What is expand 16cos^4(2)(e^{4x}*4x^4+4x^3e^{4x}) ?

The solution to expand 16cos^4(2)(e^{4x}*4x^4+4x^3e^{4x}) is 64cos^4(2)e^{4x}x^4+64cos^4(2)e^{4x}x^3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand 16cos^4(2)(e^{4x}*4x^4+4x^3e^{4x})

expand

16 cos^{4} (2) (e^{4 x} \cdot 4 x^{4} + 4 x^{3} e^{4 x})

64 cos^{4} (2) e^{4 x} x^{4} + 64 cos^{4} (2) e^{4 x} x^{3}

Solution steps

16 cos^{4} (2) (e^{4 x} \cdot 4 x^{4} + 4 x^{3} e^{4 x})

= 16 cos^{4} (2) (4 e^{4 x} x^{4} + 4 e^{4 x} x^{3})

Apply the distributive law:

a (b + c) = ab + a c

a = 16 cos^{4} (2), b = e^{4 x} \cdot 4 x^{4}, c = 4 x^{3} e^{4 x}

= 16 cos^{4} (2) e^{4 x} \cdot 4 x^{4} + 16 cos^{4} (2) \cdot 4 x^{3} e^{4 x}

= 16 \cdot 4 cos^{4} (2) e^{4 x} x^{4} + 16 \cdot 4 cos^{4} (2) e^{4 x} x^{3}

Multiply the numbers:

16 \cdot 4 = 64

= 64 cos^{4} (2) e^{4 x} x^{4} + 64 cos^{4} (2) e^{4 x} x^{3}

(A e^{t} + B e^{- t})^{2}

\frac{- 17 x - 36}{( x + 1 ) ( 5 x + 9 )}

(4, - 3) y (1, 3)

(x - 4) \cdot (x^{2} - 4)

x^{2} + (y - 3)^{2}

What is expand 16cos^4(2)(e^{4x}*4x^4+4x^3e^{4x}) ?
The solution to expand 16cos^4(2)(e^{4x}*4x^4+4x^3e^{4x}) is 64cos^4(2)e^{4x}x^4+64cos^4(2)e^{4x}x^3