What is (t-1/2)(t^2-t+1/6) ?

The solution to (t-1/2)(t^2-t+1/6) is ((2t-1)(6t^2-6t+1))/(12)

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simplify (t-1/2)(t^2-t+1/6)

simplify

(t - \frac{1}{2}) (t^{2} - t + \frac{1}{6})

\frac{( 2 t - 1 ) ( 6 t ^{2} - 6 t + 1 )}{12}

Solution steps

(t - \frac{1}{2}) (t^{2} - t + \frac{1}{6})

Simplify

t - \frac{1}{2} : \frac{2 t - 1}{2}

= \frac{2 t - 1}{2} (t^{2} - t + \frac{1}{6})

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 2 t - 1 ) ( t ^{2} - t + \frac{1}{6} )}{2}

Simplify

(2 t - 1) (t^{2} - t + \frac{1}{6}) : \frac{( 2 t - 1 ) ( t ^{2} \cdot 6 - t \cdot 6 + 1 )}{6}

= \frac{\frac{( 2 t - 1 ) ( t ^{2} \cdot 6 - t \cdot 6 + 1 )}{6}}{2}

Apply the fraction rule:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{( 2 t - 1 ) ( t ^{2} \cdot 6 - t \cdot 6 + 1 )}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{( 2 t - 1 ) ( t ^{2} \cdot 6 - t \cdot 6 + 1 )}{12}

= \frac{( 2 t - 1 ) ( 6 t ^{2} - 6 t + 1 )}{12}

\frac{1}{( x + y i ) ^{2} - 4}

- cos (v) sin (u) (2 cos (v) + cos (v) cos (u))

(1 + t) (- \frac{1}{2} ((t + 1)^{2} - t^{2}))

(sin (2 t) + 2 sin (t)) \cdot (2 sin (t) - 2 sin (2 t))

\frac{s + 1}{( s + 30 ) ( s ^{2} + s + 1 ) ( s + 15 )}

What is (t-1/2)(t^2-t+1/6) ?
The solution to (t-1/2)(t^2-t+1/6) is ((2t-1)(6t^2-6t+1))/(12)