What are the solutions to the equation -a^3+12a^2-32a=0 ?

The solutions to the equation -a^3+12a^2-32a=0 are a=0,a=4,a=8

Find the zeros of -a^3+12a^2-32a=0

The zeros of -a^3+12a^2-32a=0 are a=0,a=4,a=8

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- a^{3} + 12 a^{2} - 32 a = 0

a = 0, a = 4, a = 8

Solution steps

- a^{3} + 12 a^{2} - 32 a = 0

Factor

- a^{3} + 12 a^{2} - 32 a : - a (a - 4) (a - 8)

- a (a - 4) (a - 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

a = 0 or a - 4 = 0 or a - 8 = 0

Solve

a - 4 = 0 : a = 4

Solve

a - 8 = 0 : a = 8

The solutions are

a = 0, a = 4, a = 8

X^{3} - 3 X^{2} + 20 = 0

s (s + 3) (s^{2} + 2 s + 2) = 0

\frac{256}{625} = 100 r^{6}

n^{3} = 343

4 X^{3} = 2 X^{2}

What are the solutions to the equation -a^3+12a^2-32a=0 ?
The solutions to the equation -a^3+12a^2-32a=0 are a=0,a=4,a=8
Find the zeros of -a^3+12a^2-32a=0
The zeros of -a^3+12a^2-32a=0 are a=0,a=4,a=8