What is expand (sqrt(v)+2)^4 ?

The solution to expand (sqrt(v)+2)^4 is v^2+8vsqrt(v)+24v+32sqrt(v)+16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (sqrt(v)+2)^4

expand

(v + 2)^{4}

v^{2} + 8 v v + 24 v + 32 v + 16

Solution steps

(v + 2)^{4}

Apply binomial theorem:

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

a = v, b = 2

= i = 0 \sum 4 (i 4) (v)^{(4 - i)} \cdot 2^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (v)^{4} \cdot 2^{0} : v^{2}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (v)^{3} \cdot 2^{1} : 8 (v)^{3}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (v)^{2} \cdot 2^{2} : 24 v

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (v)^{1} \cdot 2^{3} : 32 v

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (v)^{0} \cdot 2^{4} : 16

= v^{2} + 8 (v)^{3} + 24 v + 32 v + 16

Simplify

(v)^{3} : v v

= v^{2} + 8 v v + 24 v + 32 v + 16

(sin (x) - cos (x))^{3}

(4 - x + 5)^{2}

(2 - sin (x))^{2}

2 (csc (x) + sec (x))

A (6, - 5)

What is expand (sqrt(v)+2)^4 ?
The solution to expand (sqrt(v)+2)^4 is v^2+8vsqrt(v)+24v+32sqrt(v)+16