Practice 1 Solutions

Explain the number of complex roots that can occur for a ninth degree polynomial.

A ninth degree polynomial will have nine complex roots according to the Fundamental Theorem of Algebra.

Explain if it is possible for a tenth degree polynomial to have three non-real, complex roots.

It is not possible to have three non-real, complex roots because they occur in conjugate pairs.

Sketch a graph that matches the given information.

Note

Remember a sketch does not reflect the exact scale of a graph. It is used to give a general idea of what is happening at key points on the graph.

A fifth degree polynomial function with a positive leading coefficient, with a single and a double real root.

Sample sketches:

A sixth degree polynomial function where $a < 0$ , and no real roots

Sample sketches:

A fourth degree polynomial function where $a > 0$ , and one real root

$4 - 1 = 3$ non-real, complex

This is not possible because non-real, complex roots occur in conjugate pairs.

A third degree polynomial function with a positive leading coefficient, with three real roots

Sample sketch:

What is the maximum number of turning points that can occur in a polynomial graph?

When n = the degree of the polynomial, $n - 1$ represents the maximum number of turning points.  

Explain why a polynomial graph will not always have the maximum number of turning points.

A polynomial graph may not have the maximum number of turning points when a root has a multiplicity greater than one.

Note

Remember, a polynomial with degree two or higher will have at least one turning point, even with multiplicities. (i.e., The parent function of a quadratic, x2.)

For problems 9–12: 

1. Sketch the equation. Include roots and turning points.
2. Name the interval(s) in which the function is increasing and decreasing.
3. Name the relative minimum and maximum across the given interval.

Note

Students should use technology to help create a sketch of the graph and include the roots and turning points.

$g (x) = {(x - 1)}^{3} {(x + 1)}^{2} (2 x + 1) across [- 1, 2]$

Increasing intervals: (–1, –0.702), (0.119, 1), (1, +∞) 
Decreasing intervals: (–∞, –1), (–0.702, 0.119) 
Relative minimum: (0.119, –1.06) 
Relative maximum: (–0.702, 0.177)

Note

You could also write the single interval $(0.119, + \infty)$ since the graph continues toward positive infinity over the entire interval.

$p (x) = x^{3} - 19 x + 30 across [- 10, 0]$

Increasing intervals: (–∞, –2.517), (2.517, +∞)
Decreasing intervals: (–2.517, 2.517)
Relative minimum: none
Relative maximum: (–2.517, 61.877)

Note

There is no relative minimum over the given interval. However, if the interval was extended to $[- 10, 3]$ , the relative minimum would be $(2.517, - 1.877)$ .

$f (x) = - x^{4} + x^{3} + 12 x^{2} across [- 3, 4]$

Increasing intervals: (–∞, –2.103), (0, 2.853)
Decreasing intervals: (–2.103, 0), (2.853, +∞)
Relative minimum: (0, 0)
Relative maximum: (2.853, 54.644)

$y = 6 x^{3} + x^{2} - 19 x + 6 across [- 2, 2]$

Increasing intervals: (–∞, –1.084), (0.973, +∞)
Decreasing intervals: (–1.084, 0.973)
Relative minimum: (0.973, –6.013)
Relative maximum: (–1.084, 20.128)

For problems 13–14: 

1. Sketch a graph that matches the equation.
2. Name all real and non-real, complex roots.

$k (x) = (x + 2) (2 x^{2} + 2 x + 1)$

Note

This problem has a third degree polynomial with one real root. You need to solve for the non-real, complex roots.

$2 x^{2} + 2 x + 1 = 0 x = \frac{- 2 \pm \sqrt{{(2)}^{2} - 4 (2) (1)}}{2 (2)} x = \frac{- 2 \pm \sqrt{- 4}}{2} = \frac{- 2 \pm 2 i}{2} x = - 1 \pm i$

$x = - 1 \pm i, - 2$

$f (x) = (x^{2} - 1) (x^{2} - 9)$

Note

This problem has a fourth degree polynomial with all real roots that can be determined from the graph using technology or by factoring the given function.

$x = \pm 3, \pm 1$