Question

Expand (2x-2/x)⁵ answer the question.​

Answer 1

Answer:

HEY MATE HERE'S YOUR ANSWER

Step-by-step explanation:

$</p><p></p><p>1 {(2)}^{5} {( - y)}^{0} + 5 {(2x)}^{4} {(y)}^{1} + 10 {(2x)}^{3} {( - y)}^{2} + 10 {(2x)}^{2} {(y)}^{3} + 5 {(2x)}^{1} {( - y)}^{4} + 1 {(2x)}^{0} {( - y)}^{5} = \\ {32x}^{5} - {80x}^{4} y + {80x}^{3} {y}^{2} - {40x}^{2} {y}^{3} + {10xy}^{4} - {y}^{5} \\ \\so \\ \\ {(2x - y}^{5} = {32x}^{5} - {80x}^{4} y + {80x}^{3} {y}^{2} - {40x}^{2} {y}^{3} + 10x {y}^{4} - {y}^{5}$

As you can see, the

6

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h

row has six numbers, 1, 5, 10, 10, 5 and 1 respectively.

We must plug these numbers in to the following formula. As you can see, the powers on the 2x are descending and the powers on the -y are ascending. The exponents of 2x and -y must always add up to 5, the expression's exponent. Essentially, you must multiply the numbers, in order, of the Pascal's triangle, with the exponents on the first term of the expression (2x) descending from your total exponent (5) to 0. Do the same thing with the second term (-y), except ascending, from 0 to 5.

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