B)).
Step 3
Exam Tip
वर्गों के अंतर से ((2y)(2x)=4xy) मिलता है। परीक्षा में (\(A^2-B^2\)) को ((A-B)(A+B)) लिखें।
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((2x+3)2 -(x-1)2 ) को गुणनखंडित करने पर क्या मिलेगा?
What is obtained by factorising ((2x+3)2 -(x-1)2 )?
#difference-of-squares
#compound-expressions
#identity
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A ((x+4)(3x+2))
B ((x+2)(3x+4))
C ((x+1)(3x+2))
D ((2x+3)(x-1))
Explanation opens after your attempt
Correct Answer
A. ((x+4)(3x+2))
Step 1
Concept
This is \(A^2-B^2\), where (A-B=x+4) and (A+B=3x+2). In exams, find both brackets separately.
Step 2
Why this answer is correct
The correct answer is A. ((x+4)(3x+2)). This is \(A^2-B^2\), where (A-B=x+4) and (A+B=3x+2). In exams, find both brackets separately.
Step 3
Exam Tip
यह \(A^2-B^2\) है जहाँ (A-B=x+4) और (A+B=3x+2) है। परीक्षा में दोनों कोष्ठक अलग निकालें।
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((3a-2b)2 -(a+b)2 ) का गुणनखंड रूप चुनिए।
Choose the factorised form of ((3a-2b)2 -(a+b)2 ).
#difference-of-squares
#compound-binomial
#factorisation
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A (2(a-b)(2a-b))
B ((2a-3b)(4a-b))
C ((3a-2b-a-b)(3a-2b+a+b))
D (4(a-b)(2a-b))
Explanation opens after your attempt
Correct Answer
C. ((3a-2b-a-b)(3a-2b+a+b))
Step 1
Concept
Difference of squares directly gives (A-B) and (A+B). In exams, you may simplify it to ((2a-3b)(4a-b)).
Step 2
Why this answer is correct
The correct answer is C. ((3a-2b-a-b)(3a-2b+a+b)). Difference of squares directly gives (A-B) and (A+B). In exams, you may simplify it to ((2a-3b)(4a-b)).
Step 3
Exam Tip
वर्गों के अंतर से सीधे (A-B) और (A+B) मिलता है। परीक्षा में चाहें तो इसे ((2a-3b)(4a-b)) तक सरल करें।
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\(x^2+4y^2+4xy-9z^2\) का गुणनखंड रूप क्या है?
What is the factorised form of \(x^2+4y^2+4xy-9z^2\)?
#hidden-square
#three-variables
#factorisation
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A ((x+2y-3z)(x+2y+3z))
B ((x-2y-3z)(x-2y+3z))
C ((x+2y)2 +9z-2 )
D ((x+3z)(x-3z)+4y-2 )
Explanation opens after your attempt
Correct Answer
A. ((x+2y-3z)(x+2y+3z))
Step 1
Concept
First (x-2 +4xy+4y-2 =(x+2y)2 ). In exams, rearrange terms and identify the perfect square.
Step 2
Why this answer is correct
The correct answer is A. ((x+2y-3z)(x+2y+3z)). First (x-2 +4xy+4y-2 =(x+2y)2 ). In exams, rearrange terms and identify the perfect square.
Step 3
Exam Tip
पहले (x-2 +4xy+4y-2 =(x+2y)2 ) बनता है। परीक्षा में पदों को पुनः क्रम में रखकर पूर्ण वर्ग पहचानें।
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\(16a^2-24ab+9b^2-25c^2\) को गुणनखंडित कीजिए।
Factorise \(16a^2-24ab+9b^2-25c^2\).
#three-variables
#perfect-square
#difference-of-squares
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A ((4a-3b-5c)(4a-3b+5c))
B ((4a+3b-5c)(4a+3b+5c))
C ((16a-9b-25c)(a+b+c))
D ((4a-3b-5c)2 )
Explanation opens after your attempt
Correct Answer
A. ((4a-3b-5c)(4a-3b+5c))
Step 1
Concept
The first three terms are ((4a-3b)2 ), and (25c-2 =(5c)2 ). In exams, use two-step factorisation.
Step 2
Why this answer is correct
The correct answer is A. ((4a-3b-5c)(4a-3b+5c)). The first three terms are ((4a-3b)2 ), and (25c-2 =(5c)2 ). In exams, use two-step factorisation.
Step 3
Exam Tip
पहले तीन पद ((4a-3b)2 ) हैं और (25c-2 =(5c)2 ) है। परीक्षा में दो-चरणीय गुणनखंडन करें।
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\(x^2+2xy+y^2-4z^2+4x+4y\) का गुणनखंड रूप क्या है?
What is the factorised form of \(x^2+2xy+y^2-4z^2+4x+4y\)?
#factorisation
#error-analysis
#compound-expression
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A ((x+y-2z)(x+y+2z+4))
B ((x+y+2z)(x+y-2z+4))
C ((x+y)(x+y+4)-4z-2 )
D ((x+y-2z+2)(x+y+2z+2))
Explanation opens after your attempt
Correct Answer
D. ((x+y-2z+2)(x+y+2z+2))
Step 1
Concept
Writing it as ((x+y)2 +4(x+y)-4z-2 ) does not directly factor as option (D). Verify by expansion carefully.
Step 2
Why this answer is correct
The correct answer is D. ((x+y-2z+2)(x+y+2z+2)). Writing it as ((x+y)2 +4(x+y)-4z-2 ) does not directly factor as option (D). Verify by expansion carefully.
Step 3
Exam Tip
इसे ((x+y)2 +4(x+y)-4z-2 ) लिखने से तुरंत नहीं टूटता, पर ((x+y+2)2 -(2z)2 -4) नहीं है। सही जाँच विस्तार से करें।
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\(2x^2+3xy-2y^2\) का सही गुणनखंड चुनिए।
Choose the correct factorisation of \(2x^2+3xy-2y^2\).
#two-variable-quadratic
#factorisation
#expert
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A ((2x-y)(x+2y))
B ((2x+y)(x-2y))
C ((x-y)(2x+2y))
D ((2x-2y)(x+y))
Explanation opens after your attempt
Correct Answer
A. ((2x-y)(x+2y))
Step 1
Concept
((2x-y)(x+2y)) gives \(2x^2+4xy-xy-2y^2\). In exams, check (4xy-xy=3xy).
Step 2
Why this answer is correct
The correct answer is A. ((2x-y)(x+2y)). ((2x-y)(x+2y)) gives \(2x^2+4xy-xy-2y^2\). In exams, check (4xy-xy=3xy).
Step 3
Exam Tip
((2x-y)(x+2y)) से \(2x^2+4xy-xy-2y^2\) मिलता है। परीक्षा में (4xy-xy=3xy) जाँचें।
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\(3a^2-5ab-2b^2\) को गुणनखंडित कीजिए।
Factorise \(3a^2-5ab-2b^2\).
#two-variable-quadratic
#signs
#factorisation
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A ((3a+b)(a-2b))
B ((3a-b)(a+2b))
C ((a-b)(3a+2b))
D ((3a-2b)(a+b))
Explanation opens after your attempt
Correct Answer
A. ((3a+b)(a-2b))
Step 1
Concept
((3a+b)(a-2b)) gives \(3a^2-6ab+ab-2b^2\). In exams, check that the middle terms add to (-5ab).
Step 2
Why this answer is correct
The correct answer is A. ((3a+b)(a-2b)). ((3a+b)(a-2b)) gives \(3a^2-6ab+ab-2b^2\). In exams, check that the middle terms add to (-5ab).
Step 3
Exam Tip
((3a+b)(a-2b)) से \(3a^2-6ab+ab-2b^2\) मिलता है। परीक्षा में मध्य पदों का योग (-5ab) देखें।
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\(x^2+2xy+y^2-z^2-2z-1\) का गुणनखंड रूप क्या है?
What is the factorised form of \(x^2+2xy+y^2-z^2-2z-1\)?
#hidden-square
#difference-of-squares
#advanced
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A ((x+y-z-1)(x+y+z+1))
B ((x+y+z-1)(x+y-z+1))
C ((x+y)2 -(z-1)2 )
D ((x-y-z-1)(x-y+z+1))
Explanation opens after your attempt
Correct Answer
A. ((x+y-z-1)(x+y+z+1))
Step 1
Concept
First it becomes ((x+y)2 -(z+1)2 ). In exams, understand \(-z^2-2z-1\) as (-(z+1)2 ).
Step 2
Why this answer is correct
The correct answer is A. ((x+y-z-1)(x+y+z+1)). First it becomes ((x+y)2 -(z+1)2 ). In exams, understand \(-z^2-2z-1\) as (-(z+1)2 ).
Step 3
Exam Tip
पहले ((x+y)2 -(z+1)2 ) बनता है। परीक्षा में \(-z^2-2z-1\) को (-(z+1)2 ) समझें।
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\(4x^2+4xy+y^2-6x-3y\) का गुणनखंड रूप चुनिए।
Choose the factorised form of \(4x^2+4xy+y^2-6x-3y\).
#substitution
#common-binomial
#factorisation
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A ((2x+y)(2x+y-3))
B ((2x-y)(2x+y-3))
C ((2x+y-3)2 )
D ((4x+y)(x+y-3))
Explanation opens after your attempt
Correct Answer
A. ((2x+y)(2x+y-3))
Step 1
Concept
First (4x-2 +4xy+y-2 =(2x+y)2 ), and the rest is (-3(2x+y)). In exams, take the common binomial outside.
Step 2
Why this answer is correct
The correct answer is A. ((2x+y)(2x+y-3)). First (4x-2 +4xy+y-2 =(2x+y)2 ), and the rest is (-3(2x+y)). In exams, take the common binomial outside.
Step 3
Exam Tip
पहले (4x-2 +4xy+y-2 =(2x+y)2 ) और बाकी (-3(2x+y)) है। परीक्षा में समान द्विपद को सामान्य लें।
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\(x^2-2xy+y^2-4x+4y\) का गुणनखंड रूप क्या है?
What is the factorised form of \(x^2-2xy+y^2-4x+4y\)?
#substitution
#common-binomial
#factorisation
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A ((x-y)(x-y-4))
B ((x+y)(x+y-4))
C ((x-y)2 -4)
D ((x-y-2)2 )
Explanation opens after your attempt
Correct Answer
A. ((x-y)(x-y-4))
Step 1
Concept
First (x-2 -2xy+y-2 =(x-y)2 ) and the remaining part is (-4(x-y)). In exams, treat the common binomial as one term.
Step 2
Why this answer is correct
The correct answer is A. ((x-y)(x-y-4)). First (x-2 -2xy+y-2 =(x-y)2 ) and the remaining part is (-4(x-y)). In exams, treat the common binomial as one term.
Step 3
Exam Tip
पहले (x-2 -2xy+y-2 =(x-y)2 ) और बाकी (-4(x-y)) है। परीक्षा में समान द्विपद को एक पद मानकर सामान्य लें।
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