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Expand (2\[sqrt(2)] - 3\[sqrt(4)])5

Mathematics

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Answer

188482−2697618848\sqrt{2} - 26976

Step-by-step solution

5 steps

  1. Simplify the expression inside the parenthesis.

    We have 4=2\sqrt{4} = 2.
    So, the expression becomes:

    (22−34)5=(22−3×2)5=(22−6)5(2\sqrt{2} - 3\sqrt{4})^5 = (2\sqrt{2} - 3 \times 2)^5 = (2\sqrt{2} - 6)^5

  2. Identify the coefficients from Pascal's triangle for the power of 5.

    The coefficients for (a+b)5(a+b)^5 are 1, 5, 10, 10, 5, 1.

  3. Apply the binomial theorem.

    Let a=22a = 2\sqrt{2} and b=−6b = -6. The expansion is given by:

    (a+b)5=1a5b0+5a4b1+10a3b2+10a2b3+5a1b4+1a0b5(a+b)^5 = 1a^5b^0 + 5a^4b^1 + 10a^3b^2 + 10a^2b^3 + 5a^1b^4 + 1a^0b^5

  4. Calculate each term.

    Term 1: 1⋅(22)5⋅(−6)01 \cdot (2\sqrt{2})^5 \cdot (-6)^0

    (22)5=25⋅(2)5=32⋅(21/2)5=32⋅25/2=32⋅22⋅2=32⋅4⋅2=1282(2\sqrt{2})^5 = 2^5 \cdot (\sqrt{2})^5 = 32 \cdot (2^{1/2})^5 = 32 \cdot 2^{5/2} = 32 \cdot 2^2 \cdot \sqrt{2} = 32 \cdot 4 \cdot \sqrt{2} = 128\sqrt{2} (−6)0=1(-6)^0 = 1 1⋅1282⋅1=12821 \cdot 128\sqrt{2} \cdot 1 = 128\sqrt{2}

    Term 2: 5⋅(22)4⋅(−6)15 \cdot (2\sqrt{2})^4 \cdot (-6)^1

    (22)4=24⋅(2)4=16⋅(21/2)4=16⋅22=16⋅4=64(2\sqrt{2})^4 = 2^4 \cdot (\sqrt{2})^4 = 16 \cdot (2^{1/2})^4 = 16 \cdot 2^2 = 16 \cdot 4 = 64 (−6)1=−6(-6)^1 = -6 5⋅64⋅(−6)=320⋅(−6)=−19205 \cdot 64 \cdot (-6) = 320 \cdot (-6) = -1920

    Term 3: 10⋅(22)3⋅(−6)210 \cdot (2\sqrt{2})^3 \cdot (-6)^2

    (22)3=23⋅(2)3=8⋅(21/2)3=8⋅23/2=8⋅2⋅2=162(2\sqrt{2})^3 = 2^3 \cdot (\sqrt{2})^3 = 8 \cdot (2^{1/2})^3 = 8 \cdot 2^{3/2} = 8 \cdot 2 \cdot \sqrt{2} = 16\sqrt{2} (−6)2=36(-6)^2 = 36 10⋅162⋅36=1602⋅36=5760210 \cdot 16\sqrt{2} \cdot 36 = 160\sqrt{2} \cdot 36 = 5760\sqrt{2}

    Term 4: 10⋅(22)2⋅(−6)310 \cdot (2\sqrt{2})^2 \cdot (-6)^3

    (22)2=22⋅(2)2=4⋅2=8(2\sqrt{2})^2 = 2^2 \cdot (\sqrt{2})^2 = 4 \cdot 2 = 8 (−6)3=−216(-6)^3 = -216 10⋅8⋅(−216)=80⋅(−216)=−1728010 \cdot 8 \cdot (-216) = 80 \cdot (-216) = -17280

    Term 5: 5⋅(22)1⋅(−6)45 \cdot (2\sqrt{2})^1 \cdot (-6)^4

    (22)1=22(2\sqrt{2})^1 = 2\sqrt{2} (−6)4=1296(-6)^4 = 1296 5⋅22⋅1296=102⋅1296=1296025 \cdot 2\sqrt{2} \cdot 1296 = 10\sqrt{2} \cdot 1296 = 12960\sqrt{2}

    Term 6: 1⋅(22)0⋅(−6)51 \cdot (2\sqrt{2})^0 \cdot (-6)^5

    (22)0=1(2\sqrt{2})^0 = 1 (−6)5=−7776(-6)^5 = -7776 1⋅1⋅(−7776)=−77761 \cdot 1 \cdot (-7776) = -7776

  5. Sum all the terms.

    Combine terms with 2\sqrt{2}:

    1282+57602+129602=(128+5760+12960)2=188482128\sqrt{2} + 5760\sqrt{2} + 12960\sqrt{2} = (128 + 5760 + 12960)\sqrt{2} = 18848\sqrt{2}

    Combine constant terms:

    −1920−17280−7776=−26976-1920 - 17280 - 7776 = -26976

    The expanded form is the sum of these combined terms.

    188482−2697618848\sqrt{2} - 26976
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