Ax+bn #160;cx+dnLet #160;u #160;=ax+bn, #160;v=cx+dnThen, #160;u #39;=naax+bn 1, #160;v #39;=nccx+dn 1Using #160;the #160;product #160;rul ...

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Byju's Answer

Standard VIII

Mathematics

Multiplication of Any Polynomial

a x+bnc x+dn

Question

(ax + b)ⁿ (cx + d)ⁿ

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Solution

${(a x + b)}^{n} {(c x + d)}^{n} Let u = {(a x + b)}^{n}, v = {(c x + d)}^{n} Then, u' = n a {(a x + b)}^{n - 1}, v' = n c {(c x + d)}^{n - 1} Using the product rule : \frac{d}{d x} (u v) = u v' + u' v \frac{d}{d x} [{(a x + b)}^{n} {(c x + d)}^{n}] = {(a x + b)}^{n} \times n c {(c x + d)}^{n - 1} + n a {(a x + b)}^{n - 1} \times {(c x + d)}^{n} = n {(a x + b)}^{n - 1} {(c x + d)}^{n - 1} (a c x + c b + a c x + a d) = n {(a x + b)}^{n - 1} {(c x + d)}^{n - 1} (2 a c x + c b + a d)$

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(ax + b)n (cx + d)n

ax+bn cx+dnLet u =ax+bn, v=cx+dnThen, u'=naax+bn-1, v'=nccx+dn-1Using the product rule:ddxuv=uv'+u'vddxax+bn cx+dn=ax+bn×nccx+dn-1+naax+bn-1×cx+dn =nax+bn-1cx+dn-1acx+cb+acx+ad =nax+bn-1cx+dn-12acx+cb+ad

(ax + b)ⁿ (cx + d)ⁿ