Dear Student , Please find below the solution to the asked query : 7 i 6 x+2y3 + 8 x+2y2=2 x+2y2 3x+2y + 4=2 x+2y x+2y 3x+6y+4 ANS ii 14a-3b3 - 21pa-3b=7 a-3b 2a-3b2 - 3p =7 a-3b 2 a-3b2 - 3p2 =7 a-3b 2a-3b+3p 2a-3b-3p Since x2-y2=x+yx-y =7 a-3b 2a-32b+3p 2a-32b-3p ANS Please ask single query in single thread Hope this information will clear your doubts about the topic If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible Regards

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Solve this: HCF of the Exercise Factorise the following (1 to 8) polynomials, (I) 8Xy3 + 12x2y2 3. (t) 25abc2- 15a2b2c 4. lox 5. (1) 18p2q2 — 24"2 + 30p2q + 8(r + 2y)2 (it) — 9ax2 (it) 4x• — 6x-2 (it) x2yz + ry2z • (iO 14mn + 22m — CTORISING BY GROUPING OF TERMS the grouping Of terms of the given polynomial gives rise to common fa 'en olvnomial can be factorised by the following procedure in such a way that ea

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Please find below the solution to the asked query :

7 . (i) 6 {(x + 2 y)}^{3} + 8 {(x + 2 y)}^{2} = 2 {(x + 2 y)}^{2} [3 (x + 2 y) + 4] = 2 (x + 2 y) (x + 2 y) (3 x + 6 y + 4) ANS . . . (ii) 14 {(a - 3 b)}^{3} - 21 p (a - 3 b) = 7 (a - 3 b) [2 {(a - 3 b)}^{2} - 3 p] = 7 (a - 3 b) [{(\sqrt{2} (a - 3 b))}^{2} - {(\sqrt{3} p)}^{2}] = 7 (a - 3 b) (\sqrt{2} (a - 3 b) + \sqrt{3} p) (\sqrt{2} (a - 3 b) - \sqrt{3} p) (Since x^{2} - y^{2} = (x + y) (x - y)) = 7 (a - 3 b) (\sqrt{2} a - 3 \sqrt{2} b + \sqrt{3} p) (\sqrt{2} a - 3 \sqrt{2} b - \sqrt{3} p) ANS . . . Please ask single query in single thread .

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