What is expand 4(mx-m+4)^2 ?

The solution to expand 4(mx-m+4)^2 is 4m^2x^2-8m^2x+32mx+4m^2-32m+64

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expand 4(mx-m+4)^2

expand

4 (m x - m + 4)^{2}

4 m^{2} x^{2} - 8 m^{2} x + 32 m x + 4 m^{2} - 32 m + 64

Solution steps

4 (m x - m + 4)^{2}

(m x - m + 4)^{2} = m^{2} x^{2} - 2 m^{2} x + 8 m x + m^{2} - 8 m + 16

= 4 (m^{2} x^{2} + m^{2} - 2 m^{2} x + 8 m x - 8 m + 16)

Distribute parentheses

= 4 m^{2} x^{2} + 4 (- 2 m^{2} x) + 4 \cdot 8 m x + 4 m^{2} + 4 (- 8 m) + 4 \cdot 16

Apply minus-plus rules

+ (- a) = - a

= 4 m^{2} x^{2} - 4 \cdot 2 m^{2} x + 4 \cdot 8 m x + 4 m^{2} - 4 \cdot 8 m + 4 \cdot 16

Simplify

4 m^{2} x^{2} - 4 \cdot 2 m^{2} x + 4 \cdot 8 m x + 4 m^{2} - 4 \cdot 8 m + 4 \cdot 16 : 4 m^{2} x^{2} - 8 m^{2} x + 32 m x + 4 m^{2} - 32 m + 64

= 4 m^{2} x^{2} - 8 m^{2} x + 32 m x + 4 m^{2} - 32 m + 64

e^{x + 0.6} - 2 \cdot (x + 0.7)^{2}

d (t (1 - t) (1 - 2 t)) + a (t (1 - t) (1 - 2 t))

\frac{( 12 v ^{3} + 38 v ^{2} - 16 v - 20 )}{( 6 v - 5 )}

\frac{- 3 x ^{2} - 4 x - 5}{x ( x + 1 ) ( x - 5 )}

(\frac{( 16 x + 1 )}{2})^{2}

What is expand 4(mx-m+4)^2 ?
The solution to expand 4(mx-m+4)^2 is 4m^2x^2-8m^2x+32mx+4m^2-32m+64