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

The solution to expand (x-1/2)^4 is x^4-2x^3+(3x^2)/2-x/2+1/16

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expand (x-1/2)^4

expand

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

x^{4} - 2 x^{3} + \frac{3 x ^{2}}{2} - \frac{x}{2} + \frac{1}{16}

Solution steps

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

Apply binomial theorem:

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

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

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

= x^{4} - 2 x^{3} + \frac{3 x ^{2}}{2} - \frac{x}{2} + \frac{1}{16}

3 \times 3 \times 3 \times 3 \times 3 \times 3

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(2 x + 8) (x + 1)

19 x^{2}

\frac{a ^{- 4}}{a ^{- 7}}

What is expand (x-1/2)^4 ?
The solution to expand (x-1/2)^4 is x^4-2x^3+(3x^2)/2-x/2+1/16