UP Board Solutions Class 9 Maths Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem And Factor Theorem Ex 6.2 शेषफल प्रमेय तथा गुणनखण्ड प्रमेय

Ex 6.2 Remainder Theorem And Factor Theorem अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

Question 1. बहुपद (25x\(^2\) - 1) + (1 - 5x) का एक गुणनखण्ड ज्ञात कीजिए।
Answer: हलः
\((25x^2 - 1) + (1 - 5x)^2 = 25x^2 - 1 + 1 + 25x^2 - 10x = 50x^2 - 10x = 10x (5x - 1)\)
x बहुपद का एक गुणनखण्ड है।
In simple words: To find a factor, we simplify the given expression by expanding the squared term and combining like terms. Then, we factor out common terms to identify a binomial factor.

🎯 Exam Tip: Remember the algebraic identity \((a-b)^2 = a^2 - 2ab + b^2\) for expansion and look for common factors after simplification.

Question 2. बहुपद x\(^3\) - 6x\(^2\) + 11x - 6 के गुणनखण्ड ज्ञात कीजिए।
Answer: हलः
\(x^3 - 6x^2 + 11x - 6\) में \(x = 1\) रखने पर
शेषफल \( = (1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0\)
अतः \((x - 1)\) इसका एक गुणनखण्ड है। इसी प्रकार \((x - 2)\) व \((x - 3)\) भी इसके गुणनखण्ड हैं।
In simple words: We use the Remainder Theorem by testing integer values for \(x\) (divisors of the constant term). If substituting \(x=1\) yields a remainder of 0, then \((x-1)\) is a factor. Similarly, \((x-2)\) and \((x-3)\) are also factors as they also result in a zero remainder.

🎯 Exam Tip: For polynomials of degree 3 or higher, try integer factors of the constant term (like \(\pm 1, \pm 2, \pm 3\)) to find the first factor using the Remainder Theorem.

Question 3. यदि (x + 1) बहुपद f (x) = 2x\(^2\) + kx, का एक गुणनखण्ड है तो k का मान ज्ञात कीजिए।
Answer: हलः
यदि \((x + 1)\), \(f(x) = 2x^2 + kx\) का एक गुणनखण्ड है तो \(x + 1 = 0\)
* \(x = 0 - 1 = -1\) रखने पर \(f(-1) = 0\)
\(f(-1) = 2(-1)^2 + k(-1)\)
\(0 = 2 - k\)
\(\implies k = 2\)
In simple words: Since \((x+1)\) is a factor, substituting \(x=-1\) into the polynomial \(f(x)\) must result in a remainder of 0. By setting \(f(-1)=0\), we can solve for the unknown value \(k\).

🎯 Exam Tip: The Factor Theorem states that if \((x-a)\) is a factor of a polynomial \(f(x)\), then \(f(a)=0\). Apply this principle carefully to find unknown coefficients.

Question 4. यदि (x - 2) बहुपद 4x\(^3\) + 3x\(^2\) - 4x + k का एक गुणनखण्ड है तो k का मान ज्ञात कीजिए।
Answer: हलः
यदि \((x - 2)\), \(f(x) = 4x^3 + 3x^2 - 4x + k\) का एक गुणनखण्ड है तो \(x - 2 = 0\) या \(x = 2\) रखने पर
... \(f(2) = 0\)
\(4(2)^3 + 3(2)^2 - 4(2) + k = 0\)
\(32 + 12 - 8 + k = 0\)
\(36 + k = 0\)
\(\implies k = -36\)
In simple words: Since \((x-2)\) is a factor, we know that when \(x=2\) is substituted into the polynomial, the remainder must be zero. We set the polynomial equal to zero with \(x=2\) and solve the resulting equation for \(k\).

🎯 Exam Tip: Ensure precise calculations, especially with exponents and signs, when evaluating the polynomial after substitution to correctly find the value of the unknown.

Ex 6.2 Remainder Theorem And Factor Theorem लघु उत्तरीय प्रश्न - I (Short Answer Type Questions - I)

Question 5. a का मान ज्ञात कीजिए यदि (x + 1) बहुपद 2x\(^3\) - ax\(^2\) - (2a - 3)x + 2 का एक गुणनखण्ड है।
Answer: हलः
यदि \((x + 1)\) बहुपद \(2x^3 - ax^2 - (2a - 3)x + 2\) का एक गुणनखण्ड है तो
\(x + 1 = 0\) या \(x = -1\) रखने पर ।
शेषफल \( = 0\)
\(2(-1)^3 - a(-1)^2 - (2a - 3)(-1) + 2 = 0\)
\(-2 - a + 2a - 3 + 2 = 0\)
\(a - 3 = 0\)
\(\implies a = 3\)
In simple words: Given that \((x+1)\) is a factor, we use the Factor Theorem, which means substituting \(x=-1\) into the polynomial will yield a remainder of zero. We then solve the resulting algebraic equation to find the value of \(a\).

🎯 Exam Tip: When substituting a negative value for \(x\), be meticulous with the signs, especially with terms like \((-1)^2\) (which is 1) and \((-1)^3\) (which is -1).

Question 6. k का मान ज्ञात कीजिए, यदि (x - 3) बहुपद k\(^2\)x\(^2\) - kx - 2 का गुणनखण्ड है।
Answer: हलः
यदि \((x - 3)\), \(k^2x^2 - kx - 2\) का एक गुणनखण्ड है तो \(x - 3 = 0\) या \(x = 3\) रखने पर
शेषफल \( = 0\)
\(k^2.(3)^2 - k.3 - 2 = 0\)
\(9k^2 - 3k - 2 = 0\)
\(9k^2 - (6 - 3)k - 2 = 0\)
\(9k^2 - 6k + 3k - 2 = 0\)
\(3k(3k - 2) + 1(3k - 2) = 0\)
\((3k - 2)(3k + 1) = 0\)
यदि \(3k - 2 = 0\)
\(\implies k = \frac{2}{3}\)
यदि \(3k + 1 = 0\)
\(\implies k = -\frac{1}{3}\)
In simple words: Since \((x-3)\) is a factor, by the Factor Theorem, substituting \(x=3\) into the polynomial makes it zero. This leads to a quadratic equation in terms of \(k\), which is then factored to find the two possible values for \(k\).

🎯 Exam Tip: After setting the polynomial to zero, you might get a quadratic equation. Remember to solve it by factorization or the quadratic formula to find all possible values of the unknown.

Ex 6.2 Remainder Theorem And Factor Theorem लघु उत्तरीय प्रश्न - II (Short Answer Type Questions - II)

Remainder Theorem Calculator is a free online tool that displays the quotient and remainder of division for the given polynomial expressions.

Question 7. यदि (x - 1) बहुपद x\(^4\) - 3x\(^3\) + bx\(^2\) + 8x - 4 का एक गुणनखण्ड है, तो b का मान ज्ञात कीजिए।
Answer: हलः
यदि \((x - 1)\), \(x^4 - 3x^3 + bx^2 + 8x - 4\) का एक गुणनखण्ड है तो \(x - 1 = 0\) या \(x = 1\) रखने पर
शेषफल \( = 0\)
\((1)^4 - 3(1)^3 + b(1)^2 + 8(1) - 4 = 0\)
\(1 - 3 + b + 8 - 4 = 0\)
\(b + 2 = 0\)
\(b = 0 - 2\)
\(\implies b = -2\)
In simple words: Since \((x-1)\) is a factor, setting \(x=1\) in the polynomial makes the remainder zero. We evaluate the polynomial at \(x=1\) and solve the resulting equation to find the value of \(b\).

🎯 Exam Tip: Be meticulous with arithmetic when simplifying the expression after substitution; a small error can lead to an incorrect value of \(b\).

Question 8. सिद्ध कीजिए कि (x - 3) व (x + 4) बहुपद x\(^2\) + x - 12 के गुणनखण्ड हैं।
Answer: हलः
बहुपद \(x^2 + x - 12\) के गुणनखण्ड \((x - 3)\) तथा \((x + 4)\) होंगे।
यदि \(x - 3 = 0\) या \(x = 3\) रखने पर शेषफल \( = (3)^2 + 3 - 12 = 9 + 3 - 12 = 0\)
यदि \(x + 4 = 0\) या \(x = -4\) रखने पर शेषफल \( = (-4)^2 - 4 - 12 = 16 - 16 = 0\)
\(\therefore (x - 3)\) व \((x + 4)\) बहुपद \(x^2 + x - 12\) के गुणनखण्ड हैं।
In simple words: To prove that \((x-3)\) and \((x+4)\) are factors, we substitute \(x=3\) and \(x=-4\) respectively into the polynomial \(x^2+x-12\). If both substitutions result in a remainder of zero, then they are indeed factors.

🎯 Exam Tip: The simplest way to verify factors is to use the Remainder Theorem: if \(f(a)=0\), then \((x-a)\) is a factor. Apply this for each potential factor.

Question 9. गुणनखण्ड प्रमेय का प्रयोग करके जाँचिये कि g(x), बहुपद f(x) का गुणनखण्ड है या नहीं।
(i) f(x) = x\(^3\) - 6x\(^2\) - 19x + 84 तथा g(x) = x - 7
(ii) f(x) = x\(^3\) - 3x\(^2\) + 4x - 4 तथा g(x) = x - 2
(iii) f(x) = 3x\(^4\) + 17x\(^3\) + 9x\(^2\) - 7x - 10 तथा g(x) = x + 5
(iv) f(x) = 2x\(^3\) + 4x + 6 तथा g(x) = x + 1
Answer: हलः
(i) \(f(x) = x^3 - 6x^2 - 19x + 84\) तथा \(g(x) = x - 7\)
\(g(x) = x - 7 = 0\) या \(x = 7\) का मान \(f(x)\) में रखने पर
\(f(7) = (7)^3 - 6(7)^2 - 19(7) + 84 = 343 - 294 - 133 + 84 = 427 - 427 = 0\)
अतः \(g(x)\), \(f(x)\) का एक गुणनखण्ड है।
In simple words: For part (i), we use the Factor Theorem. Since \(g(x) = x-7\), we substitute \(x=7\) into \(f(x)\). As the result is 0, \(g(x)\) is a factor of \(f(x)\).

🎯 Exam Tip: Clearly show the substitution and calculation steps for \(f(7)\) to demonstrate that the remainder is indeed zero, thus confirming \(g(x)\) as a factor.

(ii) \(f(x) = x^3 - 3x^2 + 4x - 4\) तथा \(g(x) = x - 2\)
\(g(x) = 0\) या \(x - 2 = 0\) या \(x = 2\) रखने पर
\(f(2) = (2)^3 - 3(2)^2 + 4(2) - 4 = 8 - 12 + 8 - 4 = 16 - 16 = 0\)
अतः \(g(x)\), \(f(x)\) का एक गुणनखण्ड है।
In simple words: For part (ii), substitute \(x=2\) (from \(g(x)=x-2\)) into \(f(x)\). The calculation shows \(f(2)=0\), confirming that \(g(x)\) is a factor of \(f(x)\).

🎯 Exam Tip: Pay attention to the signs and order of operations (PEMDAS/BODMAS) when evaluating the polynomial, especially with positive substitutions.

(iii) \(f(x) = 3x^4 + 17x^3 + 9x^2 - 7x - 10\) तथा \(g(x) = x + 5\)
\(g(x) = 0\) या \(x + 5 = 0\) या \(x = -5\) रखने पर
\(f(-5) = 3(-5)^4 + 17(-5)^3 + 9(-5)^2 - 7(-5) - 10\)
\( = 3 \times 625 - 17 \times 125 + 9 \times 25 + 35 - 10\)
\( = 1875 - 2125 + 225 + 25 = 2125 - 2125 = 0\)
अतः \(g(x)\), \(f(x)\) का एक गुणनखण्ड है।
In simple words: For part (iii), substitute \(x=-5\) (from \(g(x)=x+5\)) into \(f(x)\). The lengthy calculation demonstrates that \(f(-5)=0\), thereby verifying that \(g(x)\) is a factor of \(f(x)\).

🎯 Exam Tip: When dealing with negative substitutions and higher powers, be extremely careful to correctly evaluate \((-5)^4\), \((-5)^3\), and \((-5)^2\) to avoid sign and magnitude errors.

(iv) \(f(x) = 2x^3 + 4x + 6\) तथा \(g(x) = x + 1\)
\(g(x) = 0\) या \(x + 1 = 0\) या \(x = -1\) रखने पर
\(f(-1) = 2(-1)^3 + 4(-1) + 6 = -2 - 4 + 6 = 0\)
अतः \(g(x)\), \(f(x)\) का एक गुणनखण्ड है।
In simple words: For part (iv), substitute \(x=-1\) (from \(g(x)=x+1\)) into \(f(x)\). The result, \(f(-1)=0\), confirms that \(g(x)\) is a factor of \(f(x)\).

🎯 Exam Tip: Practice quickly and accurately evaluating polynomial expressions with negative inputs, as this is a frequent step in factor theorem problems.

Question 10. सिद्ध कीजिए कि 2x\(^4\) - 6x\(^3\) + 3x\(^2\) + 3x - 2; x\(^2\) - 3x + 2 से पूर्णतया विभाजित है।
Answer: हलः
\(2x^4 - 6x^3 + 3x^2 + 3x - 2\) को \(x^2 - 3x + 2\) से भाग करने पर
*: \(x^2 - 3x + 2 = (x - 2)(x - 1)\)
यदि \(x - 2 = 0\) या \(x = 2\) रखने पर
शेषफल \( = 2(2)^4 - 6(2)^3 + 3(2)^2 + 3(2) - 2\)
\( = 32 - 48 + 12 + 6 - 2\)
\( = 50 - 50 = 0\)
\(\therefore (x - 2)\) से पूर्णतया विभाजित है।
यदि \(x - 1 = 0\) या \(x = 1\) रखने पर
शेषफल \( = 2(1)^4 - 6(1)^3 + 3(1)^2 + 3(1) - 2\)
\( = 2 - 6 + 3 + 3 - 2 = 8 - 8 = 0\)
\(\therefore (x - 1)\) से पूर्णतया विभाजित है।
In simple words: To show that a polynomial is completely divisible by a quadratic expression, first factorize the quadratic divisor. Then, use the Remainder Theorem to show that the dividend is perfectly divisible by each of the linear factors of the divisor, meaning the remainder is zero for both.

🎯 Exam Tip: Complete divisibility by a polynomial means it is divisible by all its irreducible factors. Here, checking for \((x-1)\) and \((x-2)\) is sufficient because \(x^2-3x+2=(x-1)(x-2)\).

Question 11. सिद्ध कीजिए कि (x - 1), बहुपद x\(^{10}\) - 1 तथा x\(^{11}\) - 1 का गुणनखण्ड है।
Answer: हलः
\((x - 1)\), बहुपद \(x^{10} - 1\) का गुणनखण्ड होगा।
यदि \(x - 1 = 0\) या \(x = 1\) रखने पर
\(x^{10} - 1\) का शेषफल \( = (1)^{10} - 1 = 1 - 1 = 0\)
\(\therefore (x - 1)\), \(x^{10} - 1\) का गुणनखण्ड है।
\(x^{11} - 1\) का शेषफल \( = (1)^{11} - 1 = 1 - 1 = 0\)
\(\therefore (x - 1)\), \(x^{11} - 1\) का गुणनखण्ड है।
In simple words: Using the Factor Theorem, we check if \((x-1)\) is a factor of both polynomials. By substituting \(x=1\) into \(x^{10}-1\) and \(x^{11}-1\), we find that the remainder is 0 for both, thus proving \((x-1)\) is a factor for both.

🎯 Exam Tip: Remember the general rule that \((x-a)\) is a factor of \((x^n - a^n)\) for any positive integer \(n\). In this case, \(a=1\), so \((x-1)\) is a factor of \(x^{10}-1^{10}\) and \(x^{11}-1^{11}\).

Question 12. बहुपद 4x\(^3\) + 16x\(^2\) - x + 5 से क्या घटाया जाये कि ऐसा बहुपद प्राप्त हो जो (x + 5) से पूर्णतया विभाजित हो?
Answer: हलः
यदि \((x + 5)\) से \(4x^3 + 16x^2 - x + 5\) को पूर्णतया विभाजित किया जाए तो
\(x + 5 = 0\) या \(x = 0 - 5 = -5\) रखने पर
शेषफल \( = 4(-5)^3 + 16(-5)^2 - (-5) + 5\)
\( = 4(-125) + 16(25) + 5 + 5\)
\( = -500 + 400 + 10\)
\( = -90\)
In simple words: To make the polynomial perfectly divisible by \((x+5)\), we need to subtract the remainder obtained when dividing the polynomial by \((x+5)\). We calculate this remainder by substituting \(x=-5\) into the polynomial, which gives us \(-90\). Therefore, \(-90\) must be subtracted.

🎯 Exam Tip: The quantity to be subtracted from a polynomial \(P(x)\) to make it divisible by \((x-a)\) is simply the remainder \(P(a)\).

Question 13. गुणनखण्ड प्रमेय के प्रयोग से k का मान ज्ञात कीजिए यदि (x + 2), बहुपद (x + 1)\(^7\) + (2x + k)\(^3\) का एक गुणनखण्ड है।
Answer: हलः
यदि \((x + 2)\), बहुपद \((x + 1)^7 + (2x + k)^3\) का एक गुणनखण्ड है तो
\(x + 2 = 0\) या \(x = 0 - 2 = -2\) रखने पर
शेषफल \( = 0\)
\((-2 + 1)^7 + (2 \times -2 + k)^3 = 0\)
\((-1)^7 + (-4 + k)^3 = 0\)
\(-1 + (-4 + k)^3 = 0\)
\((-4 + k)^3 = 1\)
\((-4 + k)^3 = (1)^3\)
\(-4 + k = 1\)
\(k = 1 + 4\)
\(k = 5\)
In simple words: Since \((x+2)\) is a factor, substituting \(x=-2\) into the polynomial makes the expression equal to zero. This simplifies into an equation involving \(k\), which we then solve by taking the cube root of both sides.

🎯 Exam Tip: Carefully handle powers of negative numbers; \((-1)^7 = -1\). Also, remember that if \(A^3=B^3\), then \(A=B\).

Question 14. m तथा n के मान ज्ञात कीजिए यदि (x - 1) तथा (x + 2), बहुपद 2x\(^3\) + mx\(^2\) + nx - 14 के गुणनखण्ड हैं।
Answer: हलः
यदि \((x - 1)\), बहुपद \(2x^3 + mx^2 + nx - 14\) का एक गुणनखण्ड है तो \(x - 1 = 0\) या \(x = 1\) रखने पर
\(2(1)^3 + m(1)^2 + n(1) - 14 = 0\)
\(2 + m + n - 14 = 0\)
\(m + n = 12\) .............(1)
यदि \((x + 2)\), बहुपद \(2x^3 + mx^2 + nx - 14\) का एक गुणनखण्ड है तो \(x + 2 = 0\) या \(x = -2\) रखने पर
\(2(-2)^3 + m(-2)^2 + n(-2) - 14 = 0\)
\(-16 + 4m - 2n - 14 = 0\)
\(4m - 2n = 30\) ...(2)
समी० (1) में 2 से गुणा करने पर
\(2m + 2n = 24\) ...(3)
समी० (2) व (3) को जोड़ने पर
\(6m = 54\)
\(m = \frac{54}{6} = 9\)
समीकरण (1) में \(m\) का मान रखने पर
\(9 + n = 12\)
\(n = 12 - 9 = 3\)
In simple words: Since both \((x-1)\) and \((x+2)\) are factors, substituting \(x=1\) and \(x=-2\) into the polynomial will yield zero remainders, forming two linear equations with two variables, \(m\) and \(n\). These two equations are then solved simultaneously to find the values of \(m\) and \(n\).

🎯 Exam Tip: When dealing with two unknown coefficients and two factors, always set up a system of two linear equations and solve them using methods like substitution or elimination.

Question 15. a व \(\beta\) के मान ज्ञात कीजिए यदि (x + 1) तथा (x + 2), बहुपद x\(^3\) + 3x\(^2\) - 2ax + \(\beta\) के गुणनखण्ड हैं।
Answer: हलः
यदि \((x + 1)\), \(x^3 + 3x^2 - 2ax + \beta\) का एक गुणनखण्ड है तो
\(x + 1 = 0\) या \(x = 0 - 1 = -1\) रखने पर
शेषफल \( = (-1)^3 + 3(-1)^2 - 2\alpha . (-1) + \beta = 0\)
\(-1 + 3 + 2\alpha + \beta = 0\)
\(2 + 2\alpha + \beta = 0\)
\(2\alpha + \beta = -2\) ................ .(1)
यदि \((x + 2)\), \(x^3 + 3x^2 - 2ax + \beta\) का एक गुणनखण्ड है। तो .
\(x + 2 = 0\) या \(x = 0 - 2 = -2\) रखने पर
शेषफल
\(\implies (-2)^3 + 3(-2)^2 - 2\alpha (-2) + \beta = 0\)
\(-8 + 12 + 4\alpha + \beta = 0\)
\(4\alpha + \beta = -4\) ...(2)
समी० (1) से समी० (2) घटाने पर,
\( (4\alpha + \beta) - (2\alpha + \beta) = -4 - (-2) \)
\(2\alpha = -2\)
\(\implies \alpha = \frac{-2}{2} = -1\)
समीकरण (1) से
\(2x (-1) + \beta = -2\)
\(\implies \beta = -2 + 2 = 0\)
\(\alpha = -1, \beta = 0\)
In simple words: Similar to the previous problem, using the Factor Theorem for both factors \((x+1)\) and \((x+2)\) creates two simultaneous linear equations involving \(\alpha\) and \(\beta\). Solving these equations systematically (e.g., by elimination) yields the values for \(\alpha\) and \(\beta\).

🎯 Exam Tip: Pay close attention to negative signs during substitution and algebraic manipulation, especially when subtracting equations, to avoid common errors in calculations.

Question 16. गुणनखण्ड प्रमेय का प्रयोग करके सिद्ध कीजिए कि a + b, b + c, c + a बहुपद (a + b + c)\(^3\) - (a\(^3\) + b\(^3\) + c\(^3\)) के गुणनखण्ड हैं।
Answer: हलः
** \(a + b\) एक गुणनखण्ड होगा \((a + b + c)^3 - (a^3 + b^3 + c^3)\) का
यदि \(a + b = 0\) या \(a = -b\) रखने पर,
शेषफल \( = (-b + b + c)^3 - ((-b)^3 + b^3 + c^3)\)
\( = c^3 - (-b^3 + b^3 + c^3) = c^3 - c^3\) C \( = 0\)
\(\therefore (a + b)\) इसका एक गुणनखण्ड है।.
* \(b + c\) एक गुणनखण्ड है \((a + b + c)^3 - (a^3 + b^3 + c^3)\) का
यदि \(b + c = 0\) या \(b = -c\) रखने पर,
शेषफल \( = (a - c + c)^3 - (a^3 - c^3 + c^3) = a^3 - a^3 = 0\)
\(\therefore (b + c)\) इसका एक गुणनखण्ड है।
C \(c + a\) एक गुणनखण्ड है \((a + b + c)^3 - (a^3 + b^3 + c^3)\) का।
यदि \(c + a = 0\) या \(c = -a\) रखने पर,
शेषफल \( = (a + b - a)^3 - (a^3 + b^3 - a^3) = b^3 - b^3 = 0\)
\(\therefore c + a\) इसका एक गुणनखण्ड है।
In simple words: To prove that \((a+b)\), \((b+c)\), and \((c+a)\) are factors, we apply the Factor Theorem for each. For \((a+b)\), we substitute \(a=-b\) into the polynomial. For \((b+c)\), we substitute \(b=-c\). For \((c+a)\), we substitute \(c=-a\). In each case, the remainder becomes zero, confirming they are factors.

🎯 Exam Tip: This problem demonstrates a key algebraic identity. If \(a+b+c=0\), then \(a^3+b^3+c^3 = 3abc\). Although not directly used to solve, it's a related concept often tested.

Balaji Publications Mathematics Class 9 Solutions