How do you solve $z (6 z - 5) (z + 5) = 0$ $z( 6z - 5) ( z + 5) = 0$ ?

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Answer 1 · 2017-11-11T05:43:54

distribute $z$ $z$ then foil

$z (6 z - 5) (z + 5) = 0$ $z(6z -5 )(z+5) = 0$

distribute z to the first parentheses term

$6 z \cdot z = 6 z^{2}$ $6z*z = 6z^2$

$(6 z^{2} - 5 z) (z + 5)$ $(6z^2 -5z)(z+5)$

now we foil ( First - Outside - Inner - Last)

F
$6 z^{2} \cdot z = 6 z^{3}$ $6z^2*z = 6z^3$

O

$6 z^{2} \cdot 5 = 30 z^{2}$ $6z^2*5= 30z^2$

I
$- 5 z \cdot z = - 5 z^{2}$ $-5z * z = -5z^2$

L

$- 5 z \cdot 5 = - 25 z$ $-5z * 5 = -25z$

Now we just bring like terms together keep in mind the signs
and the final answer should be

$6 z^{3} + 25 z^{2} - 25 z$ $6z^3 + 25z^2 - 25z$

How do you solve z( 6z - 5) ( z + 5) = 0z(6z−5)(z+5)=0z( 6z - 5) ( z + 5) = 0?