What is expand (t-3)\H(t-1) ?

The solution to expand (t-3)\H(t-1) is t\H(t-1)-3\H(t-1)

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expand (t-3)\H(t-1)

expand

(t - 3) H (t - 1)

t H (t - 1) - 3 H (t - 1)

Solution steps

(t - 3) H (t - 1)

= H (t - 1) (t - 3)

Apply the distributive law:

a (b - c) = ab - a c

a = H (t - 1), b = t, c = 3

= H (t - 1) t - H (t - 1) \cdot 3

= t H (t - 1) - 3 H (t - 1)

\frac{- 2 x ^{2} + 3 x - 24}{( x ^{2} + 9 ) ( x - 2 )}

\frac{1}{s ( s + 2 ) ( s + 7 )}

\frac{x - 2}{( x - 4 ) ( 3 x + 2 )}

\frac{1}{x ^{2} ( x ^{2} + 9 )}

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

What is expand (t-3)\H(t-1) ?
The solution to expand (t-3)\H(t-1) is t\H(t-1)-3\H(t-1)