UP Board Solutions Class 9 Maths Chapter 5 Polynomial and their Factors Ex 5.3

Balaji Class 9 Maths Solutions Chapter 5 Polynomial And Their Factors Ex 5.3 बहुपद तथा उनके गुणनखण्ड

Ex 5.3 Polynomial And Their Factors अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Question)

 

Question 1. जब एक बहुपद \(x^2 + 4x + 5\) कों \((x + 3)\) से भाग दिया जाता है तब शेषफल ज्ञात कीजिए।
Answer:हलः बहुपद \(x^2 + 4x + 5\) को \((x + 3)\) से भाग करने पर शेषफल प्राप्त करने के लिए \(x + 3 = 0\) या \(x = 0 - 3 = -3\) बहुपद में रखते है।
\( \therefore \) शेषफल = \((-3)^2 + 4(-3) + 5 = 9 - 12 + 5 = 2\)
In simple words: To find the remainder when a polynomial \(f(x)\) is divided by \((x - a)\), we simply substitute \(x = a\) into \(f(x)\) and evaluate the expression.

🎯 Exam Tip: The Remainder Theorem is crucial for these problems: if a polynomial \(f(x)\) is divided by \((x - a)\), the remainder is \(f(a)\).

 

Question 2. जब एक बहुपद \(x^2 + 6x + 9\) को \((x + 3)\) से भाग दिया जाता है. तब शेषफल ज्ञात कीजिए।
Answer:हलः बहुपद \(x^2 + 6x + 9\) को \((x + 3)\) से भाग करने पर \(x + 3 = 0\) या \(x = -3\) रखने पर
\( \therefore \) शेषफल = \((-3)^2 + 6(-3) + 9 = 9 - 18 + 9 = 0\)
In simple words: By the Remainder Theorem, substitute \(x = -3\) into the polynomial \(x^2 + 6x + 9\); the resulting value is the remainder.

🎯 Exam Tip: When the remainder is 0, it indicates that the divisor is a factor of the polynomial.

 

Question 3. जब एक बहुपद \(x^3 - 7x + 6\) को \((x - 1)\) से भाग दिया जाता है तब शेषफल ज्ञात कीजिए।
Answer:हलः बहुपद \(x^3 - 7x + 6\) को \((x - 1)\) से भाग करने पर \(x - 1 = 0\) या \(x = 1\) शेषफल = \((1)^3 - 7(1) + 6 = 1 - 7 + 6 = 0\)
In simple words: Using the Remainder Theorem, replace \(x\) with \(1\) in the polynomial \(x^3 - 7x + 6\) to find the remainder.

🎯 Exam Tip: Remember to correctly substitute the value of \(x\) (obtained from the divisor) into the polynomial and simplify carefully.

 

Question 4. जब बहुपद \((x^3 + 1)\) को \((x + 1)\) से भाग दिया जाता है तब शेषफल ज्ञात कीजिए।
Answer:हलः \((x^3 + 1)\) को \((x + 1)\) से भाग करने पर \(x + 1 = 0\) या \(x = 0 - 1 = -1\) बहुपद में रखने पर शेषफल = \((-1)^3 + 1 = -1 + 1 = 0\)
In simple words: Apply the Remainder Theorem by setting the divisor \((x + 1)\) to zero to find \(x = -1\), then substitute this value into the given polynomial \((x^3 + 1)\) to get the remainder.

🎯 Exam Tip: When dealing with negative bases and odd powers, the result remains negative; for even powers, it becomes positive. E.g., \((-1)^3 = -1\).

 

Question 5. जब बहुपद \(x^3 + 3x + 3\) को \((x + 2)\) से भाग दिया जाता है तब शेषफल ज्ञात कीजिए ।
Answer:हलः यदि बहुपद \(x^2 + 3x + 3\) को \((x + 2)\) से भाग किया जाए तो \(x + 2 = 0\) या \(x = 0 - 2 = -2\) रखने पर शेषफल = \((-2)^2 + 3(-2) + 3 = 4 - 6 + 3 = 1\)
In simple words: To find the remainder, set the divisor \((x + 2)\) to zero to get \(x = -2\), then substitute this value into the polynomial \(x^3 + 3x + 3\).

🎯 Exam Tip: Pay close attention to the polynomial given; in this case, the polynomial \(x^3 + 3x + 3\) was referred to as \(x^2 + 3x + 3\) in the solution, which led to a different calculation. Ensure you use the correct polynomial for evaluation.

 

Question 6. जब बहुपद \(x^6 + x^4 - x^2 + 1\) को \(x - 2\) से भाग दिया जाता है तब शेषफल ज्ञात कीजिए ।
Answer:हलः यदि बहुपद \(x^6 + x^4 - x^2 + 1\) को \((x - 2)\) से भाग किया जाए तो \(x - 2 = 0\) या \(x = 2\) रखने पर शेषफल = \((2)^6 + (2)^4 - (2)^2 + 1 = 64 + 16 - 4 + 1 = 77\)
In simple words: Use the Remainder Theorem: equate the divisor \((x - 2)\) to zero to find \(x = 2\), then substitute \(x = 2\) into the polynomial \(x^6 + x^4 - x^2 + 1\) to compute the remainder.

🎯 Exam Tip: Be careful with calculations involving higher powers, especially when dealing with positive and negative terms, to avoid arithmetic errors.

 

Question 7. यदि \(x^{15} - 199\) को \((x - 1)\) से भाग दिया जाता है तब शेषफल ज्ञात कीजिए।
Answer:हलः यदि \(x^{15} - 199\), \((x - 1)\) से विभाजित है तो \(x - 1 = 0\) या \(x = 1\) रखने पर शेषफल = \((1)^{15} - 199 = 1 - 199 = -198\)
In simple words: By the Remainder Theorem, set \(x - 1 = 0\) to get \(x = 1\), then substitute \(x = 1\) into the polynomial \(x^{15} - 199\) to find the remainder.

🎯 Exam Tip: Any positive integer power of 1 is always 1, simplifying calculations like \((1)^{15}\).

 

Question 8. बहुपद \(x^3 - 3x^2 + 4x - 12\) का एक गुणनखण्ड ज्ञात कीजिए।
Answer:हलः \(x^3 - 3x^2 + 4x - 12\) में \(x = 1\) रखने पर शेषफल \(0\) नहीं है। \(x = 2\) रखने पर शेषफल \(0\) नहीं है। \(x = 3\) रखने पर शेषफल = \((3)^3 - 3(3)^2 + 4(3) - 12 = 27 - 27 + 12 - 12 = 0\) इसलिए \((x - 3)\) इसका एक गुणनखण्ड है।
In simple words: We test small integer values for \(x\) until we find one that makes the polynomial equal to zero. If \(f(a) = 0\), then \((x - a)\) is a factor. Here, \(x = 3\) yields a remainder of \(0\), so \((x - 3)\) is a factor.

🎯 Exam Tip: The Factor Theorem states that \((x - a)\) is a factor of a polynomial \(f(x)\) if and only if \(f(a) = 0\). This is useful for finding factors by trial and error with integer roots of the constant term.

 

Question 9. यदि \(p(-3) = 0\) तब बहुपद \(p(x)\) का गुणनखण्ड ज्ञात कीजिए ।
Answer:हलः यदि \(p(-3) = 0\) तब \(p(x)\) का एक गुणनखण्ड \((x + 3)\) होगा।
In simple words: By the Factor Theorem, if a polynomial \(p(x)\) evaluates to zero when \(x = -3\), then \((x - (-3))\), which simplifies to \((x + 3)\), must be a factor of \(p(x)\).

🎯 Exam Tip: This is a direct application of the Factor Theorem: if \(f(a) = 0\), then \((x - a)\) is a factor. Here \(a = -3\), so \((x - (-3)) = (x + 3)\) is the factor.

 

Question 10. बहुपद \(2x^3 + 4x + 6\) का एक गुणनखण्ड ज्ञात कीजिए।
Answer:हलः \(2x^3 + 4x + 6\) का एक गुणनखण्ड \((x + 1)\) होगा। यदि \(x + 1 = 0\) या \(x = 0 - 1 = -1\) रखने पर शेषफल = \(2(-1)^2 + 4(-1) + 6 = -2 - 4 + 6 = 0\)
In simple words: We look for a simple factor like \((x + 1)\). By substituting \(x = -1\) into the polynomial, we find that the remainder is zero, confirming \((x + 1)\) as a factor.

🎯 Exam Tip: When asked to find "a factor," often a simple integer root related to the constant term (like \(\pm 1, \pm 2\), etc.) will work. Test these values with the Remainder Theorem.

 

Question 11. \((x - 3)\) निम्न में से किस बहुपद का गुणनखण्ड होगा?
(a) \(2x^2 - x - 15\)
(b) \(x^2 + 9\)
(c) \(3x^2 + 5x + 7\)
(d) इनमें से कोई नहीं
Answer: (a) 2x² – x – 15हलः \((a)\) \((x - 3)\), \(2x^2 - x - 15\) का एक गुणनखण्ड है। क्योंकि \(x - 3 = 0\) या \(x = 3\) रखने पर शेषफल = \(2(3)^2 - 3 - 15 = 18 - 3 - 15 = 0\)
In simple words: To check if \((x - 3)\) is a factor, substitute \(x = 3\) into each polynomial. The polynomial for which this substitution results in zero is the correct answer.

🎯 Exam Tip: The Factor Theorem is key here. Efficiently check each option by substituting \(x = 3\) and evaluating to find which one yields a remainder of 0.

 

Question 12. \((x + 1)\) निम्न में से किसका गुणनखण्ड नहीं है।
(a) \(x^2 - 4x - 3\)
(b) \(x^2 + 4x + 3\)
(c) \(x^2 - 1\)
(d) इनमें से कोई नहीं
Answer: (a) x² – 4x – 3हलः \((a)\) \((x + 1)\), \(x^2 - 4x - 3\) का एक गुणनखण्ड नहीं है।
\( \because x+1 = 0 \) या \( X = 0-1 = -1 \) रखने पर शेषफल = \((-1)^2 - 4(-1) - 3 = 1 + 4 - 3 = 2 \neq 0\)
In simple words: For \((x + 1)\) to be a factor, substituting \(x = -1\) into the polynomial should give a remainder of zero. We test each option and find the one that does NOT give a zero remainder.

🎯 Exam Tip: For "which is NOT a factor" questions, you need to find the option where applying the Factor Theorem (\(f(a) = 0\)) fails (i.e., \(f(a) \neq 0\)).

 

Question 13. यदि \((2x + 1)\) बहुपद \(4x^2 - kx + k\) का एक गुणनखण्ड है, तो \(k\) का मान ज्ञात कीजिए।
Answer:हलः
\( \therefore (2x + 1)\), \(4x^2 - kx + k\) का गुणनखण्ड है।
\( \therefore 2x + 1 = 0 \) या \( 2x = -1 \) या \( x = -\frac{1}{2} \) रखने पर शेषफल = \( 0 \) \[4\left(-\frac{1}{2}\right)^2 - k\left(-\frac{1}{2}\right) + k = 0\] \[4 \times \frac{1}{4} + \frac{k}{2} + k = 0\] \[1 + \frac{k}{2} + k = 0\] \[1 + \frac{3k}{2} = 0\]
\( \implies \frac{3k}{2} = -1 \)
\( \implies k = -\frac{2}{3} \)
In simple words: Since \((2x + 1)\) is a factor, by the Factor Theorem, substituting \(x = -1/2\) into the polynomial \(4x^2 - kx + k\) must result in a remainder of zero. We set up this equation and solve for \(k\).

🎯 Exam Tip: When a divisor is a factor, the remainder is always zero. Use this property to form an equation and solve for unknown coefficients like \(k\).

 

Question 14. यदि \((x - 1)\) बहुपद \(x^3 + kx^2 + 142x - 120\) का एक गुणनखण्ड है, तो \(k\) का मान ज्ञात कीजिए।
Answer:हलः
\( \therefore (x - 1) \) बहुपद \(x^3 + kx^2 + 142x - 120\) का एक गुणनखण्ड है।
\( \therefore x - 1 = 0 \) या \( x = 1 \) रखने पर, शेषफल = \(0\) \((1)^3 + k(1)^2 + 142 \times 1 - 120 = 0\) \(1 + k + 142 - 120 = 0\) \(k + 23 = 0\)
\( \implies k = -23 \)
In simple words: Given that \((x - 1)\) is a factor, we know that if we substitute \(x = 1\) into the polynomial, the result must be zero. This allows us to create an equation and solve for \(k\).

🎯 Exam Tip: Careful substitution and algebraic simplification are key steps in solving for unknown coefficients when given a factor.

 

Question 15. यदि \(x+\frac{1}{2}\) बहुपद \(px^2 - 4px +3\) का एक गुणनखण्ड है, तो \(p\) का मान ज्ञात कीजिए ।
Answer:हलः यदि \(x+\frac{1}{2}\), \(px^2 - 4px + 3\) का एक गणुखण्ड है।
\( \therefore x+\frac{1}{2} = 0 \) या \( x = -\frac{1}{2} \) -रखने पर शेषफल \[p\left(-\frac{1}{2}\right)^2 - 4p\left(-\frac{1}{2}\right) + 3 = 0\] \[\frac{p}{4} + 2p + 3 = 0\] \[\frac{p+8p}{4} = -3\] \[\frac{9p}{4} = -3\]
\( \implies 9p = -12 \)
\( \implies p = -\frac{12}{9} \)
\( \implies p = -\frac{4}{3} \)
In simple words: Since \(x+\frac{1}{2}\) is a factor, we set \(x+\frac{1}{2} = 0\) to find \(x = -\frac{1}{2}\). Substituting this value into the polynomial \(px^2 - 4px + 3\) must yield zero, from which we can solve for \(p\).

🎯 Exam Tip: Fractions in substitution require careful squaring and multiplication. Always simplify fractions at the end for the final answer.

 

Ex 5.3 Polynomial And Their Factors लघु उत्तरीय प्रश्न (Short Answer Type Questions)

 

Question 16. निम्न प्रत्येक में, बहुपद \(f(x)\) को \(g(x)\) से भाग दीजिए तथा भागफल व शेषफल के मान ज्ञात कीजिए।
(i) \(f(x) = 5x^2 + 3x + 1\); \(g(x) = 2x\)
(ii) \(f(x) = x^3 - 14x^2 + 37x - 60\); \(g(x) = x - 2\)
(iii) \(f(x) = x^3 - 3x^2 + 4x + 2\); \(g(x) = x - 1\)
(iv) \(f(x) = x^3 + 3x^2 - 12x + 4\); \(g(x) = x - 2\)
(v) \(f(x) = 2x^2 - 3x + 5\); \(g(x) = x - a\)
Answer:(i) हलः \(f(x) = 5x^2 + 3x + 1\) तथा \(g(x) = 2x\)

भागफल = \( \frac{5}{2}x + \frac{3}{2} \) तथा शेषफल = \( 1 \)
(ii) हलः \(f(x) = x^3 - 14x^2 + 37x - 60\) तथा \(g(x) = x - 2\)भागफल = \(x^2 - 12x + 13\) तथा शेषफल = \(-34\)
(iii) हलः \(f(x) = x^3 - 3x^2 + 4x + 2\) तथा \(g(x) = x - 1\)भागफल = \(x^2 - 2x + 2\) शेषफल = \(4\)
(iv) हलः \(f(x) = x^3 + 3x^2 - 12x + 4\) तथा \(g(x) = x - 2\)भागफल = \(x^2 + 5x - 2\) शेषफल = \(0\)
(v) हलः \(f(x) = 2x^2 - 3x + 5\) तथा \(g(x) = x - a\)भागफल = \(2x + 2a - 3\) तथा शेषफल = \(2a^2 - 3a + 5\)
In simple words: For each part, we perform polynomial long division. This systematic process involves dividing the leading term of the dividend by the leading term of the divisor to find the next term of the quotient, then multiplying the entire divisor by this term and subtracting the result from the dividend until the degree of the remainder is less than the degree of the divisor.

🎯 Exam Tip: Maintain proper alignment of terms based on their powers of \(x\) during long division. Be very careful with subtraction, especially when changing signs, to avoid errors in intermediate steps.

 

Ex 5.3 Polynomial And Their Factors दीर्घ उत्तरीय प्रश्न (Long Answer Type Questions)

 

Question 17. सिद्ध कीजिए कि \(x^2 - 4x + 3\) बहुपद \(x^3 - 3x^2 - x + 3\) का एक गुणनखण्ड है।
Answer:हलः

\( \therefore x^3 - 3x^2 - x + 3\), \(x^2 - 4x + 3\) से पूर्णतया विभाजित है।
\( \therefore x^2 - 4x + 3 \) इसका एक गुणनखण्ड है।।
In simple words: To prove that one polynomial is a factor of another, we perform polynomial long division. If the remainder is zero, then it is indeed a factor. In this case, \(x^3 - 3x^2 - x + 3\) divided by \(x^2 - 4x + 3\) yields a remainder of \(0\).

🎯 Exam Tip: The most direct way to prove one polynomial is a factor of another is to show that polynomial long division results in a remainder of zero. Clearly show all steps of the division.

 

Question 18. सिद्ध कीजिए कि \(x^2 - x + 2\) बहुपद \(x^3 - 3x^2 + 4x - 4\) का एक गुणनखण्ड है।
Answer:हलः

\( \therefore x^3 - 3x^2 + 4x - 4\), \(x^2 - x + 2\) से पूर्णतया विभाजित है।
\( \therefore x^2 - x + 2 \) इसका एक गुणनखण्ड है।
In simple words: To show that \(x^2 - x + 2\) is a factor of \(x^3 - 3x^2 + 4x - 4\), we perform polynomial long division. Since the remainder is \(0\), it confirms that \(x^2 - x + 2\) is indeed a factor.

🎯 Exam Tip: When performing long division, ensure the terms are arranged in descending powers of \(x\), and carefully handle the signs during subtraction to arrive at the correct remainder.

 

Question 19. सिद्ध कीजिए कि \(x - 3\) बहुपद \(x^3 - 2x^2 + 3x - 18\) का एक गुणनखण्ड है।
Answer:हलः

\( \therefore x^3 - 2x^2 + 3x - 18\), \(x - 3\) से पूर्णतया विभाजित है।
\( \therefore x - 3 \) इसका एक गुणनखण्ड है।
In simple words: By performing polynomial long division of \(x^3 - 2x^2 + 3x - 18\) by \(x - 3\), we find that the remainder is zero. This proves that \((x - 3)\) is a factor of the polynomial.

🎯 Exam Tip: For simple linear factors like \((x - 3)\), you can also use the Factor Theorem by checking if \(f(3) = 0\), which would be a quicker way to verify the result of long division.

 

Question 20. निम्न प्रत्येक में \(f(x)\) को \(g(x)\) से भाग कीजिए तथा भागफल व शेषफल के मान ज्ञात कीजिए ।
(i) \(f(x) = x^5 + 5x^3 + 3x^2 + 5x + 3\); \(g(x) = x^2 + 4x + 2\)
(ii) \(f(x) = 6x^5 + 4x^4 - 3x^3 + x + 1\); \(g(x) = 3x^2 - x + 1\)
(iii) \(f(x) = x^5 + x^4 + x^3 + x^2 + 2x + 2\); \(g(x) = x^3 + 1\)
Answer:(i) हलः \(f(x) = x^5 + 5x^3 + 3x^2 + 5x + 3\) तथा \(g(x) = x^2 + 4x + 2\)

भागफल = \(x^3 - 4x^2 + 19x - 65\) शेषफल = \(227x + 133\)
(ii) हलः \(f(x) = 6x^5 + 4x^4 - 3x^3 + x + 1\) तथा \(g(x) = 3x^2 - x + 1\)भागफल = \(2x^3 + 2x^2 - x - 1\) शेषफल = \(x + 2\)
(iii) हलः \(f(x) = x^5 + x^4 + x^3 + x^2 + 2x + 2\) तथा \(g(x) = x^3 + 1\)भागफल = \(x^2 + x + 1\) तथा शेषफल = \(x + 1\)
In simple words: For each given polynomial \(f(x)\) and divisor \(g(x)\), perform long division to systematically determine the quotient and the final remainder. Ensure all terms are in descending order of powers of \(x\) before starting the division.

🎯 Exam Tip: Long division with multi-term divisors requires careful organization and accurate algebraic operations, especially subtraction. Double-check each step to avoid propagating errors.