Question

Find the zeros of polynomial functions and solve polynomial equations. f(x)=8x^3+27

Answer 1

$8x^3+27=0\\\\ 8x^3=-27\\\\ x^3=-\dfrac{27}{8}\\\\ x=\sqrt[3]{-\dfrac{27}{8}}\\\\ x=-\dfrac{3}{2}$

Answer 2

f(x) = 8x³ + 27

To find the zeros equal the equation to 0

8x³ + 27 = 0

8x³ = -27

x³ = -27/8

x = $\sqrt[3]{\frac{-27}{8}}$

x = $\frac{-3}{2}$

This is the real root, but there are two other irrational roots.

By rule we know:

For f(z) = x³ the three solutions are:

x = $\sqrt[3]{f(z)}$

x = $\sqrt[3]{f(z)}.\frac{-1-\sqrt{3}i}{2}$

x = $\sqrt[3]{f(z)}.\frac{-1+\sqrt{3}i}{2}$

Making some subs.

x = $\frac{-3}{2}$

x = $\frac{-3}{2}.\frac{-1-\sqrt{3}i}{2}$

x = $\frac{-3}{2}.\frac{-1+\sqrt{3}i}{2}$

So, after all we have:

x = $\frac{-3}{2}$

x = $\frac{3+3\sqrt{3}i}{4}$

x = $\frac{3-3\sqrt{3}i}{4}$