To find: Expansion of (1 + 2x – 3x²)⁴

Formula used: (i)

(ii) (a+b)ⁿ =   ⁿC₀aⁿ + ⁿC₁a^n-1b + ⁿC₂a^n-2b² + …… +ⁿC_n-1ab^n-1 + ⁿC_nbⁿ

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