The polynomial equation of degree 4 having re | vedclass

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(D) Given that the polynomial equation of degree $4$ has real coefficients. Since complex roots occur in conjugate pairs,if $1+2i$ is a root,then $1-2

Vedclass · Accepted Answer

$x^4-6x^3+14x^2-22x+5=0$

Vedclass · Answer

(D) Given that the polynomial equation of degree $4$ has real coefficients. Since complex roots occur in conjugate pairs,if $1+2i$ is a root,then $1-2i$ must also be a root. The roots are $2+\sqrt{3}$,$2-\sqrt{3}$,$1+2i$,and $1-2i$. The quadratic factor corresponding to roots $2 \pm \sqrt{3}$ is: $(x-(2+\sqrt{3}))(x-(2-\sqrt{3})) = (x-2)^2 - 3 = x^2-4x+4-3 = x^2-4x+1$. The quadratic factor corresponding to roots $1 \pm 2i$ is: $(x-(1+2i))(x-(1-2i)) = (x-1)^2 - (2i)^2 = x^2-2x+1+4 = x^2-2x+5$. The required polynomial is the product of these two factors: $(x^2-4x+1)(x^2-2x+5) = x^2(x^2-2x+5) - 4x(x^2-2x+5) + 1(x^2-2x+5)$ $= x^4-2x^3+5x^2-4x^3+8x^2-20x+x^2-2x+5$ $= x^4-6x^3+14x^2-22x+5 = 0$. Thus,the correct option is $D$.

The polynomial equation of degree $4$ having real coefficients with three of its roots as $2 \pm \sqrt{3}$ and $1+2i$ is:

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