What is expand (t^2+4t+4)*(2^{1/2}(t+2)) ?

The solution to expand (t^2+4t+4)*(2^{1/2}(t+2)) is 2^{1/2}t^3+6*2^{1/2}t^2+12*2^{1/2}t+2^{1/2}*8

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expand (t^2+4t+4)*(2^{1/2}(t+2))

expand

(t^{2} + 4 t + 4) \cdot (2^{\frac{1}{2}} (t + 2))

2^{\frac{1}{2}} t^{3} + 6 \cdot 2^{\frac{1}{2}} t^{2} + 12 \cdot 2^{\frac{1}{2}} t + 2^{\frac{1}{2}} \cdot 8

Solution steps

(t^{2} + 4 t + 4) (2^{\frac{1}{2}} (t + 2))

= (2^{\frac{1}{2}} (t + 2)) (t^{2} + 4 t + 4)

Apply the distributive law:

m (a + b + c) = ma + mb + m c

(2^{\frac{1}{2}} (t + 2)) (t^{2} + 4 t + 4) = 2^{\frac{1}{2}} (t + 2) t^{2} + 2^{\frac{1}{2}} (t + 2) \cdot 4 t + 2^{\frac{1}{2}} (t + 2) \cdot 4

= 2^{\frac{1}{2}} (t + 2) t^{2} + 2^{\frac{1}{2}} (t + 2) \cdot 4 t + 2^{\frac{1}{2}} (t + 2) \cdot 4

Expand

2^{\frac{1}{2}} (t + 2) t^{2} + 2^{\frac{1}{2}} (t + 2) \cdot 4 t + 2^{\frac{1}{2}} (t + 2) \cdot 4 : 2^{\frac{1}{2}} t^{3} + 6 \cdot 2^{\frac{1}{2}} t^{2} + 12 \cdot 2^{\frac{1}{2}} t + 2^{\frac{1}{2}} \cdot 8

= 2^{\frac{1}{2}} t^{3} + 6 \cdot 2^{\frac{1}{2}} t^{2} + 12 \cdot 2^{\frac{1}{2}} t + 2^{\frac{1}{2}} \cdot 8

\frac{( k + 1 ) ( k + 2 ) ( k + 3 )}{6}

(p + \frac{a}{v ^{2}}) (v - b)

cos ((5 - n) π)

- \frac{4 x y + 8 y - 6}{( x + 2 ) ^{2}}

\frac{x ^{2}}{( x - 1 ) ^{2} ( x - 2 )}

What is expand (t^2+4t+4)*(2^{1/2}(t+2)) ?
The solution to expand (t^2+4t+4)*(2^{1/2}(t+2)) is 2^{1/2}t^3+6*2^{1/2}t^2+12*2^{1/2}t+2^{1/2}*8