# How do you find all the zeros of P(x)=x^4+6x^2+9?

Mar 4, 2016

${x}_{1} = {x}_{3} = i \cdot \sqrt{3} , {x}_{2} = {x}_{4} = - i \cdot \sqrt{3}$

#### Explanation:

The polynomial is of a biquadratic kind so, making
${x}^{2} = y$
And equating the polynomial to zero we have
${y}^{2} + 6 y + 9 = 0$

$\Delta = - 36 - 36 = 0$
$y = \frac{- 6 \pm 0}{2}$ => ${y}_{1} = {y}_{2} = - 3$
Then
${x}_{1 , 2} = \sqrt{{y}_{1}} = \sqrt{- 3}$ => ${x}_{1} = i \cdot \sqrt{3}$, ${x}_{2} = - i \cdot \sqrt{3}$
${x}_{3 , 4} = \sqrt{{y}_{2}} = \sqrt{- 3}$ => ${x}_{3} = i \cdot \sqrt{3}$, ${x}_{4} = - i \cdot \sqrt{3}$
So
${x}_{1} = {x}_{3} = i \cdot \sqrt{3}$
${x}_{2} = {x}_{4} = - i \cdot \sqrt{3}$