The equation x^4-x^3-6x^2+4x+8=0 has two equa | vedclass

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(B) Given the equation $x^4-x^3-6x^2+4x+8=0$. By testing values,we find $x=2$ is a root. Dividing the polynomial by $(x-2)$,we get $(x-2)(x^3+x^2-4x-4

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$5$

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(B) Given the equation $x^4-x^3-6x^2+4x+8=0$. By testing values,we find $x=2$ is a root. Dividing the polynomial by $(x-2)$,we get $(x-2)(x^3+x^2-4x-4)=0$. Testing $x=2$ again in the cubic factor $x^3+x^2-4x-4$,we get $8+4-8-4=0$,so $x=2$ is a root again. Dividing $(x^3+x^2-4x-4)$ by $(x-2)$,we get $(x-2)(x^2+3x+2)=0$. Factoring the quadratic part: $(x^2+3x+2) = (x+1)(x+2)$. Thus,the roots are $x=2, 2, -1, -2$. The two equal roots are $2, 2$. The other two roots are $\alpha = -1$ and $\beta = -2$. Therefore,$\alpha^2+\beta^2 = (-1)^2 + (-2)^2 = 1 + 4 = 5$.

The equation $x^4-x^3-6x^2+4x+8=0$ has two equal roots. If $\alpha$ and $\beta$ are the other two roots of this equation,then $\alpha^2+\beta^2=$

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