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(C) Let $P(x) = x^4-4x^3+3x^2+2x-2$. By the Rational Root Theorem,possible integer roots are $\pm 1, \pm 2$. Testing $x=1$: $1-4+3+2-2 = 0$. So,$(x-1)

Vedclass · Accepted Answer

$11$

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(C) Let $P(x) = x^4-4x^3+3x^2+2x-2$. By the Rational Root Theorem,possible integer roots are $\pm 1, \pm 2$. Testing $x=1$: $1-4+3+2-2 = 0$. So,$(x-1)$ is a factor. Testing $x=-1$: $1+4+3-2-2 = 4 
eq 0$. Testing $x=2$: $16-32+12+4-2 = -2 
eq 0$. Testing $x=-2$: $16+32+12-4-2 = 54 
eq 0$. Since $x=1$ is a root,divide $P(x)$ by $(x-1)$ to get $x^3-3x^2+2$. Testing $x=1$ again in $x^3-3x^2+2$: $1-3+2 = 0$. So,$(x-1)^2$ is a factor. Dividing $x^3-3x^2+2$ by $(x-1)$ gives $x^2-2x-2$. Thus,the roots are $\alpha=1, \beta=1$ and the roots of $x^2-2x-2=0$,which are $\gamma, \delta = 1 \pm \sqrt{3}$. We need to calculate $\alpha+2\beta+\gamma^2+\delta^2$. Since $\gamma, \delta$ are roots of $x^2-2x-2=0$,$\gamma+\delta=2$ and $\gamma\delta=-2$. Then $\gamma^2+\delta^2 = (\gamma+\delta)^2 - 2\gamma\delta = (2)^2 - 2(-2) = 4+4 = 8$. Substituting the values: $1 + 2(1) + 8 = 1+2+8 = 11$.

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4-4x^3+3x^2+2x-2=0$ such that $\alpha$ and $\beta$ are integers and $\gamma, \delta$ are irrational numbers,then $\alpha+2\beta+\gamma^2+\delta^2=$

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