How do you find the complex roots of $a^{3} + 2 a^{2} - 8 a = 0$ $a^3+2a^2-8a=0$ ?

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Answer 1 · 2017-06-28T14:52:44

See below.

We can see that all of the variables have an $a$ $a$ , so we can factor this out immediately.

$a (a^{2} + 2 a - 8) = 0$ $a(a^2+2a-8)=0$

So, one solution is $a = 0$ $a=0$ . The other solutions pertain to the quadratic, which we can find using factoring.

$a^{2} + 2 a - 8 = 0$ $a^2+2a-8=0$

$a^{2} - 2 a + 4 a - 8 = 0$ $a^2-2a+4a-8=0$

$a (a - 2) + 4 (a - 2) = 0$ $a(a-2)+4(a-2)=0$

$(a + 4) (a - 2) = 0$ $(a+4)(a-2)=0$

So,

$a = 2, - 4, 0$ $a=2,-4,0$ , and there are no complex roots.

How do you find the complex roots of a^3+2a^2-8a=0a3+2a2−8a=0a^3+2a^2-8a=0?