Polynomial roots

Find the other zeroes of the polynomial 2x⁴+3x³-3x²-6x-2, if two of them are root2 & -root2

Correct answer:

x₃ =  -0.5
x₄ =  -1

Step-by-step explanation:

$$\begin{gathered} x_1=\sqrt{2}=\sqrt{2}\doteq 1.4142 \\ {x}_2=-\sqrt{2}=-\sqrt{2}\doteq -1.4142 \\ (x-x_1)(x-x_2)=x^2-2 \\ 2x^4+3x^3-3x^2-6x-2=(x^2-2)(2x^2+3x+1) \\ 2z^2+3z+1=0 \\ 2z^2+3z+1=0 \\ 2z^2+3z+1=0 \\ a=2;b=3;c=1 \\ D=b^2-4ac=3^2-4\cdot 2\cdot 1=1 \\ D>0 \\ {z}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{-3\pm \sqrt{1}}{4}} \\ {z}_{1,2}={\displaystyle \frac{-3\pm 1}{4}} \\ {z}_{1,2}=-0.75\pm 0.25 \\ {z}_1=-0.5 \\ {z}_2=-1 \\ \text{ Factored form of the equation: } \\ 2(z+0.5)(z+1)=0 \\ {x}_3=z_1=(-0.5)=-0.5 \end{gathered}$$

Our quadratic equation calculator calculates it.

$x_4=z_2=(-1)=-1$

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