Question

f (x), has (x-7) and (x+4i) as factors. F(X)=2x^5-13x^4+22x^3-187x^2-160x+336 what is the total number of real numbers of real zeros of f(X)

Answer 1

Answer: 3

Step-by-step explanation:

(x + 4i) is a root which means its conjugate (x - 4i) is also a root

(x + 4i)(x - 4i) = x² + 16

Use Long Division for 2x⁵ - 13x⁴ + 22x³ - 187x² - 160x + 336 ÷ x² + 16

            2x³ - 13x² - 10x + 21                              

x² + 16 ) 2x⁵ - 13x⁴ + 22x³ - 187x² - 160x + 336

           - (2x⁵ + 0x⁴ + 32x³)   ↓          ↓         ↓

                       -13x⁴ - 10x³ - 187x²     ↓         ↓

                    - (-13x⁴ +0x³ - 208x²)    ↓         ↓

                                  -10x³ + 21x² - 160x     ↓

                               - (-10x³  + 0x² - 160x)    ↓

                                               21x² + 0x + 336

                                            - (21x² + 0x + 336)

                                                                      0

Use Synthetic Division to divide (x - 7) by the reduced polynomial (2x³ - 13x² - 10x + 21)

7 |  2   -13   -10   21

  |  ↓    14     7   -21        

    2      1     -3     0  ← remainder is 0      

Use grouping to factor the reduced polynomial (2x² + x - 3)  

 2x² - 2x + 3x - 3

= 2x(x - 1) +1(x - 1)

= (2x + 1)(x - 1)  

The factors are: (x - 7) (2x + 1) (x - 1) (x + 4i) (x - 4i)

The first 3 are real roots and the last 2 are complex roots