What is expand (4x-1/2 y)^4 ?

The solution to expand (4x-1/2 y)^4 is (4096x^4+y^4)/(16)-128x^3y+24x^2y^2-2xy^3

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expand (4x-1/2 y)^4

expand

(4 x - \frac{1}{2} y)^{4}

\frac{4096 x ^{4} + y ^{4}}{16} - 128 x^{3} y + 24 x^{2} y^{2} - 2 x y^{3}

Solution steps

(4 x - \frac{1}{2} y)^{4}

Apply binomial theorem:

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

a = 4 x, b = - \frac{1}{2} y

= i = 0 \sum 4 (i 4) (4 x)^{(4 - i)} (- \frac{1}{2} y)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (4 x)^{4} (- \frac{1}{2} y)^{0} : 256 x^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (4 x)^{3} (- \frac{1}{2} y)^{1} : - 128 x^{3} y

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (4 x)^{2} (- \frac{1}{2} y)^{2} : 24 x^{2} y^{2}

Simplify

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

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (4 x)^{0} (- \frac{1}{2} y)^{4} : \frac{y ^{4}}{16}

= 256 x^{4} - 128 x^{3} y + 24 x^{2} y^{2} - 2 x y^{3} + \frac{y ^{4}}{16}

Join

256 x^{4} - 128 x^{3} y + 24 x^{2} y^{2} - 2 x y^{3} + \frac{y ^{4}}{16} : \frac{4096 x ^{4} + y ^{4}}{16} - 128 x^{3} y + 24 x^{2} y^{2} - 2 x y^{3}

= \frac{4096 x ^{4} + y ^{4}}{16} - 128 x^{3} y + 24 x^{2} y^{2} - 2 x y^{3}

x (\frac{3 x}{4} + \frac{10 - 3 x}{2})

3 (2 x^{3} + 1) (x^{4} + 2 x)^{\frac{1}{2}}

(X^{4} - 2 X^{2} + 2) l e f t (X^{2} - 2 X + 3)

- \frac{- sin ( θ ) + x cos ( θ )}{( cos ( θ ) + x sin ( θ ) ) ^{2}}

e^{(1 + 2 i) x}

What is expand (4x-1/2 y)^4 ?
The solution to expand (4x-1/2 y)^4 is (4096x^4+y^4)/(16)-128x^3y+24x^2y^2-2xy^3