100 results found for "binomial factor" in Class 10.
Which of the following is a binomial?
A. \(x^2+3\)
A polynomial with two terms is a binomial. \(x^2+3\) has two terms.
The correct answer is A. \(x^2+3\). A polynomial with two terms is a binomial. \(x^2+3\) has two terms.
दो पदों वाला बहुपद द्विपदी कहलाता है। \(x^2+3\) में दो पद हैं।
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Which polynomial is a binomial?
B. (5x-7)
A binomial has two terms. The expression (5x-7) has the two terms (5x) and (-7).
The correct answer is B. (5x-7). A binomial has two terms. The expression (5x-7) has the two terms (5x) and (-7).
द्विपदी में दो पद होते हैं। (5x-7) में (5x) और (-7) दो पद हैं।
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If (p(x)=x-3-4x-2-7x+10) and (x-1) is a factor, what is the remaining quadratic factor?
A. \(x^2-3x-10\)
Dividing (p(x)) by (x-1) gives \(x^2-3x-10\). Verify by multiplying ((x-1)\(x^2-3x-10\)=p(x)).
The correct answer is A. \(x^2-3x-10\). Dividing (p(x)) by (x-1) gives \(x^2-3x-10\). Verify by multiplying ((x-1)\(x^2-3x-10\)=p(x)).
(p(x)) को (x-1) से भाग देने पर \(x^2-3x-10\) मिलता है। गुणा करके जाँचें कि ((x-1)\(x^2-3x-10\)=p(x))।
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If (x-3) is a factor of (p(x)=x-3-5x-2+ax+12), what is the value of (a)?
A. (2)
By the factor theorem (p(3)=0), so (27-45+3a+12=0) and (a=2). In exams, substitute the zero from the given factor directly.
The correct answer is A. (2). By the factor theorem (p(3)=0), so (27-45+3a+12=0) and (a=2). In exams, substitute the zero from the given factor directly.
गुणनखंड प्रमेय से (p(3)=0), इसलिए (27-45+3a+12=0) और (a=2)। परीक्षा में दिए गए गुणनखंड से मूल तुरंत रखिए।
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If (p(x)=x-3-7x+6), which of the following is not a factor of (p(x))?
A. (x+3)
(p(-3)=-27+21+6=0), so (x+3) is actually a factor. This reveals an option error, so check each option using (p(a)).
The correct answer is A. (x+3). (p(-3)=-27+21+6=0), so (x+3) is actually a factor. This reveals an option error, so check each option using (p(a)).
(p(-3)=-27+21+6=0), इसलिए (x+3) वास्तव में गुणनखंड है। सही जाँच में यह प्रश्न विकल्प-त्रुटि दिखाता है, इसलिए हर विकल्प को (p(a)) से जाँचें।
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If (p(x)=x-3+3x-2-4x-12), which of the following is a factor?
A. (x+3)
(p(-3)=-27+27+12-12=0), so (x+3) is a factor. For (x+a), test (x=-a).
The correct answer is A. (x+3). (p(-3)=-27+27+12-12=0), so (x+3) is a factor. For (x+a), test (x=-a).
(p(-3)=-27+27+12-12=0), इसलिए (x+3) गुणनखंड है। (x+a) के लिए (x=-a) रखकर जाँचें।
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If (x-2) is a factor of (p(x)=x-3+kx-2-4x-4), what is the value of (k)?
A. (1)
By factor theorem (p(2)=0), so (8+4k-8-4=0) and (k=1). In exams, substitute the given zero directly.
The correct answer is A. (1). By factor theorem (p(2)=0), so (8+4k-8-4=0) and (k=1). In exams, substitute the given zero directly.
गुणनखंड प्रमेय से (p(2)=0), इसलिए (8+4k-8-4=0) और (k=1)। परीक्षा में पहले दिए गए मूल को सीधे रखिए।
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What is obtained by taking the common factor from \(4x^5+12x^4\)?
A. (4x-4(x+3))
The common factor in both terms is \(4x^4\). Hence the form is (4x-4(x+3)).
The correct answer is A. (4x-4(x+3)). The common factor in both terms is \(4x^4\). Hence the form is (4x-4(x+3)).
दोनों पदों में सामान्य गुणनखंड \(4x^4\) है। इसलिए रूप (4x-4(x+3)) है।
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What is obtained by taking the common factor from \(3x^4+9x^3\)?
A. (3x-3(x+3))
The common factor in both terms is \(3x^3\). Hence the form is (3x-3(x+3)).
The correct answer is A. (3x-3(x+3)). The common factor in both terms is \(3x^3\). Hence the form is (3x-3(x+3)).
दोनों पदों में सामान्य गुणनखंड \(3x^3\) है। इसलिए रूप (3x-3(x+3)) है।
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What is obtained by taking the common factor from \(2x^3+6x^2\)?
A. (2x-2(x+3))
The common factor in both terms is \(2x^2\). Hence the form is (2x-2(x+3)).
The correct answer is A. (2x-2(x+3)). The common factor in both terms is \(2x^2\). Hence the form is (2x-2(x+3)).
दोनों पदों में सामान्य गुणनखंड \(2x^2\) है। इसलिए रूप (2x-2(x+3)) है।
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By taking common factor, how will \(4x^2+28x=0\) be written?
A. (4x(x+7)=0)
(4x) is the common factor, so (4x(x+7)=0). In exams, take out the greatest common factor.
The correct answer is A. (4x(x+7)=0). (4x) is the common factor, so (4x(x+7)=0). In exams, take out the greatest common factor.
(4x) सामान्य गुणनखंड है, इसलिए (4x(x+7)=0) मिलता है। परीक्षा में सबसे बड़ा सामान्य गुणनखंड निकालें।
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By taking common factor, how will \(5x^2+15x=0\) be written?
A. (5x(x+3)=0)
(5x) is the common factor, so (5x(x+3)=0). In exams, take out the greatest common factor.
The correct answer is A. (5x(x+3)=0). (5x) is the common factor, so (5x(x+3)=0). In exams, take out the greatest common factor.
(5x) सामान्य गुणनखंड है, इसलिए (5x(x+3)=0) मिलता है। परीक्षा में सबसे बड़ा सामान्य गुणनखंड निकालें।
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Which factor form can represent \(x^2+11x+30=0\)?
A. ((x+5)(x+6)=0)
\(5\cdot6=30\) and (5+6=11). Therefore the correct factors are ((x+5)(x+6)).
The correct answer is A. ((x+5)(x+6)=0). \(5\cdot6=30\) and (5+6=11). Therefore the correct factors are ((x+5)(x+6)).
\(5\cdot6=30\) और (5+6=11) है। इसलिए सही गुणनखंड ((x+5)(x+6)) हैं।
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Which factor form can represent \(x^2+9x+20=0\)?
A. ((x+4)(x+5)=0)
\(4\cdot5=20\) and (4+5=9). Therefore the correct factors are ((x+4)(x+5)).
The correct answer is A. ((x+4)(x+5)=0). \(4\cdot5=20\) and (4+5=9). Therefore the correct factors are ((x+4)(x+5)).
\(4\cdot5=20\) और (4+5=9) है। इसलिए सही गुणनखंड ((x+4)(x+5)) हैं।
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Which factor form can represent \(x^2+7x+12=0\)?
A. ((x+3)(x+4)=0)
The product of (3) and (4) is (12), and their sum is (7). Therefore ((x+3)(x+4)=0) is correct.
The correct answer is A. ((x+3)(x+4)=0). The product of (3) and (4) is (12), and their sum is (7). Therefore ((x+3)(x+4)=0) is correct.
(3) और (4) का गुणनफल (12) और योग (7) है। इसलिए ((x+3)(x+4)=0) सही है।
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Which option is the factor form of \(x^2+x-12=0\)?
A. ((x+4)(x-3)=0)
Expanding ((x+4)(x-3)) gives \(x^2+x-12\). To check factors, expand them.
The correct answer is A. ((x+4)(x-3)=0). Expanding ((x+4)(x-3)) gives \(x^2+x-12\). To check factors, expand them.
((x+4)(x-3)) फैलाने पर \(x^2+x-12\) मिलता है। गुणनखंड जांचने के लिए विस्तार करें।
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How do we write \(x^2-49=0\) in factor form?
A. ((x-7)(x+7)=0)
\(x^2-49\) is a difference of squares. Hence it becomes ((x-7)(x+7)).
The correct answer is A. ((x-7)(x+7)=0). \(x^2-49\) is a difference of squares. Hence it becomes ((x-7)(x+7)).
\(x^2-49\) वर्गों का अंतर है। इसलिए यह ((x-7)(x+7)) बनता है।
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Which option shows \(x^2+10x+25=0\) in factor form?
A. ((x+5)2=0)
\(x^2+10x+25\) is a perfect square. It equals ((x+5)2).
The correct answer is A. ((x+5)2=0). \(x^2+10x+25\) is a perfect square. It equals ((x+5)2).
\(x^2+10x+25\) पूर्ण वर्ग है। यह ((x+5)2) के बराबर है।
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Which option is the factor form of \(x^2-2x-15=0\)?
A. \((x-5)(x+3)=0\)
Expanding ((x-5)(x+3)) gives \(x^2-2x-15\). To check factors, expand them.
The correct answer is A. \((x-5)(x+3)=0\). Expanding ((x-5)(x+3)) gives \(x^2-2x-15\). To check factors, expand them.
((x-5)(x+3)) फैलाने पर \(x^2-2x-15\) मिलता है। गुणनखंड जांचने के लिए विस्तार करें।
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How do we write \(x^2-16=0\) in factor form?
A. \((x-4)(x+4)=0\)
\(x^2-16\) is a difference of squares. Hence it becomes ((x-4)(x+4)).
The correct answer is A. \((x-4)(x+4)=0\). \(x^2-16\) is a difference of squares. Hence it becomes ((x-4)(x+4)).
\(x^2-16\) वर्गों का अंतर है। इसलिए यह ((x-4)(x+4)) बनता है।
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Which option shows \(x^2+6x+9=0\) in factor form?
A. \((x+3)^2=0\)
\(x^2+6x+9\) is a perfect square. It equals ((x+3)2).
The correct answer is A. \((x+3)^2=0\). \(x^2+6x+9\) is a perfect square. It equals ((x+3)2).
\(x^2+6x+9\) एक पूर्ण वर्ग है। यह ((x+3)2) के बराबर है।
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A. (x-\(2-\sqrt{3}\))
The companion zero is \(2-\sqrt{3}\), so the factor is (x-\(2-\sqrt{3}\)). In exams remember the relation between a zero and factor as \(x-\alpha\).
The correct answer is A. (x-\(2-\sqrt{3}\)). The companion zero is \(2-\sqrt{3}\), so the factor is (x-\(2-\sqrt{3}\)). In exams remember the relation between a zero and factor as \(x-\alpha\).
साथी शून्यक \(2-\sqrt{3}\) होगा, इसलिए गुणनखंड (x-\(2-\sqrt{3}\)) है। परीक्षा में शून्यक और गुणनखंड का संबंध \(x-\alpha\) याद रखें।
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B. (273)
The factors (3), (7), and (13) must be removed from the reduced denominator, so the minimum factor is \(3\cdot 7\cdot 13=273\). Factors (2) and (5) may remain.
The correct answer is B. (273). The factors (3), (7), and (13) must be removed from the reduced denominator, so the minimum factor is \(3\cdot 7\cdot 13=273\). Factors (2) and (5) may remain.
सरलतम हर से (3), (7) और (13) हटने चाहिए इसलिए न्यूनतम गुणनखंड \(3\cdot 7\cdot 13=273\) है। (2) और (5) हर में रह सकते हैं।
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B. (1539)
The factors \(3^4\) and (19) must be removed from the reduced denominator, so the minimum factor is \(81\cdot 19=1539\). Factors (2) and (5) may remain.
The correct answer is B. (1539). The factors \(3^4\) and (19) must be removed from the reduced denominator, so the minimum factor is \(81\cdot 19=1539\). Factors (2) and (5) may remain.
सरलतम हर से \(3^4\) और (19) हटने चाहिए इसलिए न्यूनतम गुणनखंड \(81\cdot 19=1539\) है। (2) और (5) हर में रह सकते हैं।
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B. (621)
The factors \(3^3\) and (23) must be removed from the reduced denominator, so the minimum factor is \(27\cdot 23=621\). Factors (2) and (5) may remain.
The correct answer is B. (621). The factors \(3^3\) and (23) must be removed from the reduced denominator, so the minimum factor is \(27\cdot 23=621\). Factors (2) and (5) may remain.
सरलतम हर से \(3^3\) और (23) हटने चाहिए इसलिए न्यूनतम गुणनखंड \(27\cdot 23=621\) है। (2) और (5) हर में रह सकते हैं।
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A. (153)
The factors \(3^2\) and (17) must be removed from the reduced denominator, so the minimum factor is \(3^2\cdot 17=153\). Factors (2) and (5) may remain.
The correct answer is A. (153). The factors \(3^2\) and (17) must be removed from the reduced denominator, so the minimum factor is \(3^2\cdot 17=153\). Factors (2) and (5) may remain.
सरलतम हर से \(3^2\) और (17) हटने चाहिए, इसलिए न्यूनतम गुणनखंड \(3^2\cdot 17=153\) है। (2) और (5) हर में रह सकते हैं।
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B. (57)
The denominator contains (2), (5), (3), and (19).
For a terminating decimal, (3) and (19) must cancel. So the minimum factor is \(3\cdot 19=57\).
Only (2) and (5) may remain in the denominator. चरण 1: हर में (2), (5), (3), और (19) हैं। चरण 2: सांत दशमलव के लिए (3) और (19) कटने चाहिए। इसलिए न्यूनतम गुणनखंड \(3\cdot 19=57\) है। चरण 3: केवल (2) और (5) हर में रह सकते हैं।
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If \(\frac{x}{540}\) has a terminating decimal expansion, what factor must (x) contain at minimum?
B. (27)
\(540=2^2\cdot 3^3\cdot 5\).
For a terminating decimal, \(3^3\) must cancel completely from the denominator. So (x) must contain (27).
(2) and (5) may remain, but (3) must not. चरण 1: \(540=2^2\cdot 3^3\cdot 5\) है। चरण 2: सांत दशमलव के लिए सरलतम हर से \(3^3\) पूरी तरह कटना चाहिए। इसलिए (x) में (27) अवश्य होना चाहिए। चरण 3: (2) और (5) रह सकते हैं, पर (3) नहीं।
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C. (231)
\(2310=2\cdot 3\cdot 5\cdot 7\cdot 11\).
For a terminating decimal, (3), (7), and (11) must cancel from the denominator. So the minimum factor is \(3\cdot 7\cdot 11=231\).
(2) and (5) may remain, but other prime factors must not. चरण 1: \(2310=2\cdot 3\cdot 5\cdot 7\cdot 11\) है। चरण 2: सांत दशमलव के लिए (3), (7), और (11) हर से कटने चाहिए। इसलिए न्यूनतम गुणनखंड \(3\cdot 7\cdot 11=231\) है। चरण 3: (2) और (5) रह सकते हैं, पर अन्य अभाज्य गुणनखंड नहीं।
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If \(\frac{m}{735}\) has a terminating decimal expansion, what factor must (m) contain at minimum?
C. (147)
\(735=3\cdot 5\cdot 7^2\).
For a terminating decimal, (3) and \(7^2\) must not remain in the reduced denominator. So (m) must contain \(3\cdot 7^2=147\).
The factor (5) may remain, but (3) and (7) must cancel. चरण 1: \(735=3\cdot 5\cdot 7^2\) है। चरण 2: सांत दशमलव के लिए सरलतम हर में (3) और \(7^2\) नहीं बचने चाहिए। इसलिए (m) में \(3\cdot 7^2=147\) अवश्य होना चाहिए। चरण 3: (5) हर में रह सकता है, पर (3) और (7) कटने चाहिए।
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If \(\frac{m}{56}\) has a terminating decimal expansion, which factor must (m) contain?
C. (7)
\(56=2^3\cdot 7\).
For a terminating decimal, (7) must not remain in the reduced denominator. Therefore (m) must contain (7).
Cancel all denominator primes other than (2) and (5). चरण 1: \(56=2^3\cdot 7\) है। चरण 2: सांत दशमलव के लिए सरलतम हर में (7) नहीं बचना चाहिए। इसलिए (m) में (7) अवश्य होना चाहिए। चरण 3: हर से (2) और (5) के अलावा बाकी अभाज्य गुणनखंड कटवाएँ।
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If the reduced denominator contains the factor (17), what type of decimal expansion will occur?
B. असमाप्त आवर्तीNon-terminating recurring
(17) is a prime other than (2) and (5).
If (17) remains in the reduced denominator, the decimal will not terminate.
Such a non-terminating decimal of a rational number is recurring. चरण 1: (17) एक ऐसा अभाज्य है जो (2) या (5) नहीं है। चरण 2: सरलतम हर में (17) होने पर दशमलव समाप्त नहीं होगा। चरण 3: परिमेय संख्या का ऐसा असमाप्त दशमलव आवर्ती होता है।
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B. असमाप्त आवर्तीNon-terminating recurring
In the reduced denominator, (3) is a prime other than (2) and (5).
So the decimal will not terminate but will repeat.
A non-terminating decimal of a rational number is recurring. चरण 1: सरलतम हर में (3) एक ऐसा अभाज्य है जो (2) या (5) नहीं है। चरण 2: इसलिए दशमलव समाप्त नहीं होगा, पर दोहराव आएगा। चरण 3: परिमेय संख्या का असमाप्त दशमलव हमेशा आवर्ती होता है।
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A. कथन और कारण दोनों सही हैं, और कारण कथन को समझाता हैBoth assertion and reason are true, and the reason explains the assertion
For a terminating decimal, the denominator in lowest form must be made only of (2) and (5).
If (7) remains, this condition fails and the decimal will not terminate.
The reason correctly explains the assertion, so the first option is correct. चरण 1: समाप्त दशमलव के लिए सरल रूप का भाजक केवल (2) और (5) से बना होना चाहिए। चरण 2: यदि (7) बचा है, तो यह शर्त पूरी नहीं होती और दशमलव समाप्त नहीं होगा। चरण 3: कारण कथन को ठीक से समझा रहा है, इसलिए पहला विकल्प सही है।
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B. असमाप्त आवर्तीNon-terminating recurring
For a terminating decimal, the denominator should have only (2) and (5).
If (3) remains, the decimal will not terminate and will be recurring because the number is rational.
The remaining factors in lowest form decide the result. चरण 1: समाप्त दशमलव के लिए भाजक में केवल (2) और (5) होने चाहिए। चरण 2: (3) बचने पर दशमलव समाप्त नहीं होगा और परिमेय होने से आवर्ती होगा। चरण 3: सरल रूप में बचे गुणनखंड ही निर्णय करते हैं।
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A. पहले \(5\mid a\), फिर (a=5k) रखने से \(5\mid b\)First \(5\mid a\), then substituting (a=5k) gives \(5\mid b\)
From \(a^2=5b^2\), \(5\mid a\).
Putting (a=5k) gives \(b^2=5k^2\), so \(5\mid b\).
Now (5) becomes a common factor and gives the contradiction. चरण 1: \(a^2=5b^2\) से \(5\mid a\) मिलता है। चरण 2: (a=5k) रखने पर \(b^2=5k^2\), इसलिए \(5\mid b\)। चरण 3: अब (5) दोनों में साझा गुणनखंड बनकर विरोधाभास देता है।
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Which statement correctly tells the role of factor (2) in the proof of \(\sqrt{2}\)?
A. यह (p) और (q) दोनों में साझा गुणनखंड बनकर विरोधाभास देता हैIt becomes a common factor of both (p) and (q) and gives contradiction
From \(p^2=2q^2\), factor (2) first appears in (p).
Later factor (2) also appears in (q).
Common factor (2) contradicts the coprime condition. चरण 1: \(p^2=2q^2\) से पहले (p) में (2) का गुणनखंड आता है। चरण 2: बाद में (q) में भी (2) का गुणनखंड मिलता है। चरण 3: दोनों में (2) साझा होना सहअभाज्य शर्त से टकराता है।
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Which statement correctly states the prime factor rule?
A. यदि अभाज्य संख्या किसी वर्ग को विभाजित करे, तो वह मूल संख्या को भी विभाजित करती हैIf a prime divides a square, it also divides the original number
Prime factors in a square occur in pairs.
If a prime divides \(p^2\), it also divides (p).
This rule is needed in the proofs of \(\sqrt{3}\) and \(\sqrt{5}\). चरण 1: अभाज्य गुणनखंड वर्ग में जोड़े में आते हैं। चरण 2: यदि कोई अभाज्य \(p^2\) को विभाजित करता है, तो वह (p) को भी विभाजित करेगा। चरण 3: \(\sqrt{3}\) और \(\sqrt{5}\) के प्रमाण में यह नियम जरूरी है।
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Which prime factor plays the main role in proving \(\sqrt{3}\) irrational?
C. (3)
Assuming \(\sqrt{3}\) rational gives \(p^2=3q^2\).
The factor (3) appears commonly in (p) and (q).
The number under the square root becomes the key factor. चरण 1: \(\sqrt{3}\) को परिमेय मानने पर \(p^2=3q^2\) मिलता है। चरण 2: यहां (3) का गुणनखंड ही (p) और (q) में साझा रूप से आता है। चरण 3: जिस संख्या का वर्गमूल हो, वही मुख्य गुणनखंड बनती है।
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In which proof is the prime factor (5) used mainly?
A. \(\sqrt{5}\) की अपरिमेयता के प्रमाण मेंIn the proof of irrationality of \(\sqrt{5}\)
Assuming \(\sqrt{5}\) rational gives \(a^2=5b^2\).
Here prime factor (5) is the key.
It leads to common factor (5) in both numbers. चरण 1: \(\sqrt{5}\) को परिमेय मानने पर \(a^2=5b^2\) मिलता है। चरण 2: यहां (5) अभाज्य गुणनखंड मुख्य है। चरण 3: इसी से दोनों संख्याओं में (5) साझा गुणनखंड मिलता है।
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In which proof is the prime factor (3) used mainly?
A. \(\sqrt{3}\) की अपरिमेयता के प्रमाण मेंIn the proof of irrationality of \(\sqrt{3}\)
Assuming \(\sqrt{3}\) rational gives \(a^2=3b^2\).
So factor (3) becomes the main base of the proof.
It shows both (a) and (b) divisible by (3). चरण 1: \(\sqrt{3}\) को परिमेय मानने पर \(a^2=3b^2\) मिलता है। चरण 2: इसलिए (3) का गुणनखंड प्रमाण का मुख्य आधार बनता है। चरण 3: इसी से (a) और (b) दोनों (3) से विभाज्य मिलते हैं।
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In which proof is the prime factor (2) used mainly?
A. \(\sqrt{2}\) की अपरिमेयता के प्रमाण मेंIn the proof of irrationality of \(\sqrt{2}\)
In the proof of \(\sqrt{2}\), we get \(a^2=2b^2\).
So the factor (2) plays the main role.
The number under the square root appears as the key factor in the proof. चरण 1: \(\sqrt{2}\) के प्रमाण में \(a^2=2b^2\) मिलता है। चरण 2: इसलिए (2) का गुणनखंड मुख्य भूमिका निभाता है। चरण 3: जिस संख्या का वर्गमूल हो, वही गुणनखंड प्रमाण में आता है।
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A. (2)
An even number is always divisible by (2).
Both are even, so (2) is their common factor.
Finding a common factor contradicts the lowest-form condition. चरण 1: सम संख्या हमेशा (2) से विभाज्य होती है। चरण 2: दोनों सम हैं, इसलिए (2) दोनों का साझा गुणनखंड है। चरण 3: साझा गुणनखंड मिलना सरलतम रूप की शर्त से टकराता है।
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Two numbers are (180) and (252). What is their HCF in prime factor form?
A. \(2^2\times3^2\)
Write both numbers in prime factor form.
\(180=2^2\times3^2\times5\) and \(252=2^2\times3^2\times7\), so the common smaller powers give \(2^2\times3^2\).
Do not include non-common factors like (5) or (7) in HCF. चरण 1: पहले दोनों संख्याओं को अभाज्य गुणनखंडों में लिखें। चरण 2: \(180=2^2\times3^2\times5\) और \(252=2^2\times3^2\times7\), इसलिए समान छोटी घातें \(2^2\times3^2\) हैं। चरण 3: महत्तम समापवर्तक में (5) या (7) जैसे असमान गुणनखंड न लें।
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What will be the greatest odd factor of \(2^3 \times 3^2 \times 5^2\)?
C. 225
An odd factor must not contain (2).
Removing \(2^3\) leaves \(3^2 \times 5^2=9 \times 25=225\).
To get the greatest odd factor, remove all powers of (2). चरण 1: विषम गुणनखंड में (2) नहीं होना चाहिए। चरण 2: \(2^3\) हटाने पर \(3^2 \times 5^2=9 \times 25=225\) बचता है। चरण 3: सबसे बड़ा विषम गुणनखंड पाने के लिए सभी (2) हटा दें।
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What is the greatest prime factor of \(2^2 \times 3^3 \times 11\)?
D. 11
Prime factors are the base numbers.
The prime bases here are (2,3,11), and the greatest is (11).
Do not treat a composite value like (9) as a prime factor. चरण 1: अभाज्य गुणनखंड आधार संख्याएं होती हैं। चरण 2: यहां अभाज्य आधार (2,3,11) हैं, जिनमें सबसे बड़ा (11) है। चरण 3: (9) जैसे संयुक्त मान को अभाज्य गुणनखंड न मानें।
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What will be the greatest odd factor of \(2^2 \times 3^3 \times 5\)?
C. 135
An odd factor must not contain (2).
After removing \(2^2\), we get \(3^3 \times 5=27 \times 5=135\).
To get the greatest odd factor, remove all powers of (2). चरण 1: विषम गुणनखंड में (2) नहीं होना चाहिए। चरण 2: \(2^2\) हटाने पर \(3^3 \times 5=27 \times 5=135\) बचता है। चरण 3: सबसे बड़ा विषम गुणनखंड पाने के लिए सभी (2) हटा दें।
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What is the greatest prime factor of \(2^3 \times 3^2 \times 5\)?
C. 5
Prime factors are the base numbers.
The prime bases here are (2,3,5), and the greatest is (5).
Do not treat a number like (9) as a prime factor. चरण 1: अभाज्य गुणनखंड आधार संख्याएं होती हैं। चरण 2: यहां अभाज्य आधार (2,3,5) हैं, जिनमें सबसे बड़ा (5) है। चरण 3: (9) जैसी संख्या को अभाज्य गुणनखंड न मानें।
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What is the greatest prime factor of \(2^3 \times 3 \times 5\)?
C. 5
The prime factors are the base numbers.
Here (2,3,5) are prime factors, and the greatest is (5).
When the greatest prime factor is asked, do not choose a composite number. चरण 1: अभाज्य गुणनखंड आधार संख्याएं हैं। चरण 2: यहां (2,3,5) अभाज्य गुणनखंड हैं, इनमें सबसे बड़ा (5) है। चरण 3: सबसे बड़ा अभाज्य गुणनखंड पूछे जाने पर संयुक्त संख्या को विकल्प न चुनें।
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What is the smallest prime factor of \(2^3 \times 3^2\)?
A. 2
The bases in prime factorisation are the prime factors.
Here the bases are (2) and (3), and the smallest is (2).
To find the smallest prime factor, do not focus on exponents. चरण 1: अभाज्य गुणनखंडन में आधार संख्याएं अभाज्य गुणनखंड होती हैं। चरण 2: यहां आधार (2) और (3) हैं, इनमें सबसे छोटा (2) है। चरण 3: सबसे छोटा अभाज्य गुणनखंड खोजते समय घातों को न देखें।
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What is the greatest odd factor of \(2^5 \times 3\)?
A. 3
An odd factor must not contain (2).
Removing \(2^5\) leaves only (3), so the greatest odd factor is (3).
For the greatest odd factor, remove all powers of (2). चरण 1: विषम गुणनखंड में (2) नहीं होना चाहिए। चरण 2: \(2^5\) को हटाने पर केवल (3) बचता है, इसलिए सबसे बड़ा विषम गुणनखंड (3) है। चरण 3: सबसे बड़े विषम गुणनखंड के लिए सभी (2) हटा दें।
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What is the greatest odd factor of \(2^3 \times 3^2 \times 5\)?
A. 45
An odd factor must not contain (2).
Remove \(2^3\) and keep \(3^2 \times 5=45\).
For the greatest odd factor, remove all powers of (2). चरण 1: विषम गुणनखंड में (2) नहीं होना चाहिए। चरण 2: सबसे बड़ा विषम गुणनखंड पाने के लिए \(2^3\) हटाकर \(3^2 \times 5=9 \times 5=45\) लें। चरण 3: विषम गुणनखंडों में सभी (2) हटा दें।
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If \(N=2^4 \times 3^3 \times 7\), what is the smallest prime factor of (N)?
A. (2)
The bases in prime factorisation are the prime factors.
Among (2,3,7), the smallest prime number is (2).
For the smallest prime factor, look at the base, not the exponent. चरण 1: अभाज्य गुणनखंडन में लिखी सभी संख्याएं अभाज्य गुणनखंड होती हैं। चरण 2: (2,3,7) में सबसे छोटी अभाज्य संख्या (2) है। चरण 3: सबसे छोटे गुणनखंड के लिए घात नहीं, आधार संख्या देखें।
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Which of the following was not a factor that increased conflict in the Balkans?
D. सभी समुदायों का एक ही राष्ट्रीय लक्ष्य पर पूर्ण सहमत होनाComplete agreement of all communities on one national goal
Ethnic nationalism, Ottoman decline, and great power rivalry increased conflict.
Complete agreement among all communities would have reduced conflict.
Therefore, the fourth option was not a factor increasing conflict. चरण 1: जातीय राष्ट्रवाद, ओटोमन पतन और बड़ी शक्तियों की होड़ ने संघर्ष बढ़ाया। चरण 2: सभी समुदायों की पूर्ण सहमति होती तो संघर्ष कम होता। चरण 3: इसलिए चौथा विकल्प संघर्ष बढ़ाने वाला कारण नहीं है।
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What is obtained by simplifying ((3x+2)(x-4))?
A. \(3x^2-10x-8\)
((3x+2)(x-4)=3x-2-12x+2x-8=3x-2-10x-8). Multiply each term by each term.
The correct answer is A. \(3x^2-10x-8\). ((3x+2)(x-4)=3x-2-12x+2x-8=3x-2-10x-8). Multiply each term by each term.
((3x+2)(x-4)=3x-2-12x+2x-8=3x-2-10x-8)। प्रत्येक पद का हर पद से गुणा करें।
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What is the polynomial form of ((x-5)(x+2))?
A. \(x^2-3x-10\)
((x-5)(x+2)=x-2+2x-5x-10=x-2-3x-10). Combine the middle terms with signs.
The correct answer is A. \(x^2-3x-10\). ((x-5)(x+2)=x-2+2x-5x-10=x-2-3x-10). Combine the middle terms with signs.
((x-5)(x+2)=x-2+2x-5x-10=x-2-3x-10)। मध्य पदों को संकेत सहित जोड़ें।
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Which option correctly describes the type of \(x^3-1\)?
C. त्रिघात द्विपदीCubic binomial
\(x^3-1\) has degree (3) and (2) terms. So it is a cubic binomial.
The correct answer is C. त्रिघात द्विपदी / Cubic binomial. \(x^3-1\) has degree (3) and (2) terms. So it is a cubic binomial.
\(x^3-1\) की घात (3) है और इसमें (2) पद हैं। इसलिए यह त्रिघात द्विपदी है।
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What is obtained by simplifying ((2x-1)(x+5))?
A. \(2x^2+9x-5\)
((2x-1)(x+5)=2x-2+10x-x-5=2x-2+9x-5). Multiply each term by each term.
The correct answer is A. \(2x^2+9x-5\). ((2x-1)(x+5)=2x-2+10x-x-5=2x-2+9x-5). Multiply each term by each term.
((2x-1)(x+5)=2x-2+10x-x-5=2x-2+9x-5)। हर पद का हर पद से गुणा करें।
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How many terms are there in \(7x^2+5\)?
B. (2)
The terms are \(7x^2\) and (5). Therefore it has (2) terms.
The correct answer is B. (2). The terms are \(7x^2\) and (5). Therefore it has (2) terms.
पद \(7x^2\) और (5) हैं। इसलिए इसमें (2) पद हैं।
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Which of the following is a polynomial with (2) terms?
B. \(5x^3-4\)
\(5x^3-4\) has (2) terms. A polynomial with two terms is called a binomial.
The correct answer is B. \(5x^3-4\). \(5x^3-4\) has (2) terms. A polynomial with two terms is called a binomial.
\(5x^3-4\) में (2) पद हैं। दो पदों वाला बहुपद द्विपदी कहलाता है।
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How many terms are there in (p(x)=7x-2-1)?
B. (2)
There are two terms \(7x^2\) and (-1). So it is also a binomial.
The correct answer is B. (2). There are two terms \(7x^2\) and (-1). So it is also a binomial.
\(7x^2\) और (-1) दो पद हैं। इसलिए यह द्विपदी भी है।
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What is the expansion of ((5x+2)2)?
A. \(25x^2+20x+4\)
Take (a=5x) and (b=2) in ((a+b)2). The middle term is \(2\cdot5x\cdot2=20x\).
The correct answer is A. \(25x^2+20x+4\). Take (a=5x) and (b=2) in ((a+b)2). The middle term is \(2\cdot5x\cdot2=20x\).
((a+b)2) में (a=5x) और (b=2) लें। मध्य पद \(2\cdot5x\cdot2=20x\) है।
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What is the expansion of ((4x-5)2)?
A. \(16x^2-40x+25\)
Use ((a-b)2=a-2-2ab+b-2). Here (a=4x) and (b=5).
The correct answer is A. \(16x^2-40x+25\). Use ((a-b)2=a-2-2ab+b-2). Here (a=4x) and (b=5).
((a-b)2=a-2-2ab+b-2) लगाएँ। यहाँ (a=4x) और (b=5) हैं।
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What is the expansion of ((4x+5)(x-3))?
A. \(4x^2-7x-15\)
By distribution, \(4x^2-12x+5x-15=4x^2-7x-15\). Combine like terms at the end.
The correct answer is A. \(4x^2-7x-15\). By distribution, \(4x^2-12x+5x-15=4x^2-7x-15\). Combine like terms at the end.
वितरण से \(4x^2-12x+5x-15=4x^2-7x-15\) है। समान पदों को अंत में जोड़ें।
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What is the expansion of ((x-7)(x+3))?
A. \(x^2-4x-21\)
Multiplying gives \(x^2+3x-7x-21=x^2-4x-21\). Pay attention to signs.
The correct answer is A. \(x^2-4x-21\). Multiplying gives \(x^2+3x-7x-21=x^2-4x-21\). Pay attention to signs.
गुणा करने पर \(x^2+3x-7x-21=x^2-4x-21\) है। चिह्नों पर ध्यान दें।
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What is the expansion of ((x+6)(x+4))?
A. \(x^2+10x+24\)
By distribution, \(x^2+4x+6x+24=x^2+10x+24\). The middle term forms by combining like terms.
The correct answer is A. \(x^2+10x+24\). By distribution, \(x^2+4x+6x+24=x^2+10x+24\). The middle term forms by combining like terms.
वितरण नियम से \(x^2+4x+6x+24=x^2+10x+24\) मिलता है। मध्य पद समान पद जोड़कर बनता है।
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What is the expansion of ((4x+3)2)?
A. \(16x^2+24x+9\)
Take (a=4x) and (b=3) in ((a+b)2). The middle term is \(2\cdot4x\cdot3=24x\).
The correct answer is A. \(16x^2+24x+9\). Take (a=4x) and (b=3) in ((a+b)2). The middle term is \(2\cdot4x\cdot3=24x\).
((a+b)2) में (a=4x) और (b=3) लें। मध्य पद \(2\cdot4x\cdot3=24x\) है।
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What is the expansion of ((3x-4)2)?
A. \(9x^2-24x+16\)
Use ((a-b)2=a-2-2ab+b-2). Here (a=3x) and (b=4).
The correct answer is A. \(9x^2-24x+16\). Use ((a-b)2=a-2-2ab+b-2). Here (a=3x) and (b=4).
((a-b)2=a-2-2ab+b-2) लगाएँ। यहाँ (a=3x) और (b=4) हैं।
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What is the expansion of ((3x+1)(x-4))?
A. \(3x^2-11x-4\)
By distribution, \(3x^2-12x+x-4=3x^2-11x-4\). Combine the middle like terms.
The correct answer is A. \(3x^2-11x-4\). By distribution, \(3x^2-12x+x-4=3x^2-11x-4\). Combine the middle like terms.
वितरण से \(3x^2-12x+x-4=3x^2-11x-4\) है। बीच के समान पद जोड़ें।
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What is the expansion of ((x-5)(x+2))?
A. \(x^2-3x-10\)
Multiplying gives \(x^2+2x-5x-10=x^2-3x-10\). Pay attention to signs.
The correct answer is A. \(x^2-3x-10\). Multiplying gives \(x^2+2x-5x-10=x^2-3x-10\). Pay attention to signs.
गुणा करने पर \(x^2+2x-5x-10=x^2-3x-10\) है। चिह्नों पर ध्यान दें।
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What is the expansion of ((3x+2)2)?
A. \(9x^2+12x+4\)
Take (a=3x) and (b=2) in ((a+b)2). The middle term is \(2\cdot3x\cdot2=12x\).
The correct answer is A. \(9x^2+12x+4\). Take (a=3x) and (b=2) in ((a+b)2). The middle term is \(2\cdot3x\cdot2=12x\).
((a+b)2) में (a=3x) और (b=2) लें। मध्य पद \(2\cdot3x\cdot2=12x\) है।
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What is the expansion of ((2x-3)2)?
A. \(4x^2-12x+9\)
Use ((a-b)2=a-2-2ab+b-2). Here (a=2x) and (b=3).
The correct answer is A. \(4x^2-12x+9\). Use ((a-b)2=a-2-2ab+b-2). Here (a=2x) and (b=3).
((a-b)2=a-2-2ab+b-2) लगाएँ। यहाँ (a=2x) और (b=3) हैं।
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What is the expansion of ((2x+3)(x-2))?
A. \(2x^2-x-6\)
By distribution, \(2x^2-4x+3x-6=2x^2-x-6\). Combine like terms at the end.
The correct answer is A. \(2x^2-x-6\). By distribution, \(2x^2-4x+3x-6=2x^2-x-6\). Combine like terms at the end.
वितरण से \(2x^2-4x+3x-6=2x^2-x-6\) है। समान पदों को अंत में जोड़ें।
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What is the expansion of ((x-4)(x+1))?
A. \(x^2-3x-4\)
Multiplying gives \(x^2+x-4x-4=x^2-3x-4\). Pay close attention to signs.
The correct answer is A. \(x^2-3x-4\). Multiplying gives \(x^2+x-4x-4=x^2-3x-4\). Pay close attention to signs.
गुणा करने पर \(x^2+x-4x-4=x^2-3x-4\) मिलता है। चिह्नों पर विशेष ध्यान दें।
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What is the expanded form of ((2x+1)(x-4)=0)?
A. \(2x^2-7x-4=0\)
Expanding gives \(2x^2-8x+x-4=0\). Combining like terms gives \(2x^2-7x-4=0\).
The correct answer is A. \(2x^2-7x-4=0\). Expanding gives \(2x^2-8x+x-4=0\). Combining like terms gives \(2x^2-7x-4=0\).
विस्तार करने पर \(2x^2-8x+x-4=0\) मिलता है। समान पद जोड़कर \(2x^2-7x-4=0\) बनता है।
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What is the expanded form of ((x+4)(x+5)=0)?
A. \(x^2+9x+20=0\)
Expanding gives \(x^2+5x+4x+20=0\). Combining like terms gives \(x^2+9x+20=0\).
The correct answer is A. \(x^2+9x+20=0\). Expanding gives \(x^2+5x+4x+20=0\). Combining like terms gives \(x^2+9x+20=0\).
विस्तार करने पर \(x^2+5x+4x+20=0\) मिलता है। समान पद जोड़कर \(x^2+9x+20=0\) बनता है।
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What is the expanded form of ((x+2)(x+3)=0)?
A. \(x^2+5x+6=0\)
Expansion gives \(x^2+3x+2x+6=0\). Combining like terms gives \(x^2+5x+6=0\).
The correct answer is A. \(x^2+5x+6=0\). Expansion gives \(x^2+3x+2x+6=0\). Combining like terms gives \(x^2+5x+6=0\).
विस्तार से \(x^2+3x+2x+6=0\) मिलता है। समान पद जोड़कर \(x^2+5x+6=0\) बनता है।
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Which factor mainly contributes to the broken nature of the Eastern Ghats?
A. प्रायद्वीपीय नदियों द्वारा कटावCutting by peninsular rivers
The Eastern Ghats are cut by rivers such as Mahanadi, Godavari, Krishna, and Cauvery. For exams, consider the Eastern Ghats a discontinuous range.
The correct answer is A. प्रायद्वीपीय नदियों द्वारा कटाव / Cutting by peninsular rivers. The Eastern Ghats are cut by rivers such as Mahanadi, Godavari, Krishna, and Cauvery. For exams, consider the Eastern Ghats a discontinuous range.
पूर्वी घाट महानदी गोदावरी कृष्णा और कावेरी जैसी नदियों से कटते हैं। परीक्षा में पूर्वी घाट को असतत पर्वत श्रृंखला मानें।
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What decisive factor separates Narmada and Tapi from east flowing Peninsular rivers?
A. दरार घाटियों से पश्चिम की ओर अरब सागर तक बहनाFlowing westward through rift valleys to the Arabian Sea
Narmada and Tapi are west flowing rivers linked with rift valleys. For exams remember them as exceptions.
The correct answer is A. दरार घाटियों से पश्चिम की ओर अरब सागर तक बहना / Flowing westward through rift valleys to the Arabian Sea. Narmada and Tapi are west flowing rivers linked with rift valleys. For exams remember them as exceptions.
नर्मदा और तापी पश्चिममुखी और दरार घाटी से जुड़ी नदियां हैं। परीक्षा में इन्हें अपवाद के रूप में याद रखें।
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Agricultural usefulness of doabs is most linked with which physical factor?
A. दो नदियों के बीच जलोढ़ मिट्टीAlluvial soil between two rivers
A doab is fertile alluvial land between two rivers. For exams take Ganga Yamuna Doab as an example.
The correct answer is A. दो नदियों के बीच जलोढ़ मिट्टी / Alluvial soil between two rivers. A doab is fertile alluvial land between two rivers. For exams take Ganga Yamuna Doab as an example.
दोआब दो नदियों के बीच की उपजाऊ जलोढ़ भूमि है। परीक्षा में गंगा यमुना दोआब को उदाहरण मानें।
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Which factor along with rock cut architecture increases Petra's historical importance?
B. नबातियन जल प्रबंधन और व्यापार मार्गNabataean water management and trade routes
Petra is an outstanding example of desert water management and trade control. For exams remember Nabataean and Jordan.
The correct answer is B. नबातियन जल प्रबंधन और व्यापार मार्ग / Nabataean water management and trade routes. Petra is an outstanding example of desert water management and trade control. For exams remember Nabataean and Jordan.
पेट्रा रेगिस्तान में जल प्रबंधन और व्यापार नियंत्रण का उत्कृष्ट उदाहरण है। परीक्षा में नबातियन और जॉर्डन याद रखें।
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Which factor was very important in opposition to the Pahlavi regime during the Iranian Revolution?
B. राजनीतिक दमन सांस्कृतिक असंतोष और धार्मिक नेतृत्व का गठबंधनCombination of political repression cultural discontent and religious leadership
The Iranian Revolution grew through many discontents and organized religious leadership. In exams, understand it as a multi-causal revolution.
The correct answer is B. राजनीतिक दमन सांस्कृतिक असंतोष और धार्मिक नेतृत्व का गठबंधन / Combination of political repression cultural discontent and religious leadership. The Iranian Revolution grew through many discontents and organized religious leadership. In exams, understand it as a multi-causal revolution.
ईरानी क्रांति कई असंतोषों और धार्मिक नेतृत्व के संगठित प्रभाव से बढ़ी। परीक्षा में इसे एक बहुकारक क्रांति के रूप में समझें।
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Why can civilian morale in the Battle of Britain be treated as a military factor?
A. क्योंकि मनोबल ने सरकार की लड़ाई जारी रखने की क्षमता को सहारा दियाBecause morale supported the government's ability to continue fighting
In modern war civilian morale becomes linked with the state's war capacity. For exams connect the home front with strategy.
The correct answer is A. क्योंकि मनोबल ने सरकार की लड़ाई जारी रखने की क्षमता को सहारा दिया / Because morale supported the government's ability to continue fighting. In modern war civilian morale becomes linked with the state's war capacity. For exams connect the home front with strategy.
आधुनिक युद्ध में नागरिक मनोबल राज्य की युद्ध क्षमता से जुड़ जाता है। परीक्षा में गृह मोर्चे को रणनीति से जोड़ें।
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Which geographical factor was important in the rise of Magadha?
A. लोहे और उपजाऊ भूमि की उपलब्धताAvailability of iron and fertile land
Magadha benefited from iron and the fertile Ganga plains. For exams, connect geography with political power.
The correct answer is A. लोहे और उपजाऊ भूमि की उपलब्धता / Availability of iron and fertile land. Magadha benefited from iron and the fertile Ganga plains. For exams, connect geography with political power.
मगध को लोहे और उपजाऊ गंगा मैदान का लाभ मिला। परीक्षा में भूगोल को राजनीतिक शक्ति से जोड़ें।
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If (p(x)=x-3-1), which of the following is a linear factor?
A. (x-1)
(x-3-1=(x-1)\(x^2+x+1\)). Therefore (x-1) is a linear factor.
The correct answer is A. (x-1). (x-3-1=(x-1)\(x^2+x+1\)). Therefore (x-1) is a linear factor.
(x-3-1=(x-1)\(x^2+x+1\)) है। इसलिए (x-1) रैखिक गुणनखंड है।
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If (p(x)=x-3-3x-2-4x+12), what is the complete factor form?
A. ((x-3)(x-2)(x+2))
By grouping, (x-2(x-3)-4(x-3)=(x-3)\(x^2-4\)). Thus the factors are ((x-3)(x-2)(x+2)).
The correct answer is A. ((x-3)(x-2)(x+2)). By grouping, (x-2(x-3)-4(x-3)=(x-3)\(x^2-4\)). Thus the factors are ((x-3)(x-2)(x+2)).
समूहीकरण से (x-2(x-3)-4(x-3)=(x-3)\(x^2-4\)) मिलता है। इसलिए गुणनखंड ((x-3)(x-2)(x+2)) हैं।
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If (p(x)=x-2+2x-24), which is a factor of (p(x))?
A. (x-4)
(x-2+2x-24=(x-4)(x+6)). Therefore (x-4) is a factor.
The correct answer is A. (x-4). (x-2+2x-24=(x-4)(x+6)). Therefore (x-4) is a factor.
(x-2+2x-24=(x-4)(x+6)) है। इसलिए (x-4) एक गुणनखंड है।
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If (p(x)=x-3+x-2-10x+8) and (x=1) is a zero, what is the remaining quadratic factor?
A. \(x^2+2x-8\)
Since (x=1) is a zero, (x-1) is a factor. Division gives \(x^2+2x-8\).
The correct answer is A. \(x^2+2x-8\). Since (x=1) is a zero, (x-1) is a factor. Division gives \(x^2+2x-8\).
(x=1) शून्यक होने से (x-1) गुणनखंड है। भाग देने पर \(x^2+2x-8\) मिलता है।
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If (x+2) is a factor of (p(x)=x-3+mx-2-4x-4), what is (m)?
A. (1)
Putting (p(-2)=0) gives (-8+4m+8-4=0). Thus (4m-4=0), so (m=1).
The correct answer is A. (1). Putting (p(-2)=0) gives (-8+4m+8-4=0). Thus (4m-4=0), so (m=1).
(p(-2)=0) रखने पर (-8+4m+8-4=0) मिलता है। इससे (4m-4=0) और (m=1) है।
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Which is a factor of \(2x^3-9x^2+13x-6\)?
A. (x-1)
(p(1)=2-9+13-6=0), so (x-1) is a factor. In option checking, try small values first.
The correct answer is A. (x-1). (p(1)=2-9+13-6=0), so (x-1) is a factor. In option checking, try small values first.
(p(1)=2-9+13-6=0) है इसलिए (x-1) गुणनखंड है। विकल्प जांच में छोटे मान पहले लगाएं।
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Which one is not a linear factor of \(x^4-1\)?
C. \(x^2+1\)
\(x^2+1\) is not linear because its degree is (2). Over real numbers, it is not a linear factor.
The correct answer is C. \(x^2+1\). \(x^2+1\) is not linear because its degree is (2). Over real numbers, it is not a linear factor.
\(x^2+1\) रैखिक नहीं है क्योंकि उसकी घात (2) है। वास्तविक संख्याओं में इसे रैखिक गुणनखंड नहीं माना जाता।
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If (x+1) is a factor of \(2x^3+kx^2-5x+2\), what is the correct value of (k)?
A. (-5)
Putting (p(-1)=0) gives (-2+k+5+2=0). Therefore, (k=-5) is correct.
The correct answer is A. (-5). Putting (p(-1)=0) gives (-2+k+5+2=0). Therefore, (k=-5) is correct.
(p(-1)=0) रखने पर (-2+k+5+2=0) मिलता है। इसलिए (k=-5) सही है।
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If (x+1) is a factor of \(2x^3+kx^2-5x+2\), what is (k)?
A. (5)
If (x+1) is a factor, then (p(-1)=0). From (-2+k+5+2=0), we get (k=-5).
The correct answer is A. (5). If (x+1) is a factor, then (p(-1)=0). From (-2+k+5+2=0), we get (k=-5).
(x+1) गुणनखंड होने पर (p(-1)=0) होगा। (-2+k+5+2=0) से (k=-5) नहीं बल्कि (k=-5) मिलता है।
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If (p(x)=x-3-3x-2-4x+12), which is a factor of (p(x))?
A. (x-2)
Since (p(2)=8-12-8+12=0), (x-2) is a factor. Use the factor theorem.
The correct answer is A. (x-2). Since (p(2)=8-12-8+12=0), (x-2) is a factor. Use the factor theorem.
(p(2)=8-12-8+12=0) है इसलिए (x-2) गुणनखंड है। गुणनखंड प्रमेय का प्रयोग करें।
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If (7) is a root of a quadratic polynomial, which factor is certain?
B. (x-7)
If the root is (r), the factor is (x-r). For (r=7), the factor is (x-7).
The correct answer is B. (x-7). If the root is (r), the factor is (x-r). For (r=7), the factor is (x-7).
मूल (r) होने पर गुणनखंड (x-r) होता है। (r=7) के लिए गुणनखंड (x-7) होगा।
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If (x-9) is a factor of a quadratic polynomial, which root is certain?
A. (9)
Solving (x-9=0) gives (x=9). Therefore the certain root is (9).
The correct answer is A. (9). Solving (x-9=0) gives (x=9). Therefore the certain root is (9).
(x-9=0) करने पर (x=9) मिलता है। इसलिए निश्चित मूल (9) है।
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If (-3) is a root of a quadratic polynomial, which factor is certain?
B. (x+3)
If the root is (r), the factor is (x-r). For (r=-3), the factor is (x+3).
The correct answer is B. (x+3). If the root is (r), the factor is (x-r). For (r=-3), the factor is (x+3).
मूल (r) होने पर गुणनखंड (x-r) होता है। (r=-3) के लिए गुणनखंड (x+3) होगा।
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If (x+5) is a factor of a quadratic polynomial, which root is certain?
B. (-5)
Solving (x+5=0) gives (x=-5). Therefore the certain root is (-5).
The correct answer is B. (-5). Solving (x+5=0) gives (x=-5). Therefore the certain root is (-5).
(x+5=0) करने पर (x=-5) मिलता है। इसलिए निश्चित मूल (-5) है।
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If (2) is a root of a quadratic polynomial then which factor is certain?
B. (x-2)
If the root is (2) the corresponding factor is (x-2). For root (r) remember the factor (x-r).
The correct answer is B. (x-2). If the root is (2) the corresponding factor is (x-2). For root (r) remember the factor (x-r).
मूल (2) होने पर संबंधित गुणनखंड (x-2) होता है। मूल (r) के लिए गुणनखंड (x-r) याद रखें।
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If (x-a) is a factor of a quadratic polynomial then what is (a)?
B. मूलRoot
If (x-a) is a factor then substituting (x=a) makes the value (0). So (a) is a root.
The correct answer is B. मूल / Root. If (x-a) is a factor then substituting (x=a) makes the value (0). So (a) is a root.
गुणनखंड (x-a) होने पर (x=a) रखने से मान (0) होता है। इसलिए (a) मूल है।
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