Q. 104.5( 4 Votes )

Using binomial theorem, expand each of the following:

(1 + 2x – 3x2)4

Answer :

To find: Expansion of (1 + 2x – 3x2)4


Formula used: (i)


(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn


We have, (1 + 2x – 3x2)4


Let (1+2x) = a and (-3x2) = b … (i)


Now the equation becomes (a + b)4


[4C0(a)4-0] + [4C1(a)4-1(b)1] + [4C2(a)4-2(b)2] + [4C3(a)4-3(b)3]+ [4C4(b)4]


[4C0(a)4] + [4C1(a)3(b)1] + [4C2(a)2(b)2] + [4C3(a)(b)3]+ [4C4(b)4]


(Substituting value of b from eqn. i )




(Substituting value of b from eqn. i )


… (ii)


We need the value of a4,a3 and a2, where a = (1+2x)


For (1+2x)4, Applying Binomial theorem


(1+2x)4





1 + 8x + 24x2 + 32x3 + 16x4


We have (1+2x)4 = 1 + 8x + 24x2 + 32x3 + 16x4 … (iii)


For (a+b)3 , we have formula a3+b3+3a2b+3ab2


For, (1+2x)3 , substituting a = 1 and b = 2x in the above formula


13+ (2x) 3+3(1)2(2x) +3(1) (2x) 2


1 + 8x3 + 6x + 12x2


8x3 + 12x2 + 6x + 1 … (iv)


For (a+b)2 , we have formula a2+2ab+b2


For, (1+2x)2 , substituting a = 1 and b = 2x in the above formula


(1)2 + 2(1)(2x) + (2x)2


1 + 4x + 4x2


4x2 + 4x + 1 … (v)


Putting the value obtained from eqn. (iii),(iv) and (v) in eqn. (ii)






1 + 8x + 24x2 + 32x3 + 16x4 - 96x5 - 144x4 - 72x3 - 12x2 + 216x6 + 216x5 + 54x4 -108x6 - 216x7 + 81x8


On rearranging


Ans) 81x8 - 216x7 + 108x6 + 120x5 - 74x4 - 40x3 + 12x2 +8x+ 1


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