Solving Polynomial Equations

To solve the equation, we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Plugging in the values $a = 1$, $b = -4$, and $c = 3$, we get $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(3)}}{2(1)} = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm \sqrt{4}}{2} = \frac{4 \pm 2}{2}$. Therefore, the solutions are $x = 1$ and $x = 3$.

To find the roots of the polynomial, we can use synthetic division or the rational root theorem. Using synthetic division, we find that $x = 1$ is a root. Dividing the polynomial by $x - 1$, we get $x^2 - x - 6$. Factoring this quadratic, we get $(x - 3)(x + 2)$. Therefore, the roots of the polynomial are $x = 1, 2, 3$.

To solve the equation, we can use the rational root theorem or Descartes' rule of signs. Using the rational root theorem, we find that $x = 1$ is a root. Dividing the polynomial by $x - 1$, we get $2x^3 - 3x^2 + x - 2$. Factoring this cubic, we get $(2x - 1)(x^2 - x + 2)$. Solving the quadratic $x^2 - x + 2 = 0$, we get $x = \frac{1 \pm \sqrt{1 - 4(1)(2)}}{2(1)} = \frac{1 \pm \sqrt{-7}}{2}$. Therefore, the roots of the polynomial are $x = 1, 2, -1, -2$.

To find the roots of the polynomial, we can use the rational root theorem or Descartes' rule of signs. Using the rational root theorem, we find that $x = 1$ and $x = -1$ are roots. Dividing the polynomial by $x - 1$, we get $x^4 - 2x^3 - x^2 + x + 2$. Dividing this polynomial by $x + 1$, we get $x^3 - 3x^2 + x - 2$. Factoring this cubic, we get $(x - 2)(x^2 - x + 1)$. Solving the quadratic $x^2 - x + 1 = 0$, we get $x = \frac{1 \pm \sqrt{1 - 4(1)(1)}}{2(1)} = \frac{1 \pm \sqrt{-3}}{2}$. Therefore, the roots of the polynomial are $x = 1, -1, 2, -2, i$.

To solve the equation, we can use the rational root theorem or Descartes' rule of signs. Using the rational root theorem, we find that $x = 1$ and $x = -1$ are roots. Dividing the polynomial by $x - 1$, we get $x^5 - x^4 - 2x^3 + x^2 + 3x - 7$. Dividing this polynomial by $x + 1$, we get $x^4 - 2x^3 - 3x^2 + 4x + 7$. Factoring this quartic, we get $(x^2 - x + 7)(x^2 - x - 1)$. Solving the quadratic $x^2 - x + 7 = 0$, we get $x = \frac{1 \pm \sqrt{1 - 4(1)(7)}}{2(1)} = \frac{1 \pm \sqrt{-27}}{2}$. Solving the quadratic $x^2 - x - 1 = 0$, we get $x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}$. Therefore, the roots of the polynomial are $x = 1, -1, 2, -2, i, -i$.