What is expand t^2μ(t-1)+tcos(5t) ?

The solution to expand t^2μ(t-1)+tcos(5t) is t^3μ-t^2μ+tcos(5t)

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expand t^2μ(t-1)+tcos(5t)

expand

t^{2} μ (t - 1) + t cos (5 t)

t^{3} μ - t^{2} μ + t cos (5 t)

Solution steps

t^{2} μ (t - 1) + t cos (5 t)

Apply the distributive law:

a (b - c) = ab - a c

t^{2} μ (t - 1) = t^{2} μ t - t^{2} μ \cdot 1

= t^{2} μ t - t^{2} μ \cdot 1 + t cos (5 t)

Simplify

t^{2} μ t - t^{2} μ \cdot 1 : t^{3} μ - t^{2} μ

= t^{3} μ - t^{2} μ + t cos (5 t)

(y + \frac{1}{3})^{2}

(x^{2} - 1) x

π x^{2} (6.4 - x)

(- 2 - x) \cdot (- 3 - x) \cdot (\frac{2}{3} - x)

\frac{- 4 s + 6}{5 ( s ^{2} - 2 s + 2 )}

What is expand t^2μ(t-1)+tcos(5t) ?
The solution to expand t^2μ(t-1)+tcos(5t) is t^3μ-t^2μ+tcos(5t)