Question 83034
<pre><font size = 5><b>
Solution to the problem: x^3-5x^2-7x+51

answer choices:

4+i, 4-i
4i, -4i
4, -4
4+i, 4-i

-----------------------------------------------------

x³ - 5x² - 7x + 51

The posible rational zeros are ± the factors of 51.

They are ±1, ±3, ±17, ±51

We'll start trying them with synthetic division.
We'll try 1, and it'll leave a remainder 40 (no good!)
Then we'll try -1 and it'll leave a remainder of 52,
(also no good!), then we'll try -3, and get this:

    -3| 1  -5  -7  51
      |<u>    -3  24 -51</u>
        1  -8  17   0

Whoopee! A zero remainder!

So the cubic polynomial

x³ - 5x² - 7x + 51

factors as

(x + 3)(x² - 8x + 17)

Now we can find the zeros of the polynomial

    x² - 8x + 17

by setting it = 0

    x² - 8x + 17 = 0

and solving it by the quadratic formula:
                   ______
             -b ± <font face = "symbol">Ö</font>b²-4ac
        x = ——————————————
                 2a 

where a = 1; b = -8; c = 17

                      ______________ 
             -(-8) ± <font face = "symbol">Ö</font>(-8)²-4(1)(17)
        x = —————————————————————————
                      2(1) 
                  _____ 
             8 ± <font face = "symbol">Ö</font>64-68
        x = ————————————
                 2
                  __ 
             8 ± <font face = "symbol">Ö</font>-4
        x = —————————
                2
                   _
             8 ± i<font face = "symbol">Ö</font>4
        x = —————————
                2
                   
             8 ± 21
        x = —————————
                2
                    
             8     2i
        x = ——— ± —————
             2      2
                 
        x = 4 ± i  
                      
So the zeros are 4 + i and 4 - 1.  

Edwin</pre>