What is expand (2x-3)^4 ?

The solution to expand (2x-3)^4 is 16x^4-96x^3+216x^2-216x+81

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expand (2x-3)^4

expand

(2 x - 3)^{4}

16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81

Solution steps

(2 x - 3)^{4}

Apply binomial theorem:

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

a = 2 x, b = - 3

= i = 0 \sum 4 (i 4) (2 x)^{(4 - i)} (- 3)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (2 x)^{4} (- 3)^{0} : 16 x^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (2 x)^{3} (- 3)^{1} : - 96 x^{3}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (2 x)^{2} (- 3)^{2} : 216 x^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (2 x)^{1} (- 3)^{3} : - 216 x

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (2 x)^{0} (- 3)^{4} : 81

= 16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81

\frac{d}{d x} (e^{x} ln (x^{5}))

x^{^{'}} + x = e^{t}

\int_{0}^{15} x 15 - x d x

\int 10 x d x

\int - \frac{8}{x} d x

What is expand (2x-3)^4 ?
The solution to expand (2x-3)^4 is 16x^4-96x^3+216x^2-216x+81