What is expand (2x-y)^4 ?

The solution to expand (2x-y)^4 is 16x^4-32x^3y+24x^2y^2-8xy^3+y^4

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expand (2x-y)^4

expand

(2 x - y)^{4}

16 x^{4} - 32 x^{3} y + 24 x^{2} y^{2} - 8 x y^{3} + y^{4}

Solution steps

(2 x - y)^{4}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 2 x, b = - y

= i = 0 \sum 4 (i 4) (2 x)^{(4 - i)} (- y)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (2 x)^{4} (- y)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (2 x)^{3} (- y)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (2 x)^{2} (- y)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (2 x)^{1} (- y)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (2 x)^{0} (- y)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (2 x)^{4} (- y)^{0} : 16 x^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (2 x)^{3} (- y)^{1} : - 32 x^{3} y

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (2 x)^{2} (- y)^{2} : 24 x^{2} y^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (2 x)^{1} (- y)^{3} : - 8 x y^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (2 x)^{0} (- y)^{4} : y^{4}

= 16 x^{4} - 32 x^{3} y + 24 x^{2} y^{2} - 8 x y^{3} + y^{4}

x (2 - x)^{2}

\frac{x}{x ^{2} - 4} + \frac{1}{x - 2}

2^{l o g_{2} (3) - l o g_{4} (25)}

(3 x + 5) (2 x^{2} + x + 5)

What is expand (2x-y)^4 ?
The solution to expand (2x-y)^4 is 16x^4-32x^3y+24x^2y^2-8xy^3+y^4