# Using binomial theorem, expand each of the following:(1 + 2x – 3x2)4

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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