Question 27 of 37advanced🔧 ApplyShort Answer11 marks

निम्नलिखित बीजीय व्यंजकों का गुणनखंड कीजिए—

(i)

4y2+1+116y24y^2 + 1 + \frac{1}{16y^2}

Answer

(2y+14y)2(2y + \frac{1}{4y})^2

(ii)

9m2−125n29m^2 - \frac{1}{25n^2}

Answer

(3m−15n)(3m+15n)(3m - \frac{1}{5n})(3m + \frac{1}{5n})

(iii)

27b3−164b327b^3 - \frac{1}{64b^3}

Answer

(3b−14b)(9b2+34+116b2)(3b - \frac{1}{4b})(9b^2 + \frac{3}{4} + \frac{1}{16b^2})

(iv)

x2+5x6+16x^2 + \frac{5x}{6} + \frac{1}{6}

Answer

16(3x+2)(2x+1)\frac{1}{6}(3x + 2)(2x + 1)

(v)

27u3−1125−27u25+9u2527u^3 - \frac{1}{125} - \frac{27u^2}{5} + \frac{9u}{25}

Answer

(3u−15)3(3u - \frac{1}{5})^3

(vi)

64y3+1125z364y^3 + \frac{1}{125}z^3

Answer

(4y+15z)(16y2−45yz+125z2)(4y + \frac{1}{5}z)(16y^2 - \frac{4}{5}yz + \frac{1}{25}z^2)

(vii)

p3+27q3+r3−9pqrp^3 + 27q^3 + r^3 - 9pqr

Answer

(p+3q+r)(p2+9q2+r2−3pq−3qr−rp)(p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - rp)

(viii)

9m2−12m+49m^2 - 12m + 4

Answer

(3m−2)2(3m - 2)^2

(ix)

9x3−83y3+z33+6xyz9x^3 - \frac{8}{3}y^3 + \frac{z^3}{3} + 6xyz

Answer

13(3x−2y+z)(9x2+4y2+z2+6xy+2yz−3zx)\frac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3zx)

(x)

4x2+9y2+36z2+12xy+36yz+24zx4x^2 + 9y^2 + 36z^2 + 12xy + 36yz + 24zx

Answer

(2x+3y+6z)2(2x + 3y + 6z)^2

(xi)

27u3−1216−9u22+u427u^3 - \frac{1}{216} - \frac{9u^2}{2} + \frac{u}{4}

Answer

(3u−16)3(3u - \frac{1}{6})^3

Explanation

इन बीजीय व्यंजकों के गुणनखंडन के लिए हम विभिन्न बीजीय सर्वसमिकाओं का उपयोग करते हैं, जैसे (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2, a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b), a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2 + ab + b^2), a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a+b)(a^2 - ab + b^2), (a+b)3=a3+b3+3a2b+3ab2(a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2, (a−b)3=a3−b3−3a2b+3ab2(a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2, और a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)।

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