UP Board Solutions Class 9 Maths Chapter 5 Polynomial and their Factors Ex 5.6

Balaji Class 9 Maths Solutions Chapter 5 Polynomial And Their Factors Ex 5.6 बहुपद तथा उनके गुणनखण्ड

Ex 5.6 Polynomial And Their Factors अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Question)

Question 1. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए। x\(^2\) + 7x + 12
Answer: हल:
\(x^2 + 7x + 12 = x^2 + (3 + 4)x + 12 \)
\(12 = 3 \times 4\)
\( = x^2 + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)\)
In simple words: We factorize the quadratic expression by splitting the middle term. Here, \(7x\) is split into \(3x\) and \(4x\), such that their product \(3x \times 4x = 12x^2\) equals the product of the first and last terms, \(x^2 \times 12 = 12x^2\).

🎯 Exam Tip: Master the technique of splitting the middle term in quadratic expressions for efficient factorization.

Question 2. x\(^2\) - 14x + 48
Answer: हल: \(x^2 - 14x + 48 = x^2 - (6 + 8)x + 48 \)
\((48 = 6 \times 8)\)
\( = x^2 - 6x - 8x + 48 \)
\( = x(x - 6) - 8(x - 6) = (x - 6)(x - 8)\)
In simple words: To factorize, we split the middle term \(-14x\) into \(-6x\) and \(-8x\), whose sum is \(-14x\) and product is \(48x^2\), matching the product of the first and last terms.

🎯 Exam Tip: Always check that the product of the two split terms equals the product of the first and last terms of the quadratic.

Question 3. x\(^2\) - 7x - 18
Answer: हल: \(x^2 - 7x - 18 = x^2 - (9 - 2)x - 18 \)
\( = x^2 - 9x + 2x - 18 \)
\( = x(x - 9) + 2(x - 9) = (x - 9)(x + 2)\)
In simple words: We factorize by splitting \(-7x\) into \(-9x\) and \(2x\). Their sum is \(-7x\) and their product \(-18x^2\) equals \(x^2 \times (-18)\).

🎯 Exam Tip: Pay attention to the signs when splitting terms to ensure the correct factorization.

Question 4. x\(^2\) - 25x + 84
Answer: हल: \(x^2 - 25x + 84 = x^2 - (21 + 4)x + 84 \)
\((84 = 4 \times 21)\)
\( = x^2 - 21x - 4x + 84\)
\( = x(x - 21)- 4(x - 21)= (x - 21)(x - 4)\)
In simple words: We split \(-25x\) into \(-21x\) and \(-4x\) to factorize the quadratic. Their product is \(84x^2\), matching \(x^2 \times 84\).

🎯 Exam Tip: Practice identifying factor pairs of the constant term that sum up to the coefficient of the middle term.

Question 5. 2x\(^2\) + 7x + 6
Answer: हल: \(2x^2 + 7x + 6 = 2x^2 + (3 + 4)x + 6 \)
\((2 \times 6 = 12 \implies 12 = 3 \times 4)\)
\( = 2x^2 + 3x + 4x + 6 \)
\( = x(2x + 3) + 2(2x + 3) = (2x + 3)(x + 2)\)
In simple words: Here, the product of the first and last term is \(12x^2\). We split the middle term \(7x\) into \(3x\) and \(4x\) whose product is \(12x^2\).

🎯 Exam Tip: When the leading coefficient is not 1, multiply it by the constant term and find factors that sum to the middle term's coefficient.

Question 6. 2x\(^2\) - 13x + 15
Answer: हलः \(2x^2 - 13x + 15 = 2x^2 - (3 + 10)x + 15 \)
\((2 \times 15 = 30 \implies 30 = 3 \times 10)\)
\( = 2x^2 - 3x - 10x + 15 \)
\( = x(2x - 3) - 5(2x - 3) = (2x - 3)(x - 5)\)
In simple words: The product of first and last terms is \(30x^2\). We split \(-13x\) into \(-3x\) and \(-10x\), whose product is \(30x^2\).

🎯 Exam Tip: Remember to factor out common terms from each pair after splitting the middle term.

Question 7. 3x\(^2\) - 14x + 8
Answer: हल: \(3x^2 - 14x + 8 = 3x^2 - (2 + 12)x + 8 \)
\((3 \times 8 = 24 \implies 24 = 12 \times 2)\)
\( = 3x^2 - 2x - 12x + 8\)
\( = x(3x - 2) - 4(3x - 2) = (3x - 2)(x - 4)\)
In simple words: We split \(-14x\) into \(-2x\) and \(-12x\). Their sum is \(-14x\) and product is \(24x^2\), which matches the product of \(3x^2\) and \(8\).

🎯 Exam Tip: Correctly identifying the factors for splitting the middle term is key for accurate factorization.

Question 8. 3x\(^2\) + 10x - 8
Answer: हलः \(3x^2 + 10x - 8 = 3x^2 + (12 - 2)x - 8 \)
\((3 \times 8 = 24 \implies 24 = 2 \times 12)\)
\( = 3x^2 + 12x - 2x - 8 \)
\( = 3x(x + 4) - 2(x + 4)= (x + 4)(3x - 2)\)
In simple words: We split \(10x\) into \(12x\) and \(-2x\) so their product \(-24x^2\) equals the product of \(3x^2\) and \(-8\).

🎯 Exam Tip: When the constant term is negative, the split terms will have opposite signs. Ensure the larger number takes the sign of the middle term.

Ex 5.6 Polynomial And Their Factors लघु उत्तरीय प्रश्न - I (Short Answer Type Questions - I)

Question 9. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) 8(x + 2)\(^2\) + 2(x + 2) - 15
(ii) 12(x\(^2\) + 7x)\(^2\) - 8(x\(^2\) + 7x)(2x - 1) - 15(2x - 1)\(^2\)
(iii) (x\(^2\) - 2x)\(^2\) - 23(x\(^2\) - 2x) + 120

(iv) (x + 2y)\(^2\) + 5(x + 2y)(2x + y) + 6(2x + y)\(^2\)
Answer: हल:
(i) 8(x + 2)\(^2\) + 2(x + 2) - 15
माना \(x + 2 = y\)
\( = 8y^2 + 2y - 15\)
\( = 8y^2 +(12 - 10)y - 15 \)
\((8 \times 15 = 120 \implies 120 = 12 \times 10)\)
\( = 8y^2 + 12y - 10y - 15 \)
\( = 4y(2y + 3) - 5(2y + 3) \)
\( = (2y + 3)(4y - 5)\)
\( = [2(x + 2) + 3] [4(x + 2) - 5]\)
\( = [2x + 4 + 3][4x + 8 - 5)\)
\( = (2x + 7) (4x + 3)\)
(ii) 12(x\(^2\) + 7x)\(^2\) - 8(x\(^2\) + 7x)(2x - 1) - 15(2x - 1)\(^2\)
हल:
माना \(x^2 + 7x = y\) तथा \(2x - 1 = z\)
\( = 12y^2 - 8yz - 15z^2\)
\( = 12y^2 - (18 - 10)yz - 15z^2 \)
\((12 \times 15 = 180 \implies 180 = 2 \times 2 \times 3 \times 3 \times 5)\)
\( = 12y^2 - 18yz + 10yz - 15z^2 \)
\( = 6y (2y - 3z) + 5z(2y - 3z) \)
\( = (2y - 3z)(6y + 5z) \)
\( = [2(x^2 + 7x) - 3(2x-1)][6(x^2 + 7x) + 5(2x - 1)] \)
\( = [2x^2 + 14x - 6x + 3] [6x^2 + 42x + 10x - 5] \)
\( = [2x^2 + 8x + 3][6x^2 + 52x - 5]\)
(iii) (x\(^2\) - 2x)\(^2\) - 23(x\(^2\) - 2x) + 120
हल:
माना \(x^2 - 2x = y\)
\( = y^2 - 23y + 120\)
\( = y^2 - (8 + 15)y + 120 \)
\((120 = 8 \times 15)\)
\( = y^2 - 8y - 15y + 120 \)
\( = y(y - 8) - 15(y - 8) \)
\( = (y - 15)(y - 8) \)
\( = (x^2 - 2x - 15)(x^2 - 2x - 8)\)
\((15 = 3 \times 5 \text{ व } 8 = 4 \times 2)\)
\( = [x^2 - (5 - 3)x - 15][x^2 - (4 - 2)x - 8]\)
\( = [x^2 - 5x + 3x - 15][x^2 - 4x + 2x - 8]\)
\( =[x(x - 5) + 3(x - 5)][x(x - 4) + 2(x - 4)] \)
\( = [(x + 3)(x - 5)][(x - 4)(x + 2)]\)
(iv) (x + 2y)\(^2\) + 5(x + 2y)(2x + y) + 6(2x + y)\(^2\)
हल:
माना \(x + 2y = m\) तथा \(2x + y = n\)
\( = m^2 + 5mn + 6n^2\)
\( = m^2 +(2 + 3)mn +6n^2\)
\( = m^2 + 2mn + 3mn + 6n^2\)
\( = m(m + 2n) + 3n(m + 2n)\)
\( = (m + 2n)(m + 3n)\)
यहाँ \([x + 2y + 2(2x + y)][x + 2y + 3(2x + y)]\)
\( =[x + 2y + 4x + 2y][x + 2y + 6x + 3y]\)
\( = [5x + 4y][7x + 5y]\)
In simple words: We simplify complex expressions by using substitution to reduce them to standard quadratic form, then factorize and substitute back the original terms.

🎯 Exam Tip: Substitution of variables (e.g., \(x+2=y\)) is a powerful technique for simplifying complex factorization problems. Always substitute back the original variables at the end.

Question 10. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) \( \frac{1}{3} x^{2} - 2x - 9 \)
(ii) \( \frac{1}{4} x^{2} + x - 3 \)
(iii) \(8x^3 - 2x^2y - 15xy^2\)
(iv) \(9x^3y +41x^2y^2 + 20xy^3\)
Answer: हल:
(i) \( \frac{1}{3} x^2 - 2x - 9 \)
\[ = \frac{x^2 - 6x - 27}{3} \] \[ = \frac{1}{3} [x^2 - 6x - 27] \] \[ = \frac{1}{3} [x^2 - (9 - 3)x - 27] \] \[ = \frac{1}{3} [x^2 - 9x + 3x - 27] \] \[ = \frac{1}{3} [x(x - 9) + 3(x - 9)] \] \[ = \frac{1}{3} [(x - 9)(x + 3)] \] (ii) \( \frac{1}{4} x^2 + x - 3 \)
\[ = \frac{x^2 + 4x - 12}{4} \] \[ = \frac{1}{4} [x^2 + 4x - 12] \] \[ = \frac{1}{4} [x^2 + (6 - 2)x - 12] \] \[ = \frac{1}{4} [x^2 + 6x - 2x - 12] \] \[ = \frac{1}{4} [x(x + 6) - 2(x + 6)] \] \[ = \frac{1}{4} [(x + 6)(x - 2)] \] (iii) \(8x^3 - 2x^2y - 15xy^2\)
\( = x[8x^2 - 2xy - 15y^2]\)
\( = x[8x^2 - (12 - 10)xy - 15y^2] \)
\( (8 \times 15 = 120 \implies 120 = 12 \times 10) \)
\( = x[8x^2 - 12xy + 10xy - 15y^2] \)
\( = x[4x(2x - 3y) + 5y(2x - 3y)] \)
\( = x(2x - 3y)(4x + 5y)\)
(iv) \(9x^3y + 41x^2y^2 + 20xy^3\)
\( = xy[9x^2 + 41xy + 20y^2]\)
\( = xy[9x^2 + (36 + 5)xy + 20y^2] \)
\( (9 \times 20 = 180 \implies 180 = 2 \times 2 \times 3 \times 3 \times 5)\)
\( = xy[9x^2 + 36xy + 5xy + 20y^2]\)
\( = xy[9x(x + 4y) + 5y (x + 4y)] \)
\( = xy(9x + 5y)(x + 4y)\)
In simple words: For expressions with fractional coefficients or multiple variables, first simplify by finding a common denominator or factoring out common variables, then apply the middle term splitting method.

🎯 Exam Tip: When dealing with fractional coefficients, factor out the fraction or find a common denominator before applying factorization techniques to the integer part of the polynomial.

Question 11. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) \(x^2 + 4x - 21\)
(ii) \(x^2 - 7x + 12\)
(iii) \(x^2 - 21x + 108\)
(iv) \(x^2 + 5x - 36\)
Answer: हल:
(i) \(x^2 + 4x - 21 = x^2 + (7 - 3)x - 21 \)
\((21 = 3 \times 7)\)
\( = x^2 + 7x - 3x - 21 \)
\( = x(x + 7) - 3(x + 7) = (x + 7)(x - 3)\)
(ii) \(x^2 - 7x + 12 = x^2 - (3 + 4)x + 12 \)
\((12 = 2 \times 2 \times 3)\)
\( = x^2 - 3x - 4x + 12 \)
\( = x(x - 3) - 4(x - 3) = (x - 3)(x - 4)\)
(iii) \(x^2 - 21x + 108 = x^2 - (12 + 9)x + 108 \)
\((108 = 2 \times 2 \times 3 \times 3 \times 3 = 12 \times 9)\)
\( = x^2 - 12x - 9x + 108 \)
\( = x(x - 12)- 9(x - 12) = (x - 12)(x - 9)\)
(iv) \(x^2 + 5x - 36 = x^2 + (9 - 4)x - 36 \)
\((36 = 2 \times 2 \times 3 \times 3)\)
\( = x^2 + 9x - 4x - 36 \)
\( = x(x + 9)- 4(x + 9) = (x + 9)(x - 4)\)
In simple words: These are standard quadratic factorizations where the middle term is split based on the factors of the constant term that sum up to the middle term's coefficient.

🎯 Exam Tip: Always look for factor pairs of the constant term that either add up or subtract to the middle term's coefficient.

Question 12. निम्न व्यंजकों के गुणनखण्ड इनके मध्य पद को विभक्त करके कीजिए
(i) \(x^4 + 3x^2 - 28\)
(ii) \(x^4 - 5x^2 + 4\)
Answer: हल:
(i) \(x^4 + 3x^2 - 28 = x^2 + (7 - 4)x^2 - 28 \)
\( (28 = 2 \times 2 \times 7)\)
\( = x^4 + 7x^2 - 4x^2 - 28 \)
\( = x^2(x^2 + 7) - 4(x^2 + 7)\)
\( = (x^2 + 7)(x^2 - 4) \)
\( = (x^2 + 7)[(x)^2 - (2)^2] \)
\( = (x^2 + 7)(x + 2)(x - 2)\)
(ii) \(x^4 - 5x^2 + 4 = x^4 - (1 + 4)x^2 + 4 \)
\( (: 4 = 1 \times 4)\)
\( = x^4 - x^2 - 4x^2 + 4 \)
\( = x^2(x^2 - 1) - 4(x^2 - 1) \)
\( = (x^2 - 1)(x^2 - 4)\)
\( = [(x)^2 - (1)^2][(x)^2 - (2)^2] \)
\( = (x + 1)(x - 1)(x + 2)(x - 2)\)
In simple words: These are biquadratic expressions that can be factorized by treating \(x^2\) as a single variable and applying the middle term splitting method, followed by further factorization using algebraic identities.

🎯 Exam Tip: Recognize biquadratic expressions (like \(ax^4 + bx^2 + c\)) and solve them by letting \(y = x^2\), factorizing the quadratic in \(y\), and then substituting back.

Question 13. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) \(x^2 + 4 \sqrt{2} x + 6\)
(ii) \(x^2 + 5 \sqrt{3} x + 12\)
(iii) \(x^2 + 5 \sqrt{5} x + 30\)
(iv) \(x^2 + 6 \sqrt{6} x + 48\)
Answer: हल:
(i) \(x^2 + 4\sqrt{2}x + 6 = x^2 + (3\sqrt{2} + \sqrt{2})x + 6\)
\( (6 = 2 \times 3 = \sqrt{2} \times \sqrt{2} \times 3)\)
\( = x^2 + 3\sqrt{2}x + \sqrt{2}x+\sqrt{2}x\sqrt{2}x3\)
\( = x(x+3\sqrt{2}) + \sqrt{2}(x+3\sqrt{2})\)
\( = (x+3\sqrt{2})(x+\sqrt{2})\)
(ii) \(x^2 + 5\sqrt{3}x + 12 = x^2 + (4\sqrt{3}+\sqrt{3})x+12\)
\( (12 = 2 \times 2 \times 3 = 2 \times 2 \times \sqrt{3} \times \sqrt{3})\)
\( = x^2 + 4\sqrt{3}x + \sqrt{3}x+2 \times 2 \times \sqrt{3} \times \sqrt{3}\)
\( = x(x + 4\sqrt{3}) + \sqrt{3}(x+4\sqrt{3})\)
\( = (x+4\sqrt{3})(x+\sqrt{3})\)
(iii) \(x^2 + 5\sqrt{5}x + 30 = x^2 + (2\sqrt{5} + 3\sqrt{5})x + 30\)
\( (30 = 2 \times 3 \times 5 = 2 \times 3 \times \sqrt{5} \times \sqrt{5})\)
\( = x^2 + 2\sqrt{5}x + 3\sqrt{5}x+2 \times 3 \times \sqrt{5} \times \sqrt{5}\)
\( = x(x + 2\sqrt{5}) + 3\sqrt{5}(x + 2\sqrt{5})\)
\( =(x+2\sqrt{5})(x+3\sqrt{5})\)
(iv) \(x^2 + 6\sqrt{6}x + 48\)
\( = x^2 + (2\sqrt{6}+4\sqrt{6})x + 48\)
\( (48 = 2 \times 2 \times 2 \times \sqrt{6} \times \sqrt{6})\)
\( = x^2 + 2\sqrt{6}x + 4\sqrt{6}x+2 \times 2 \times 2 \times \sqrt{6} \times \sqrt{6}\)
\( = x(x + 2\sqrt{6}) + 4\sqrt{6}(x + 2\sqrt{6}) = (x+2\sqrt{6})(x+4\sqrt{6})\)
In simple words: When expressions involve square roots, split the middle term by finding factors of the constant term that, when combined with the square root terms, sum up to the middle term's coefficient.

🎯 Exam Tip: To handle square root terms effectively, express the constant term as a product of terms involving the square root, then find suitable factors for splitting the middle term.

Question 14. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) \( \left(5 x-\frac{1}{x}\right)^{2}+5\left(5 x-\frac{1}{x}\right)+6 \)
(ii) \((p + q)^2 - 20(p + q) - 125\)
(iii) \((a^2 - a)^2 - 8(a^2 - a) + 12\)
(iv) \((x^2 - 4x)(x^2 - 4x - 1) - 20\)
(v) \((x^2 + x)^2 + 4(x^2 + x) - 12\)
(vi) \((3x - 4)^2 - (3x - 4) - 42\)
Answer: हल:
(i) \( \left(5x - \frac{1}{x}\right)^2 + 5\left(5x - \frac{1}{x}\right) + 6 \)
माना \(5x - \frac{1}{x} = y\)
\( \implies y^2 + 5y + 6 \)
\( = y^2 + (2+3)y +6\)
\( = y^2 + 2y + 3y +6\)
\( = y(y+2) + 3(y+2)\)
\( = (y + 2)(y + 3)\)
\( = \left(5x - \frac{1}{x} + 2\right)\left(5x - \frac{1}{x} + 3\right)\)
(ii) \((p + q)^2 - 20(p + q) - 125\)
माना \(P + q = x\)
\( = x^2 - 20x - 125\)
\( = x^2 - (25 - 5)x - 125\)
\( = x^2 - 25x + 5x - 125\)
\( = x(x - 25) + 5(x - 25)\)
\( = (x + 5)(x - 25)\)
\( = (p + q - 25)(p + q + 5)\)
(iii) \((a^2 - a)^2 - 8(a^2 - a) + 12\)
माना \(a^2 - a = x\)
\( \implies x^2 - 8x + 12\)
\( = x^2 - (2 + 6)x + 12\)
\( = x^2 - 2x - 6x +12\)
\( = x(x - 2) - 6(x - 2) = (x - 2)(x - 6)\)
\( = (a^2 - a - 2)(a^2 - a - 6)\)
(iv) \((x^2 - 4x)(x^2 - 4x - 1) - 20\)
माना \(x^2 - 4x = y\)
\( \implies y (y - 1) - 20\)
\( = y^2 - y - 20\)
\( = y^2 - (5 - 4)y - 20\)
\( = y^2 - 5y + 4y - 20\)
\( = y(y - 5) + 4(y - 5)\)
\( = (y - 5)(y + 4)\)
\( = (x^2 - 4x - 5)(x^2 - 4x + 4)\)
\( = [x^2 - (5 - 1)x - 5][x^2 - (2 + 2)x + 4]\)
\( = [x^2 - 5x + x - 5][x^2 - 2x - 2x + 4]\)
\( = [x(x - 5) + x - 5)][x(x - 2) - 2(x - 2)]\)
\( = (x - 5) (x + 1) (x - 2) (x - 2) \)
\( = (x - 5)(x + 1)(x - 2)^2\)
(v) \((x^2 + x)^2 + 4(x^2 + x) - 12\)
माना \(x^2 + x = y\)
\( = y^2 + 4y - 12\)
\( = y^2 + (6 - 2)y - 12\)
\( = y^2 + 6y - 2y -12\)
\( = y(y + 6) - 2(y + 6)\)
\( = (y + 6)(y - 2)\)
\( = (x^2 + x + 6) (x^2 + x - 2)\)
\( = (x^2 + x + 6)[x^2 + (2 - 1)x - 2]\)
\( = (x^2 + x + 6)[x^2 + 2x - x - 2]\)
\( = (x^2 + x + 6)[x(x + 2) - 1(x + 2)]\)
\( = (x^2 + x + 6)(x - 1)(x + 2)\)
(vi) \((3x - 4)^2 - (3x - 4) - 42\)
माना \(3x - 4 = y\)
\( \implies y^2 - y - 42 \)
\( (42 = 2 \times 3 \times 7 = 6 \times 7)\)
\( = y^2 - (7 - 6)y - 42\)
\( = y^2 - 7y + 6y - 42\)
\( = y(y - 7) + 6(y - 7)\)
\( = (y - 7)(y + 6)\)
\( = (3x - 4 - 7)(3x - 4 + 6)\)
\( = (3x - 11)(3x + 2)\)
In simple words: These problems demonstrate factorization by substitution, where a common part of the expression is replaced by a temporary variable to simplify the factoring process.

🎯 Exam Tip: Always look for opportunities to simplify complex-looking expressions using substitution to reduce them to familiar quadratic forms, making factorization straightforward.

Ex 5.6 Polynomial And Their Factors लघु उत्तरीय प्रश्न - II (Short Answer Type Questions - II)

Question 15. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) \(x^4 - x^2 - 12\)
(ii) \(m^8 - 11m^4n^4 - 80n^8\)
Answer: हल:
(i) \(x^4 - x^2 - 12\)
\( = x^4 - (4 - 3)x^2 - 12\)
\( = x^4 - 4x^2 + 3x^2 - 12\)
\( = x^2(x^2 - 4) + 3(x^2 - 4)\)
\( = (x^2 - 4)(x^2 + 3)\)
\( = (x)^2 - (2)^2(x^2 + 3)\)
\( = (x + 2)(x - 2)(x^2 + 3)\)
(ii) \(m^8 - 11m^4n^4 - 80n^8\)
\( (80 = 2 \times 2 \times 2 \times 2 \times 5)\)
\( = m^8 - (16 - 5)m^4n^4 - 80n^8\)
\( = m^8 - 16m^4n^4 + 5m^4n^4 - 80n^8\)
\( = m^4 (m^4 - 16n^4) + 5n^4(m^4 - 16n^4)\)
\( = (m^4 + 5n^4)(m^4 - 16n^4)\)
\( = (m^4 + 5n^4)(m^2 + 4n^2)(m^2 - 4n^2)\)
\( = (m^4 + 5n^4)(m^2 + 4n)(m + 2n)(m - 2n)\)
In simple words: This problem shows successive factorization, where after the initial splitting of the middle term, the resulting factors can be further factorized using algebraic identities like \(a^2 - b^2 = (a-b)(a+b)\).

🎯 Exam Tip: Always inspect factors obtained after initial factorization for any further simplification using algebraic identities, especially for higher degree polynomials.

Question 16. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) \(2x^2 + 13x + 20\)
(ii) \(6x^2 + 11x + 3\)
(iii) \(9x^2 + 27x + 20\)
(iv) \(2x^2 - 11x - 63\)
(v) \(10x^2 - 9x - 7\)
(vi) \(21x^2 + 5x - 6\)
Answer: हल:
(i) \(2x^2 + 13x + 20 \)
\((::: 2 \times 20 = 2 \times 2 \times 2 \times 5)\)
\( = 2x^2 + (5 + 8)x + 20\)
\( = 2x^2 + 5x + 8x + 20\)
\( = x(2x + 5) + 4(2x + 5)\)
\( = (2x + 5)(x + 4)\)
(ii) \(6x^2 + 11x +3 \)
\((::: 6 \times 3 = 18 = 2 \times 3 \times 3)\)
\( = 6x^2 + (2 + 9)x + 3\)
\( = 6x^2 + 2x + 9x + 3\)
\( = 2x(3x + 1) + 3(3x + 1)\)
\( = (3x + 1)(2x +3)\)
(iii) \(9x^2 + 27x + 20 \)
\((::: 9 \times 20 = 3 \times 3 \times 2 \times 2 \times 5)\)
\( = 9x^2 + (12 + 15)x + 20\)
\( = 9x^2 + 12x + 15x + 20\)
\( = 3x(3x + 4) + 5(3x + 4)\)
\( = (3x + 4)(3x + 5)\)
(iv) \(2x^2 - 11x - 63 \)
\((::: 2 \times 63 = 2 \times 3 \times 3 \times 7)\)
\( = 2x^2 - (18 - 7)x - 63\)
\( = 2x^2 - 18x + 7x - 63\)
\( = 2x(x - 9) + 7(x - 9)\)
\( = (x - 9)(2x + 7)\)
(v) \(10x^2 - 9x - 7 \)
\((::: 10 \times 7 = 2 \times 5 \times 7)\)
\( = 10x^2 - (14 - 5)x - 7\)
\( = 10x^2 - 14x + 5x -7\)
\( = 2x(5x - 7) + 1(5x - 7)\)
\( = (5x - 7)(2x + 1)\)
(vi) \(21x^2 + 5x - 6 \)
\((::: 21 \times 6 = 3 \times 7 \times 2 \times 3)\)
\( = 21x^2 + (14 - 9)x - 6\)
\( = 21x^2 + 14x - 9x - 6\)
\( = 7x (3x + 2) - 3(3x + 2)\)
\( = (3x + 2)(7x - 3)\)
In simple words: This series of problems reinforces the middle-term splitting method for factorizing quadratic polynomials with varying coefficients and signs.

🎯 Exam Tip: Practice factorizing quadratics extensively, especially those with non-unity leading coefficients. Careful sign management is crucial for correct factorization.

Question 17. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) \( \frac{1}{2} x^{2} + 4x + 6 \)
(ii) \( 2x^2 - x + \frac{1}{8} \)
Answer: हल:
(i) \( \frac{1}{2} x^2 + 4x + 6 \)
\[ = \frac{x^2 + 8x + 12}{2} \] \[ = \frac{1}{2} [x^2 + 8x + 12] \] \( (12 = 2 \times 2 \times 3)\)
\[ = \frac{1}{2} [x^2 + (2+6)x + 12] \] \[ = \frac{1}{2} [x^2 + 2x + 6x + 12] \] \[ = \frac{1}{2} [x(x + 2) + 6(x + 2)] \] \[ = \frac{1}{2} (x + 2)(x + 6) \] \[ = \left( \frac{x}{2} + 1 \right)(x + 6) \] (ii) \( 2x^2 - x + \frac{1}{8} \)
\[ = \frac{16x^2 - 8x + 1}{8} \] \[ = \frac{1}{8} [16x^2 - 8x + 1] \] \[ = \frac{1}{8} [(4x)^2 - 2 \times 4x \times 1 + (1)^2] \] \[ = \frac{1}{8} [4x - 1]^2 \] \[ = \frac{1}{8} (4x - 1)(4x - 1) \] \[ = \left( \frac{x}{2} - \frac{1}{8} \right) (4x - 1) \]In simple words: For expressions with fractions, first remove the fractional part by taking a common factor or finding a common denominator, then factorize the resulting integer polynomial.

🎯 Exam Tip: When dealing with fractional coefficients, convert the entire expression to a common denominator and factor out the fractional part to simplify the factorization process.

Question 18. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) \(7(x + 2y)^2 - 25(x + 2y) + 12\)
(ii) \(8(a + 1)^2 +2(a + 1)(b + 2) - 15(b + 2)^2\)
(iii) \(12(x^2 + 7x)^2 - 8(x^2 + 7x)(2x - 1) + (2x - 1)^2\)
(iv) \(2(y^2 + 2y)^2 - 5(y^2 + 2y) + 3\)
(v) \(6(x^2 + 4x)^2 - 11(x^2 + 4x)- 10\)
Answer: हल:
(i) \(7(x + 2y)^2 - 25(x + 2y) + 12\)
माना \(x + 2y = z\)
\( = 7z^2-25z + 12\)
\( = 7z^2 - (21 + 4)z + 12 \)
\( (7 \times 12 = 7 \times 2 \times 2 \times 3)\)
\( =7z^2 - 21z - 4z + 12\)
\( = 7z(z - 3) - 4(z - 3)\)
\( = (z - 3)(7z - 4)\)
\( =(x + 2y - 3)[7(x + 2y) - 4]\)
\( = (x + 2y - 3)(7x + 14y - 4)\)
(ii) \(8(a + 1)^2 + 2(a + 1)(b + 2) - 15(b + 2)^2\)
माना \((a + 1) = x\) तथा \((b + 2) = y\)
\( = 8x^2 + 2xy - 15y^2\)
\( = 8x^2 +(12 - 10)xy - 15y^2 \)
\( (15 \times 8 = 120 = 2 \times 2 \times 2 \times 3 \times 5)\)
\( = 8x^2 + 12xy - 10xy - 15y^2\)
\( = 4x(2x + 3y) - 5y (2x + 3y)\)
\( = (2x + 3y)(4x - 5y)\)
\( \implies [2(a + 1) + 3(b + 2)][4(a + 1) - 5(b + 2)]\)
\( = [2a + 2+ 3b + 6][4a + 4 - 5b - 10]\)
\( = [2a + 3b + 8][4a - 5b - 6]\)
(iii) \(12(x^2 + 7x)^2 - 8(x^2 + 7x)(2x - 1) + (2x - 1)^2\)
माना \(x^2 + 7x = m\) तथा \(2x - 1 = n\)
\( = 12m^2 - 8mn +n^2\)
\( (12 \times 1 = 12 = 2 \times 2 \times 3)\)
\( = 12m^2 - 6mn - 2mn + n^2\)
\( = 6m(2m - n) - n(2m - n)\)
\( = (6m - n)(2m - n)\)
\( = [6(x^2 + 7x) - (2x + 1)][2(x^2 + 7x) - (2x + 1)]\)
\( = (6x^2 + 42x - 2x + 1)(2x^2 + 14x - 2x + 1)\)
\( = (6x^2 + 40x + 1)(2x^2 + 12x + 1)\)
(iv) \(2(y^2 + 2y)^2 - 5(y^2 + 2y) + 3\)
माना \(y^2 + 2y = m\)
\( = 2m^2 - 5m + 3\)
\( = 2m^2 - (2 + 3)m + 3 \)
\( (2 \times 3 = 6)\)
\( = 2m^2 - 2m - 3m + 3\)
\( = 2m(m - 1) - 3(m - 1)\)
\( = [(2m - 3)(m - 1)]\)
\( = [2(y^2 + 2y) - 3][y^2 + 2y - 1]\)
\( =[2y^2 + 4y - 3][y^2 + 2y - 1]\)
(v) \(6(x^2 + 4x)^2 - 11(x^2 + 4x) - 10\)
माना \(x^2 + 4x = m\)
\( = 6m^2 - 11m - 10\)
\( = 6m^2 -(15 - 4)m - 10 \)
\( (10 \times 6 = 60 = 2 \times 2 \times 3 \times 5)\)
\( = 6m^2 - 15m + 4m - 10\)
\( = 3m(2m - 5) + 2(2m - 5)\)
\( = (2m - 5)(3m + 2)\)
\( = [2(x^2 + 4x) - 5][3(x^2 + 4x) + 2]\)
\( = [2x^2 + 8x - 5][3x^2 + 12x + 2]\)
In simple words: These problems are solved by substituting common algebraic expressions with temporary variables to simplify the polynomial into a factorable quadratic form, and then back-substituting the original expressions.

🎯 Exam Tip: When an expression involves multiple terms or functions of variables, strategic substitution can often transform it into a standard quadratic or biquadratic form, making factorization easier.

Question 19. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) \( 2\left(5x - \frac{1}{x}\right)^2 - 3\left(5x - \frac{1}{x}\right) - 2 \)
(ii) \( 6\left(2x - \frac{3}{x}\right)^2 + 7\left(2x - \frac{3}{x}\right) - 20 \)
(iii) \( 12\left(4x - \frac{1}{4x}\right)^2 + 5\left(4x - \frac{1}{4x}\right) - 2 \)
(iv) \( 10\left(3x + \frac{1}{x}\right)^2 - \left(3x + \frac{1}{x}\right) - 3 \)
Answer: हल:
(i) \( 2\left(5x - \frac{1}{x}\right)^2 - 3\left(5x - \frac{1}{x}\right) - 2 \)
माना \(5x - \frac{1}{x} = y\)
\( = 2y^2-3y-2\)
\( = 2y^2-(4-1)y-2\)
\( = 2y^2-4y+y-2\)
\( = 2y(y-2)+1(y-2)\)
\( = (y-2) (2y + 1)\)
\( = \left(5x - \frac{1}{x} - 2\right)\left(2\left(5x - \frac{1}{x}\right) + 1\right)\)
\( = \left(5x - \frac{1}{x} - 2\right)\left(10x - \frac{2}{x} + 1\right)\)
(ii) \( 6\left(2x - \frac{3}{x}\right)^2 + 7\left(2x - \frac{3}{x}\right) - 20 \)
माना \(2x - \frac{3}{x} = y\)
\( = 6y^2+7y-20\)
\( = 6y^2 + (15-8)y - 20\)
\( = 6y^2+15y-8y - 20 \)
\( (20 \times 6 = 120 = 2 \times 2 \times 2 \times 3 \times 5)\)
\( = 3y(2y + 5)-4(2y + 5)\)
\( = (2y + 5)(3y - 4)\)
\( = \left(2\left(2x - \frac{3}{x}\right) + 5\right)\left(3\left(2x - \frac{3}{x}\right) - 4\right)\)
\( = \left(4x - \frac{6}{x} + 5\right)\left(6x - \frac{9}{x} - 4\right)\)
(iii) \( 12\left(4x - \frac{1}{4x}\right)^2 + 5\left(4x - \frac{1}{4x}\right) - 2 \)
माना \(4x - \frac{1}{4x} = y\)
\( = 12y^2 + 5y - 2\)
\( = 12y^2 + (8-3)y - 2 \)
\( (12 \times 2 = 24 = 2 \times 2 \times 2 \times 3)\)
\( = 12y^2 + 8y - 3y - 2\)
\( = 4y(3y + 2) -1(3y + 2)\)
\( = (3y + 2)(4y - 1)\)
\( = \left(3\left(4x - \frac{1}{4x}\right) + 2\right)\left(4\left(4x - \frac{1}{4x}\right) - 1\right)\)
\( = \left(12x - \frac{3}{4x} + 2\right)\left(16x - \frac{4}{4x} - 1\right)\)
\( = \left(12x - \frac{3}{4x} + 2\right)\left(16x - \frac{1}{x} - 1\right)\)
(iv) \( 10\left(3x + \frac{1}{x}\right)^2 - \left(3x + \frac{1}{x}\right) - 3 \)
माना \(3x + \frac{1}{x} = y\)
\( = 10y^2 - y - 3\)
\( = 10y^2 - (6-5)y - 3 \)
\( (10 \times 3 = 30 = 2 \times 3 \times 5)\)
\( = 10y^2 - 6y + 5y - 3\)
\( = 2y(5y-3)+1(5y-3)\)
\( = (5y-3)(2y +1)\)
\( = \left(5\left(3x + \frac{1}{x}\right) - 3\right)\left(2\left(3x + \frac{1}{x}\right) + 1\right)\)
\( = \left(15x + \frac{5}{x} - 3\right)\left(6x + \frac{2}{x} + 1\right)\)
In simple words: These complex expressions are simplified through substitution, transforming them into basic quadratic equations which are then factorized by splitting the middle term.

🎯 Exam Tip: When dealing with fractional terms within parentheses, treat the entire parenthetical expression as a single variable for substitution. Be meticulous with distributing terms after back-substitution.

Question 20. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) \( \frac{x^2}{3} + \frac{8x}{3} + 5 \)
(ii) \( \frac{x^2}{2} + \frac{9}{2} x + 10 \)
(iii) \( \frac{x^2}{2} + \frac{6x}{2} + 18 \)
(iv) \( \frac{3x^2}{5} - \frac{19x}{5} + 4 \)
(v) \( \frac{2x^2}{3} - \frac{17x}{3} - 28 \)
(vi) \( \frac{a^2}{4b^2} - \frac{1}{3} + \frac{b^2}{9a^2} \)
Answer: हल:
(i) \( \frac{x^2}{3} + \frac{8x}{3} + 5 \)
\[ = \frac{x^2 + 8x + 15}{3} \] \[ = \frac{1}{3} [x^2 + (3+5)x + 15] \] \( (: 15 = 3 \times 5)\)
\[ = \frac{1}{3} [x^2 + 3x + 5x +15] \] \[ = \frac{1}{3} [x(x + 3) + 5(x + 3)] \] \[ = \frac{1}{3} [(x + 3)(x + 5)] \] (ii) \( \frac{x^2}{2} + \frac{9}{2} x + 10 \)
\[ = \frac{x^2 + 9x + 20}{2} \] \[ = \frac{1}{2} [x^2 + (4+5)x + 20] \] \( (20 = 2 \times 2 \times 5)\)
\[ = \frac{1}{2} [x^2 + 4x + 5x + 20] \] \[ = \frac{1}{2} [x(x + 4) + 5(x + 4)] \] \[ = \frac{1}{2} [(x + 5) (x + 4)] \] \[ = \left(x + \frac{5}{2}\right) (x + 4) \] (iii) \( \frac{x^2}{2} + \frac{6x}{2} + 18 \)
\[ = \frac{x^2 + 12x + 36}{2} \] \[ = \frac{1}{2} [x^2 + 2x \times x \times 6 + (6)^2] \] \[ = \frac{1}{2} [(x + 6)^2] \] \[ = \frac{1}{2} (x + 6)(x + 6) \] \[ = \left(\frac{x}{2} + 3\right) (x + 6) \] (iv) \( \frac{3x^2}{5} - \frac{19x}{5} + 4 \)
\[ = \frac{3x^2 - 19x + 20}{5} \] \[ = \frac{1}{5} [3x^2 - (4+15)x + 20] \] \( (: 3 \times 20 = 60 = 4 \times 15)\)
\[ = \frac{1}{5} [3x^2 - 4x - 15x + 20] \] \[ = \frac{1}{5} [x(3x - 4) - 5(3x - 4)] \] \[ = \frac{1}{5} [(3x - 4)(x - 5)] \] \[ = \left( \frac{3x}{5} - \frac{4}{5} \right) (x - 5) \] (v) \( \frac{2x^2}{3} - \frac{17x}{3} - 28 \)
\[ = \frac{2x^2 - 17x - 84}{3} \] \[ = \frac{1}{3} [2x^2 - 17x - 84] \] \[ = \frac{1}{3} [2x^2 - (24 - 7)x - 84] \] \( (2 \times 1 = 2) \)
\( (84 = 2 \times 2 \times 3 \times 7)\)
\[ = \frac{1}{3} [2x^2 - 24x + 7x - 84] \] \[ = \frac{1}{3} [2x(x - 12) + 7(x - 12)] \] \[ = \frac{1}{3} [(2x + 7)(x - 12)] \] \[ = \left(\frac{2x}{3} + \frac{7}{3}\right) (x - 12) \] (vi) \( \frac{a^2}{4b^2} - \frac{1}{3} + \frac{b^2}{9a^2} \)
\[ = \left(\frac{a}{2b}\right)^2 - 2 \left(\frac{a}{2b}\right) \left(\frac{b}{3a}\right) + \left(\frac{b}{3a}\right)^2 \] \[ = \left(\frac{a}{2b} - \frac{b}{3a}\right)^2 \] \[ = \left(\frac{3a^2 - 2b^2}{6ab}\right)^2 \] \[ = \left(\frac{3a^2 - 2b^2}{6ab}\right) \left(\frac{3a^2 - 2b^2}{6ab}\right) \]In simple words: For polynomials with fractional coefficients, convert them to a common denominator to factor out the fractional part and then factorize the integer polynomial using middle-term splitting or identities.

🎯 Exam Tip: Always look for common factors, including fractional ones, to simplify polynomials before attempting to factorize using splitting the middle term or algebraic identities.

Question 21. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
Answer: हल:
(i) \(2x^2 + 7\sqrt{2}x + 12\)
\( = 2x^2 + (3\sqrt{2} + 4\sqrt{2})x + 12 \)
\( (2 = 2 \times 1) \)
\( (12 = 2 \times 2 \times 3)\)
\( = 2x^2 + 3\sqrt{2}x + 4\sqrt{2}x + 12\)
\( = \sqrt{2}x(\sqrt{2}x + 3) + 4(\sqrt{2}x + 3)\)
\( = (\sqrt{2}x + 3)(\sqrt{2}x + 4)\)
(ii) \(5\sqrt{5}x^2 + 30x + 8\sqrt{5}\)
\( = 5\sqrt{5}x^2 + (20+10)x + 8\sqrt{5} \)
\( (5\sqrt{5} \times 8\sqrt{5} = 200)\)
\( = 5\sqrt{5}x^2 + 20x + 10x + 8\sqrt{5}\)
\( = 5x(\sqrt{5}x + 4)+ 2\sqrt{5}(\sqrt{5}x + 4)\)
\( = (\sqrt{5}x + 4)(5x + 2\sqrt{5})\)
(iii) \(8\sqrt{3}x^2 + 62x + 5\sqrt{3}\)
\( = 8\sqrt{3}x^2 + (60+2)x + 5\sqrt{3} \)
\( (8\sqrt{3} \times 5\sqrt{3} = 120 = 2 \times 2 \times 2 \times 3 \times 5)\)
\( = 8\sqrt{3}x^2 + 60x - 2x + 5\sqrt{3}\)
\( = 4\sqrt{3}x(2x - 5\sqrt{3}) - 1(2x - 5\sqrt{3})\)
\( = (2x - 5\sqrt{3})(4\sqrt{3}x - 1)\)
(iv) \(\sqrt{3}x^2 + 10x + 3\sqrt{3}\)
\( = \sqrt{3}x^2 + (9+1)x + 3\sqrt{3}\)
\( = \sqrt{3}x^2 + 9x + x + 3\sqrt{3}\)
\( = \sqrt{3}x(x + 3\sqrt{3}) +1(x + 3\sqrt{3})\)
\( = (x + 3\sqrt{3})(\sqrt{3}x + 1)\)
In simple words: These problems involve quadratic factorization with irrational coefficients; the method of splitting the middle term is adapted by finding factors of the product of the first and last terms that sum to the middle term, including the radical part.

🎯 Exam Tip: When dealing with expressions containing square roots, focus on how to express the constant term and the coefficient of the middle term in terms of the radicals to facilitate splitting.

Question 22. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए
(i) \(9a^3b + 41a^2b^2 + 20ab^3\)
(ii) \(ax^2 + (4a^2 - 3b)x - 12ab\)
(iii) \(25x^2 + 10xy - 8y^2\)
(iv) \(4x^2 + 20xy + 25y^2\)
Answer: हल:
(i) \(9a^3b + 41a^2b^2 + 20ab^3\)
\( = ab[9a^2 + 41ab + 20b^2] \)
\( (: 9 \times 20 = 180 = 2 \times 2 \times 3 \times 3 \times 5)\)
\( = ab[9a^2 + (36 + 5)ab + 20b^2] \)
\( = ab[9a^2 + 36ab + 5ab + 20b^2] \)
\( = ab[9a(a + 4b) + 5b(a + 4b)] \)
\( = ab(9a + 5b)(a + 4b)]\)
(ii) \(ax^2 + (4a^2 - 3b)x - 12ab\)
\( = ax^2 + 4a^2x - 3bx - 12ab\)
\( = ax(x + 4a) - 3b(x + 4a)\)
\( = (x + 4a)(ax - 3b)\)
(iii) \(25x^2 + 10xy - 8y^2\)
\( = 25x^2 + (20 - 10)xy - 8y^2 \)
\( (: 25 \times 8 = 200 = 2 \times 2 \times 2 \times 5 \times 5)\)
\( = 25x^2 + 20xy - 10xy - 8y^2\)
\( = 5x(5x + 4y) - 2y(5x + 4y)\)
\( = (5x + 4y)(5x - 2y)\)
(iv) \(4x^2 + 20xy + 25y^2\)
\( = 4x^2 + (10 + 10)xy + 25y^2 \)
\( (4 \times 25 = 100 = 10 \times 10)\)
\( = 4x^2 + 10xy + 10xy + 25y^2\)
\( = 2x(2x + 5y) + 5y (2x + 5y)\)
\( = (2x + 5y)(2x + 5y)\)
In simple words: These problems cover various polynomial factorizations, including those with multiple variables and complex coefficients, often using common factor extraction and then applying the splitting the middle term method.

🎯 Exam Tip: For expressions with multiple variables, factor out common monomial factors first, then use the splitting the middle term method for the remaining polynomial, treating one variable's power as the main degree. Recognize perfect squares for efficient factorization.