What is factor ρzcos(φ)+(-z^2sin(2φ)) ?

The solution to factor ρzcos(φ)+(-z^2sin(2φ)) is z(ρcos(φ)-zsin(2φ))

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factor ρzcos(φ)+(-z^2sin(2φ))

factor

ρ z cos (φ) + (- z^{2} sin (2 φ))

z (ρ cos (φ) - z sin (2 φ))

Solution steps

ρ z cos (φ) + (- z^{2} sin (2 φ))

Remove parentheses:

(- a) = - a

= ρ z cos (φ) - z^{2} sin (2 φ)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

z^{2} = zz

= ρ z cos (φ) - zz sin (2 φ)

Factor out common term

z

= z (ρ cos (φ) - z sin (2 φ))

- 4 sin^{2} (8 x + 2)

(9)^{\frac{5}{6}}

e^{x^{2} y} + 2 e^{x^{2} y} x^{2} y

\frac{24 + 8 2}{7}

5760 (π)

What is factor ρzcos(φ)+(-z^2sin(2φ)) ?
The solution to factor ρzcos(φ)+(-z^2sin(2φ)) is z(ρcos(φ)-zsin(2φ))