Part 3
1. Be sure your work is shown and steps are in order.
2. List all the possible rational roots.
3. Use synthetic division to test the possible rational roots and factor possible.
4. Identify complex roots if possible.
5. Sketch a graph of the function, please show:
a. the end behavior correctly
b. the shape of the near x-intercepts
c. (if possible) anything you can learn by considering symmetry and transformations.

3a. p(x)= 2x^5 - 9x^4 + 6x^3 + 22x^2 - 20x - 25

Answer:

   possible rational roots: ±1/2, ±1, ±5/2, ±5, ±25/2, ±25

   p(x) = (x +1)^2(2x -5)(x^2 -4x +5)

   complex roots: 2±i

   see the last attachment for a graph

Step-by-step explanation:

2. The leading coefficient of p(x) is 2, and the constant term is 25. The Rational Root Theorem tells you possible rational roots will be of the form ...

  ±(divisor of 25)/(divisor of 2)

That is, they are ...

  ±1/2, ±1, ±5/2, ±5, ±25/2, ±25

__

3. Before we get into synthetic division, we choose to see if we can reduce this list any. We note that p(0) = -25. The value of p(1) is the sum of the coefficients:

  p(1) = 2 -9 +6 +22 -20 -25 = 30 -54 = -24

Similarly, the value of p(-1) is the same sum with odd-degree coefficients negated:

  p(-1) = -2 -9 -6 +22 +20 -25 = 42 -42 = 0

So, we found our first root: -1. Using synthetic division, we can reduce the polynomial and start over. See the first attachment for this division.

__

The reduced polynomial is ...

  p1(x) = 2x^4 -11x^3 +17x^2 +5x -25

We already know that +1 is not a of it. Checking -1, we have ...

  p1(-1) = 2 +11 +17 -5 -25 = 0

So, we found our second root: -1. Using synthetic division, we can reduce the polynomial and start over. See the second attachment for this division.

__

The reduced polynomial is ...

  p2(x) = 2x^3 -13x^2 +30x -25

The alternating signs tell us there are no more negative real roots. They also tell us there are 1 or 3 positive real roots. We know p2(0) = -25. Then ...

  p2(1) = 2 -13 +30 -25 = 32 -38 = -6

The average rate of change between these points is (-6 -(-25))/(1 -0) = 19. At this rate, we expect a root between x=1 and x=2. Testing x=2 using synthetic division, we get a remainder of -1. (See the 3rd attachment.) Then the rate of change between x=1 and x=2 is (-1 -(-6))/(2-1) = 5, suggesting x=5/2 might be a worthwhile test value.

The synthetic division is shown in the 4th attachment. You will note that we divide the polynomial p2(x) by its leading coefficient, so the coefficients used for p2(x) in the synthetic division are 1, -13/2, 15, -25/2. The remainder of 0 tells us that (x -5/2) is a factor of p2(x)/2, or (2x -5) is a factor of p2(x).

__

The reduced polynomial is ...

  p3(x) = x^2 -4x +5

This can be written in vertex form as ...

  p3(x) = (x -2)^2 +1

The positive leading coefficient means the graph opens upward, and the vertex at (2, 1) means there are no real solutions.

The real solutions to p(x) are x = -1, -1, and 5/2.

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4. The complex solutions will be the solutions to ...

  (x -2)^2 +1 = 0

  (x -2)^2 = -1

  x -2 = ±√(-1) = ±i

  x = 2 ±i . . . . complex roots of p(x)

__

5. The graph is shown in the last attachment. The odd degree and positive leading coefficient of p(x) means the overall shape will be from lower left to upper right (/). That is, the sign of the end value of p(x) will match the sign of x.

The graph will touch the x-axis from below at x = -1, and will cross at x = 2.5. There is no particular symmetry.

The final quadratic factor is graphed and its vertex shown. The vertex matches that of the vertex-form equation for p3(x), above.

Answer:

possible rational roots: ±1/2, ±1, ±5/2, ±5, ±25/2, ±25

real solutions to p(x) are x = -1, -1, and 5/2.

(x -2)^2 +1 = 0

(x -2)^2 = -1

x -2 = ±√(-1) = ±i

x = 2 ±i complex roots

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O A.
A/WIN/Holx
y
16
23
30
37
44
O B.
Х
0
1
2
3
4
y
30
36
42
48
54
O c.
х
0
1
2
3
4
11
12
14
18
26
OD.
Х
0
1
2
3
4
y
10
18
26
34
42

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