Question

Solve the given equation
Q. $x^4-3x^3-2x^2+3x+1=0$

Answer 1

Dear student
$$\begin{gathered} \mathrm{We}\mathrm{have} \\ {\mathrm{x}}^4-3{\mathrm{x}}^3-2{\mathrm{x}}^2+3\mathrm{x}+1=0 \\ \Rightarrow \left({\mathrm{x}}^2-\mathrm{x}+\mathrm{x}-1\right)\left({\mathrm{x}}^2-3\mathrm{x}-1\right)=0 \\ \Rightarrow \left(\mathrm{x}\left(\mathrm{x}-1\right)+1\left(\mathrm{x}-1\right)\right)\left({\mathrm{x}}^2-3\mathrm{x}-1\right)=0 \\ \Rightarrow \left(\mathrm{x}+1\right)\left(\mathrm{x}-1\right)\left({\mathrm{x}}^2-3\mathrm{x}-1\right)=0 \\ \Rightarrow \mathrm{x}+1=0\Rightarrow \mathrm{x}=-1 \\ \mathrm{or}\mathrm{x}-1=0\Rightarrow \mathrm{x}=1 \\ \mathrm{or}{\mathrm{x}}^2-3\mathrm{x}-1=0 \\ \Rightarrow \mathrm{x}\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^2-4\left(1\right)\left(-1\right)}}{2\left(1\right)}\left[\mathrm{using}\mathrm{quadratic}\mathrm{formula}\right] \\ \Rightarrow \mathrm{x}\frac{3\pm \sqrt{9+4}}{2} \\ \Rightarrow \mathrm{x}=\frac{3\pm \sqrt{3}}{2} \end{gathered}$$
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