Practice 2 Solutions

Given the root(s), determine if there are any missing roots and write an equation of the polynomial in standard form with integer coefficients.

$x = - 5, \frac{6}{7}$

$(x - (- 5)) = 0, x - \frac{6}{7} = 0 (x - (- 5)) (x - \frac{6}{7}) = 0 (x + 5) (7 x - 6) = 0 7 x^{2} - 6 x + 35 x - 30 = 0$

$7 x^{2} + 29 x - 30 = 0$

$x = 2 - i$

$\begin{array}{ll} x = 2 - i, & x = 2 + i \\ x - (2 - i) = 0, & x - (2 + i) = 0 \end{array} (x - (2 - i)) (x - (2 + i)) = 0 (x - 2 + i) (x - 2 - i) = 0 x^{2} - 2 x - i x - 2 x + 4 + 2 i + i x - 2 i - i^{2} = 0 x^{2} - 4 x + 4 - i^{2} = 0 x^{2} - 4 x + 4 - (- 1) = 0$

$x^{2} - 4 x + 5 = 0$

$x = 1, \sqrt{13}$

$\begin{array}{rcl} \end{array} \begin{array}{rcl} \begin{matrix} x = - \sqrt{13}, & x = \sqrt{13}, & x = 1 \end{matrix} \begin{matrix} (x - (- \sqrt{13})) = 0, & (x - \sqrt{13}) = 0, & x - 1 = 0 \end{matrix} \begin{array}{rcl} (x - (- \sqrt{13})) (x - \sqrt{13}) (x - 1) & = & 0 \\ (x + \sqrt{13}) (x - \sqrt{13}) (x - 1) & = & 0 \\ (x^{2} - x \sqrt{13} + x \sqrt{13} - \sqrt{169}) (x - 1) & = & 0 \\ (x^{2} - 13) (x - 1) & = & 0 \end{array} \end{array}$

$x^{3} - x^{2} - 13 x + 13 = 0$

$x = \pm i$

$\begin{matrix} x - (- i) = 0, & x - i = 0 \end{matrix} (x - (- i)) (x - i) = 0 (x + i) (x - i) = 0 x^{2} - i x + i x - i^{2} = 0 x^{2} - i^{2} = 0 x^{2} - (- 1) = 0$

$x^{2} + 1 = 0$

$x = \frac{1}{2}, 7 i$

$\begin{matrix} x = - 7 i, & x = 7 i, & x = \frac{1}{2} \\ x - (- 7 i) = 0, & x - 7 i = 0, & x - \frac{1}{2} = 0 \end{matrix} (x - (- 7 i)) (x - 7 i) (x - \frac{1}{2}) = 0 (x + 7 i) (x - 7 i) (2 x - 1) = 0 (x^{2} - 7 i x + 7 i x - 49 i^{2}) (2 x - 1) = 0 (x^{2} - 49 (- 1)) (2 x - 1) = 0 (x^{2} + 49) (2 x - 1) = 0$

$2 x^{3} - x^{2} + 98 x - 49 = 0$

$x = 2 i, \sqrt{3}$

$\begin{matrix} x = - 2 i, & x = 2 i, & x = - \sqrt{3}, & x = \sqrt{3} \\ x - (- 2 i) = 0, & x - 2 i = 0, & x - (- \sqrt{3}) = 0, & x - \sqrt{3} = 0 \end{matrix} (x - (- 2 i)) (x - 2 i) (x - (- \sqrt{3})) (x - \sqrt{3}) = 0 (x + 2 i) (x - 2 i) (x + \sqrt{3}) (x - \sqrt{3}) = 0 (x^{2} - 2 i x + 2 i x - 4 i^{2}) (x^{2} - x \sqrt{3} + x \sqrt{3} - \sqrt{9}) = 0 (x^{2} - 4 i^{2}) (x^{2} - 3) = 0 (x^{2} - 4 (- 1)) (x^{2} - 3) = 0 (x^{2} + 4) (x^{2} - 3) = 0 x^{4} - 3 x^{2} + 4 x^{2} - 12 = 0$

$x^{4} + x^{2} - 12 = 0$

Solve.

$4 x^{2} - 4 x \sqrt{5} + 5 = 0$

$\begin{array}{rcl} (2 x - \sqrt{5}) (2 x - \sqrt{5}) & = & 0 \\ 2 x - \sqrt{5} & = & 0 \end{array}$

$x = \frac{\sqrt{5}}{2}$

Note

The Conjugate Root Theorem is not necessary here because the middle coefficient is irrational.

${(2 x + 5)}^{2} = 3$

$\begin{array}{rcl} \sqrt{{(2 x + 5)}^{2}} & = & \pm \sqrt{3} \\ 2 x + 5 & = & \pm \sqrt{3} \\ 2 x & = & - 5 \pm \sqrt{3} \end{array}$

$x = \frac{- 5 \pm \sqrt{3}}{2}$

$x^{4} - 3 x^{2} - 70 = 0$

$\begin{array}{rcl} (x^{2} - 10) (x^{2} + 7) & = & 0 \\ (x^{2} - 10) (x^{2} - (- 7)) & = & 0 \\ (x + \sqrt{10}) (x - \sqrt{10}) (x + i \sqrt{7}) (x - i \sqrt{7}) & = & 0 \end{array}$

$x = \pm \sqrt{10}, \pm i \sqrt{7}$

$x^{3} - x^{2} + 16 x - 16 = 0$

$\begin{array}{rcl} x^{2} (x - 1) + 16 (x - 1) & = & 0 \\ (x^{2} + 16) (x - 1) & = & 0 \\ (x^{2} - (- 16)) (x - 1) & = & 0 \\ (x + 4 i) (x - 4 i) (x - 1) & = & 0 \end{array}$

$x = \pm 4 i, 1$

${(7 x - 5)}^{2} + 10 = 0$

$\begin{array}{rcl} {(7 x - 5)}^{2} & = & - 10 \\ \sqrt{{(7 x - 5)}^{2}} & = & \pm \sqrt{- 10} \\ 7 x - 5 & = & \pm i \sqrt{10} \\ 7 x & = & 5 \pm i \sqrt{10} \end{array}$

$x = \frac{5 \pm i \sqrt{10}}{7}$

$x^{2} + 2 x \sqrt{7} + 7 = 0$

$(x + \sqrt{7}) (x + \sqrt{7}) = 0$

$x = - \sqrt{7}$

Note

The Conjugate Root Theorem is not necessary here because the middle coefficient is irrational. This is an example of a repeated root.

$4 x^{3} - 8 x^{2} + 100 x - 200 = 0$

$\begin{array}{rcl} 4 (x^{3} - 2 x^{2} + 25 x - 50) & = & 0 \\ 4 (x^{2} (x - 2) + 25 (x - 2)) & = & 0 \\ 4 (x^{2} + 25) (x - 2) & = & 0 \\ 4 (x^{2} - (- 25)) (x - 2) & = & 0 \\ 4 (x + 5 i) (x - 5 i) (x - 2) & = & 0 \end{array}$

$x = \pm 5 i, 2$

$x^{3} - 10 x^{2} = 0$

$x^{2} (x - 10) = 0 x^{2} = 0, x - 10 = 0 \sqrt{x^{2}} = \pm \sqrt{0}$

$x = 0, 10$

$4 {(x - 3)}^{2} + 8 = 0$

$\begin{array}{rcl} 4 {(x - 3)}^{2} & = & - 8 \\ {(x - 3)}^{2} & = & - 2 \\ \sqrt{{(x - 3)}^{2}} & = & \pm \sqrt{- 2} \\ x - 3 & = & \pm i \sqrt{2} \end{array}$

$x = 3 \pm i \sqrt{2}$

${(4 x + 9)}^{2} - 16 = 0$

$\begin{array}{rcl} {(4 x + 9)}^{2} & = & 16 \\ \sqrt{{(4 x + 9)}^{2}} & = & \pm \sqrt{16} \\ 4 x + 9 & = & \pm 4 \\ 4 x & = & - 9 \pm 4 \\ x & = & \frac{- 9 \pm 4}{4} \end{array}$

$x = - \frac{13}{4}, - \frac{5}{4}$

$x^{3} - 13 x^{2} + 4 x - 52 = 0$

$\begin{array}{rcl} x^{2} (x - 13) + 4 (x - 13) & = & 0 \\ (x^{2} + 4) (x - 13) & = & 0 \\ (x^{2} - (- 4)) (x - 13) & = & 0 \\ (x + 2 i) (x - 2 i) (x - 13) & = & 0 \end{array}$

$x = \pm 2 i, 13$

$x^{4} + 20 x^{2} + 64 = 0$

$\begin{array}{rcl} (x^{2} + 16) (x^{2} + 4) & = & 0 \\ (x^{2} - (- 16)) (x^{2} - (- 4)) & = & 0 \\ (x + 4 i) (x - 4 i) (x + 2 i) (x - 2 i) & = & 0 \end{array}$

$x = \pm 4 i, \pm 2 i$

$x^{2} - \frac{8}{9} = 0$

$\begin{array}{rcl} x^{2} & = & \frac{8}{9} \\ \sqrt{x^{2}} & = & \pm \sqrt{\frac{8}{9}} \end{array}$

$x = \pm \frac{2 \sqrt{2}}{3}$

$3 x^{3} - 12 x^{2} - 12 x + 48 = 0$

$\begin{array}{rcl} 3 (x^{3} - 4 x^{2} - 4 x + 16) & = & 0 \\ 3 (x^{2} (x - 4) - 4 (x - 4)) & = & 0 \\ 3 (x^{2} - 4) (x - 4) & = & 0 \\ 3 (x + 2) (x - 2) (x - 4) & = & 0 \end{array}$

$x = \pm 2, 4$

$5 x^{2} - 25 = 0$

$\begin{array}{rcl} 5 x^{2} & = & 25 \\ x^{2} & = & 5 \\ \sqrt{x^{2}} & = & \pm \sqrt{5} \end{array}$

$x = \pm \sqrt{5}$

${(x - 11)}^{2} + 8 = 0$

$\begin{array}{rcl} {(x - 11)}^{2} & = & - 8 \\ \sqrt{{(x - 11)}^{2}} & = & \pm \sqrt{- 8} \\ x - 11 & = & \pm i \sqrt{8} \\ x - 11 & = & \pm 2 i \sqrt{2} \end{array}$

$x = 11 \pm 2 i \sqrt{2}$