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Exercise 7.1 · Q3

Q.Expand the expression (2x−3)6(2x - 3)^6.

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Use the Binomial Theorem to expand (2x−3)6(2x - 3)^6 as a sum of seven terms, treating it as (2x+(−3))6(2x + (-3))^6 and applying the formula (6k)(2x)6−k(−3)k\binom{6}{k}(2x)^{6-k}(-3)^k for k=0,1,…,6k = 0, 1, \ldots, 6.

The Binomial Theorem tells us how to expand any expression of the form (a+b)n(a + b)^n without multiplying it out the long way. The key insight is that each term in the expansion comes from choosing either aa or bb from each of the nn factors, and the coefficient counts how many ways we can make that choice. For (a+b)n(a + b)^n, the expansion is:

∑k=0n(nk)an−kbk\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

Here we have (2x−3)6(2x - 3)^6, which we rewrite as (2x+(−3))6(2x + (-3))^6 so that a=2xa = 2x, b=−3b = -3, and n=6n = 6. The expansion will have 77 terms (from k=0k = 0 to k=6k = 6).

(6k)(2x)6−k(−3)k\binom{6}{k}(2x)^{6-k}(-3)^k

Now we compute each term systematically.

  1. Term with k=0k = 0:

(60)(2x)6(−3)0=1⋅64x6⋅1=64x6\binom{6}{0}(2x)^6(-3)^0 = 1 \cdot 64x^6 \cdot 1 = 64x^6

  1. Term with k=1k = 1:

(61)(2x)5(−3)1=6⋅32x5⋅(−3)=−576x5\binom{6}{1}(2x)^5(-3)^1 = 6 \cdot 32x^5 \cdot (-3) = -576x^5

  1. Term with k=2k = 2:

(62)(2x)4(−3)2=15⋅16x4⋅9=2160x4\binom{6}{2}(2x)^4(-3)^2 = 15 \cdot 16x^4 \cdot 9 = 2160x^4

  1. Term with k=3k = 3:

(63)(2x)3(−3)3=20⋅8x3⋅(−27)=−4320x3\binom{6}{3}(2x)^3(-3)^3 = 20 \cdot 8x^3 \cdot (-27) = -4320x^3

  1. Term with k=4k = 4:

(64)(2x)2(−3)4=15⋅4x2⋅81=4860x2\binom{6}{4}(2x)^2(-3)^4 = 15 \cdot 4x^2 \cdot 81 = 4860x^2

  1. Term with k=5k = 5:

(65)(2x)1(−3)5=6⋅2x⋅(−243)=−2916x\binom{6}{5}(2x)^1(-3)^5 = 6 \cdot 2x \cdot (-243) = -2916x

  1. Term with k=6k = 6:

(66)(2x)0(−3)6=1⋅1⋅729=729\binom{6}{6}(2x)^0(-3)^6 = 1 \cdot 1 \cdot 729 = 729

Watch out

Watch the signs carefully! Since b=−3b = -3, odd powers of (−3)(-3) are negative while even powers are positive. A common mistake is to forget the negative sign or apply it inconsistently.

Collecting all seven terms in descending powers of xx:

(2x−3)6=64x6−576x5+2160x4−4320x3+4860x2−2916x+729(2x - 3)^6 = 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729

✓Final answer

The expansion is 64x6−576x5+2160x4−4320x3+4860x2−2916x+729\boxed{64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729}.

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