Question

Find all the roots $\square$in of polynomial (one root is already given)

localid="1653638332854" ${\mathit{x}}^{\mathbf{4}}\mathbf{-}\mathbf{2}{\mathit{x}}^{\mathbf{3}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathit{x}\mathbf{-}\mathbf{6}$; root $\mathbf{1}\mathbf{+}\mathit{i}$

Short answer

Roots polynomial for$x^4-2x^3-x^2+6x-6$are $1+i,1-i,\sqrt{3},-\sqrt{3}$

Step-by-step solution

Use the Lemma 4.29

It says that if $f\left(x\right)$is a polynomial in $\square \left[x\right]$and $x+iy$is a root of $f\left(x\right)$in set of complex numbers$-$, then localid="1653639484455" $x-iy$is also a root of $f\left(x\right)$.

By using this lemma, we can say that $1-i$ is also a root of$f\left(x\right)$

Therefore, we can write$x-\left(1-i\right)=0$or$x-\left(1+i\right)=0$

Now, quadratic polynomial with the roots $1+i$and$1-i$

$$\begin{gathered} x^2-\left[\left(1+i\right)+\left(1-i\right)\right]x+\left(1+i\right)\left(1-i\right)=x^2-2x+{\left(1\right)}^2-{\left(i\right)}^2 \\ =x^2-2x+2 \end{gathered}$$

Therefore,$\left(x^2-2x+2\right)$be the factor of $x^4-2x^3-x^2+6x-6$

Let$\left(x^2+ax+b\right)$be the other factor

Then$\underline{,}$$\left(x^2+ax+b\right)\left(x^2-2x+2\right)=x^4-2x^3-1x^2+6x-6$

Find the value of a and b

Comparing the constant terms,

$2b=-6\Rightarrow b=-3$

Comparing the coefficient of $x$terms,

$$\begin{gathered} 2a-2b=6 \\ \Rightarrow 2a=0 \\ \Rightarrow a=0 \end{gathered}$$

Therefore,$\left(x^2+ax+b\right)\Rightarrow x^2-3$

Now, the roots of above polynomials are $\sqrt{3},-\sqrt{3}$

Therefore, roots of the polynomial $x^4-2x^3-x^2+6x-6$are $1+i,1-i,\sqrt{3},-\sqrt{3}$