What is expand (2x-2)(\H(t-1)-\H(t-4)) ?

The solution to expand (2x-2)(\H(t-1)-\H(t-4)) is 2x\H(t-1)-2x\H(t-4)-2\H(t-1)+2\H(t-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (2x-2)(\H(t-1)-\H(t-4))

expand

(2 x - 2) (H (t - 1) - H (t - 4))

2 x H (t - 1) - 2 x H (t - 4) - 2 H (t - 1) + 2 H (t - 4)

Solution steps

(2 x - 2) (H (t - 1) - H (t - 4))

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = 2 x, b = - 2, c = H (t - 1), d = - H (t - 4)

= 2 x H (t - 1) + 2 x (- H (t - 4)) + (- 2) H (t - 1) + (- 2) (- H (t - 4))

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= 2 x H (t - 1) - 2 x H (t - 4) - 2 H (t - 1) + 2 H (t - 4)

(- e^{x} + e^{- x}) \cdot (e^{- x} + e^{- 2 x} + e^{- 3 x})

\frac{s ^{2} + 1}{s ( s + 1 )}

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

\frac{( k + 1 ) ( 2 k + 1 ) ( 2 k + 3 )}{3}

2 (e^{4} - e)

What is expand (2x-2)(\H(t-1)-\H(t-4)) ?
The solution to expand (2x-2)(\H(t-1)-\H(t-4)) is 2x\H(t-1)-2x\H(t-4)-2\H(t-1)+2\H(t-4)