Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
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Which is the factorisation of \(y^2-10y+25\)?
C. ((y-5)2)
\(25=5^2\) और \(-10y=-2\cdot5y\) है। इसलिए (y-2-10y+25=(y-5)2) है। / \(25=5^2\) and \(-10y=-2\cdot5y\). So (y-2-10y+25=(y-5)2).
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Choose the correct factorisation of \(p^2-q^2\).
D. ((p-q)(p+q))
यह वर्गों का अंतर है। इसलिए (p-2-q-2=(p-q)(p+q)) होगा। / This is a difference of squares. Therefore (p-2-q-2=(p-q)(p+q)).
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What will be the factorisation of (3x+3y)?
A. (3(x+y))
दोनों पदों में (3) समान है। इसलिए (3x+3y=3(x+y)) है। / (3) is common in both terms. So (3x+3y=3(x+y)).
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After taking out the common factor in \(7a^2-14a\), what remains?
B. (7a(a-2))
दोनों पदों में (7a) समान है। इसलिए (7a-2-14a=7a(a-2)) है। / (7a) is common in both terms. Therefore (7a-2-14a=7a(a-2)).
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Which is the factorisation of \(x^2+5x+6\)?
B. ((x+2)(x+3))
(2+3=5) और \(2\cdot3=6\) है। परीक्षा में ऐसे दो संख्याएं खोजें जिनका योग मध्य पद और गुणन अंतिम पद दे। / (2+3=5) and \(2\cdot3=6\). Exam tip: find two numbers whose sum gives the middle term and product gives the last term.
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What are the factors of \(x^2+7x+12\)?
C. ((x+3)(x+4))
(3+4=7) और \(3\cdot4=12\) है। परीक्षा में अंतिम पद के कारकों का योग जांचें। / (3+4=7) and \(3\cdot4=12\). Exam tip: check the sum of the factors of the last term.
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Choose the correct factorisation of \(x^2+9x+20\).
A. ((x+4)(x+5))
(4+5=9) और \(4\cdot5=20\) है। परीक्षा में सही कारक वही होंगे जिनसे दोनों शर्तें पूरी हों। / (4+5=9) and \(4\cdot5=20\). Exam tip: correct factors must satisfy both conditions.
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What is the factorisation of \(x^2+8x+15\)?
B. ((x+3)(x+5))
(3+5=8) और \(3\cdot5=15\) है। परीक्षा में छोटे कारकों से शुरुआत करें। / (3+5=8) and \(3\cdot5=15\). Exam tip: start with smaller factors.
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Which is the factorisation of \(x^2-25\)?
C. ((x-5)(x+5))
\(25=5^2\) है और यह वर्गों का अंतर है। इसलिए (x-2-25=(x-5)(x+5)) होगा। / \(25=5^2\) and this is a difference of squares. Therefore (x-2-25=(x-5)(x+5)).
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What is the correct factorisation of \(9x^2-1\)?
D. ((3x-1)(3x+1))
(9x-2=(3x)2) और \(1=1^2\) है। इसलिए वर्गों का अंतर ((3x-1)(3x+1)) बनेगा। / (9x-2=(3x)2) and \(1=1^2\). So the difference of squares becomes ((3x-1)(3x+1)).
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