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Write out the terms using the binomial theorem.

Mathematics

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Answer

16x4−96x3y+216x2y2−216xy3+81y416x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4

Step-by-step solution

4 steps

To express (2x−3y)4(2x - 3y)^4, we use the binomial theorem, which states that (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k.
Here, a=2xa = 2x, b=−3yb = -3y, and n=4n = 4.

  1. Write out the terms using the binomial theorem.

    The expansion will have n+1=5n+1 = 5 terms:

    (40)(2x)4(−3y)0+(41)(2x)3(−3y)1+(42)(2x)2(−3y)2+(43)(2x)1(−3y)3+(44)(2x)0(−3y)4\binom{4}{0}(2x)^4(-3y)^0 + \binom{4}{1}(2x)^3(-3y)^1 + \binom{4}{2}(2x)^2(-3y)^2 + \binom{4}{3}(2x)^1(-3y)^3 + \binom{4}{4}(2x)^0(-3y)^4

  2. Calculate the binomial coefficients (nk)\binom{n}{k}.

    (40)=1\binom{4}{0} = 1 (41)=4\binom{4}{1} = 4 (42)=4×32×1=6\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 (43)=4\binom{4}{3} = 4 (44)=1\binom{4}{4} = 1

  3. Expand each term.

    • For k=0k=0: 1⋅(2x)4⋅(−3y)0=1⋅(16x4)⋅1=16x41 \cdot (2x)^4 \cdot (-3y)^0 = 1 \cdot (16x^4) \cdot 1 = 16x^4
    • For k=1k=1: 4⋅(2x)3⋅(−3y)1=4⋅(8x3)⋅(−3y)=−96x3y4 \cdot (2x)^3 \cdot (-3y)^1 = 4 \cdot (8x^3) \cdot (-3y) = -96x^3y
    • For k=2k=2: 6⋅(2x)2⋅(−3y)2=6⋅(4x2)⋅(9y2)=216x2y26 \cdot (2x)^2 \cdot (-3y)^2 = 6 \cdot (4x^2) \cdot (9y^2) = 216x^2y^2
    • For k=3k=3: 4⋅(2x)1⋅(−3y)3=4⋅(2x)⋅(−27y3)=−216xy34 \cdot (2x)^1 \cdot (-3y)^3 = 4 \cdot (2x) \cdot (-27y^3) = -216xy^3
    • For k=4k=4: 1⋅(2x)0⋅(−3y)4=1⋅1⋅(81y4)=81y41 \cdot (2x)^0 \cdot (-3y)^4 = 1 \cdot 1 \cdot (81y^4) = 81y^4

  4. Combine all the terms.

    (2x−3y)4=16x4−96x3y+216x2y2−216xy3+81y4(2x - 3y)^4 = 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4

    The expanded expression is:

    16x4−96x3y+216x2y2−216xy3+81y416x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4
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