Solution Bank Polynomials Target D
Answers to the Practice Problems are below in random order.
Solution Bank for Questions 1-16
Answers to the Practice Problems are below in random order.
Solution Bank for Questions 1-16
\(\dfrac{p}{q}:\pm\dfrac{1}{2},\:1,\:\frac{3}{2},\:2,\:3,\:6\) |
Julia is correct; the Rational Roots Theorem only gives values that are possible rational roots. Using synthetic division, \(x=-1\) is the only \(\dfrac{p}{q}\) root that gives a remainder of zero. |
\(x=\pm2,\:\pm2i\) |
No |
Yes |
Yes |
\(\frac{p}{q}:\pm\dfrac{1}{3},\:1,\:3\) |
\(x=\pm\sqrt{5},\:\pm2i\sqrt{2}\) |
\(x=0, \:\pm\sqrt{3}, \:\pm i\sqrt{6}\) |
\(x=\pm\dfrac{5}{3}i\) |
\(x=0,\:\pm\sqrt{3},\:\pm{i}\sqrt{3}\) |
\(x=\pm{i}\sqrt{2},\:\pm\sqrt{7}\) |
\(\frac{p}{q}:\pm\dfrac{1}{6},\:\dfrac{1}{3},\:\dfrac{1}{2},\:\dfrac{5}{6},\:1,\:\dfrac{3}{2}, \:\dfrac{5}{3},\:\dfrac{5}{2},\:3,\:5,\:\dfrac{15}{2},\:15\) |
\(x=-3,\:\dfrac{3}{2}\pm\dfrac{3i\sqrt{3}}{2}\) |
\(x=-3,\:0,\:2\) |
\(x=2,\:-1\pm{i}\sqrt{3}\) |
Solution Bank for Questions 17-30
\(x=-5,3\pm\sqrt{11}\) |
\(x=-3,\:\dfrac{5\pm\sqrt{17}}{2}\) |
\(x=-\dfrac{5}{2},\:5\pm\sqrt{22}\) |
\(x=-3,1\pm3i\) |
\(x=1,\:-3\pm\sqrt{5}\) |
\(\dfrac{8x^{21}}{27y^9z^{15}}\) |
\(x=-2,\:x=\pm i\sqrt{2},\:x=\pm\sqrt{6}\) |
\(x=-1,\:3,\:\pm{i}\sqrt{5}\) |
\(\dfrac{1}{x}\) |
\(x=-3,\:\dfrac{3}{10}\pm{i}\dfrac{\sqrt{11}}{10}\) |
\(x=-2,\:4,\:\dfrac{-3\pm\sqrt{21}}{2}\) |
Degree: 4; LC: -2 End Behavior: As \(\:x\ \longrightarrow\ -\infty,\ f(x)\ \longrightarrow\ -\infty,\) as \(\: x\ \longrightarrow\ +\infty,\ f(x)\ \longrightarrow\ -\infty\) X-intercepts: (-2,0) (-1,0) (1,0); Y-intercept: (0,8) |
\(x=-\dfrac{2}{3},\:\dfrac{3}{2}\pm\dfrac{1}{2}{i}\) |
\(x=-4,\dfrac{1}{2},\:3\) |