1. If A = 4 + 2i and B = 7 -3i, A – B = ? 2. The graph of f(x) = x2 + 6x + 10 hits the x-axis how many times? 3. If A = 3 + 2i and B = 4 – i, then AB would be?

Question

1. If A = 4 + 2i and B = 7 -3i, A – B = ?
2. The graph of f(x) = x2 + 6x + 10 hits the x-axis how many times?
3. If A = 3 + 2i and B = 4 – i, then AB would be?
4. What are the possible rational roots of f(x) = x4 -7x + 8?
5. If I know a real root of f(x) = x3 -6x2 + 11x – 6 is neither 1, even, or negative, a good guess would be?

Answer 1

1)

 

A - B = (4 + 2i) - (7 - 3i)

 

 

Now subtract.

 

 

2)

 

When the graph hits the x-axis, we have x-intercepts.  To find the x-intercepts, we set f(x) equal to zero.

 

0 = x² + 6x + 10

 

 

Since we cannot factor the right side of the equation, we must use the quadratic formula:

 

x = (-b ± √(b² - 4ac)) / 2a

 

where:

a = 1

b = 6

c = 10

 

Plug in these values into the formula to solve for x.  You will have two solutions because of the plus/minus sign. 

 

Note: Only real solutions can be x-intercepts. The number of distinct (non-repeating) solutions determines the number of times f(x) hits the x-axis.

 

 

3)

 

AB = (3 + 2i)(4 - i)

 

Use FOIL to multiply.  Also, use the fact that   i² = -1

 

 

4)

 

The possible real rational roots are determined by examining the factors of the function's last term.  The last term of f(x) is 8.  So the possible rational roots are ±1 , ±2 , ±4 , and  ±8.

 

 

5)
 
According to the rational root theorem, the only possible roots are  ±1 , ±2 , ±3 , and  ±6.   However, the conditions are that the root is not 1, even, nor negative.  So the only root we can have is 3. 
 
If we evaluate f(3), we get a result of 0.  This makes 3 a root.