Question

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

${\mathit{x}}^{\mathbf{4}}\mathbf{-}\mathbf{4}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{14}\mathit{x}\mathbf{+}\mathbf{26}$; root $\mathbf{3}\mathbf{+}\mathbf{2}\mathit{i}$

Short answer

Roots of the polynomial$x^4-4x^3+3x^2+14x+26$ are$3+2i,3-2i,-1+i,-1-i$

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 a set of complex numbers$-$, then $x-iy$is also a root of $f\left(x\right)$.

By using this lemma, we can say that $3-2i$ is also a root of$f\left(x\right)$

Therefore, we can write$x-\left(3-2i\right)=0$or$x-\left(3-2i\right)=0$

Now, the quadratic polynomial with the roots $3+2i$and $3-2i$

$$\begin{gathered} x^2-\left[\left(3+2i\right)+\left(3-2i\right)\right]x+\left(3+2i\right)\left(3-2i\right)=x^2-6x+{\left(3\right)}^2-{\left(2i\right)}^2 \\ =x^2-6x+13 \end{gathered}$$

Therefore,$x^2-6x+13$be the factor of $x^4-4x^3+3x^2+14x+26$.

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

Then$,$$\left(x^2+ax+b\right)\left(x^2-6x+13\right)=x^4-4x^3+3x^2+14x+26$.

Find the value of a and b

Comparing the constant terms,

$13b=26\Rightarrow b=2$

Comparing the coefficient of terms,

$$\begin{gathered} 13a-6b=14 \\ \Rightarrow 13a=26 \\ \Rightarrow a=2 \end{gathered}$$

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

Now, roots of the above polynomial$\underline{s}$are $-1+i,-1-i$

Therefore, the roots of the polynomial $x^4-4x^3+3x^2+14x+26$are: $3+2i,3-2i,-1+i,-1-i$.