Test 18 (Lessons 35–36): The Fundamental Theorem of Algebra and Variation Solutions

Explain the number of roots and maximum number of turning points for $q (x) = x^{3} (x^{2} + 4) (x - 8) .$

n = 3 + 2 + 1 = 6

n − 1 = 6 − 1 = 5

There will be 6 roots because the degree of q(x) is 6. The maximum number of possible turning points is 5 because $6 - 1 = 5 .$

Sketch a fourth degree polynomial with a double root and two non-real, complex roots when $a < 0 .$

Sample:

Note

double root = bounce
non-real, complex = does not intersect x-axis

For problems 3–5, use technology and the equation: $f (x) = (x^{3} - x) (x^{2} - 2)$

Sketch a graph of the equation.

Label all roots and turning points on the graph.

$x (x^{2} - 1) (x^{2} - 2) = 0 \begin{matrix} x^{2} = 1 & x^{2} = 2 \\ \sqrt{x^{2}} = \sqrt{1} & \sqrt{x^{2}} = \sqrt{2} \\ x = 0 & x = \pm 1 & x = \pm \sqrt{2} \end{matrix}$

Determine the relative minimum and maximum across the interval $[- 1.5, 0] .$

Across the interval $[- 1.5, 1],$ the relative maximum point is $(- 1.241, 0.308)$ and the relative minimum point is $(- 0.51, - 0.657) .$

Fill in the blanks.

g varies jointly as the square of h and j, and inversely as the square of m, can be modeled by the equation $g = \frac{k h^{2} j}{m^{2}} .$

In the equation $Q = \frac{π r^{2} v}{t}$ , Q varies jointly as $r^{2} v,$ and inversely as t with the constant of variation $π or pi .$

Determine the constant of variation if y varies inversely as x when $y = \frac{5}{2}$ and $x = \frac{12}{65} .$

$y = \frac{k}{x} k = x y k = (\frac{12}{65}) (\frac{5}{2}) = \frac{6}{13}$

$k = \frac{6}{13}$

When the rate is constant, the distance traveled is directly related to the time. If you drive 24.75 miles in three-quarters of an hour, how long will it take you to travel 82.5 miles?

$\begin{matrix} d = k t o r d = r t \\ 24.75 = r (0.75) & d = 33 t \\ r = 33 & 82.5 = 33 t \\ t = 2.5 \end{matrix}$

It will take 2.5 hours to travel 82.5 miles at a rate of 33 mph.

y varies jointly as x, and the square of z.
If $y = - 40$ when $x = 5,$ and $z = 4,$ find y when $x = 0.3$ and $z = - 5 .$

$\begin{matrix} y = k x z^{2} \\ - 40 = k (5) {(4)}^{2} & y = - 0.5 x z^{2} \\ k = \frac{- 40}{(5) (16)} & y = (- 0.5) (0.3) {(- 5)}^{2} \\ k = - \frac{1}{2} = - 0.5 & y = - 3.75 \end{matrix}$

y = –3.75