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Use the rational zeros theorem to find all the real zeros of the polynomial function.use the zeros to factor f over the real numbers.

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3 Answers

f(x)=x3-5x2-61x-55.  The rational roots are formed by the ratios of ±1 ±11  ±5 /±1
f(-1)=-1-5+61-55=0  -1 is a zero.  If we divide by x-(-1)=x+1 we get a quadratic
-1|1    -5   -61   -55
          -1      6     55
     1   -6   -55       0        x2-6x-55, which factors to  (x-11)(x+5), whose zeroes are 11 and -5
 
Note: -125-125+305-55=0  i.e. f(-5)=0
f(11)=1331-605-671-55=0
 
 
The polynomial of odd degree always has at least one real zero, since complex roots of a polynomial with rational coefficients must always appear in conjugate pairs (like a+bi and a-bi, where i=√-1).
 
In this case x=-1 is the root. Indeed,
(-1)3-5(-1)2-61*(-1)-55=-1-5+61-55=0
 
Thus, we can factor (x+1) out using synthetic division.
 
x3-5x2-61x-55=(x+1)(x2-6x-55).
 
Now let us solve for other roots,
x2-6x-55=0; It is immediately obvious that, since the product of two roots must be -55 and their sum must be 6, those roots are 11 and -5. Therefore we obtain:
 
x3-5x2-61x-55=(x+1)(x+5)(x-11)
 
f(x) =  X3 - 5X2 -61X - 55
 
         Possible rational roots : ±1, ±5 ,±11, ±55
 
 
       try : X=-1
 
              ( -1)3 -5 (-1)2 -61( -1) - 55 =  -1-5 + 61 -55 = 0
 
           X= -1 is one of the solution.
 
            X3 - 5X2 - 61 X - 55 , should be divisible by (X +1). You can do the long division and find the
                                            quadratic.
            
          Also : any polynomial  of any degree  axn + bXn -1 + cXn-2 + ......
 
            If the irrational and complex roots are counted, the total number of roots equal to the degree of
            polynomial .  
              Sum of the roots : X1 + X2 + X3 + ....+ Xn = -b/a
                                          X1 . X2 . X3 .....Xn = X0 /a
 
            Like in the case of  f(x) = X3 - 5X2 - 61X -55
                
                                                 X1 + X2 + X3 = 5
                                                  X1 . X2  . 3 = 55
 
                                        X1 = -1        X2 + X3 = 6
                                                           X2 . X3 =-55
 
                              Quadratic  with roots of X2 , X2 will be:
 
                             X2 - 6 X - 55 =0
                              (X - 11)( X+5 ) =0      X = 11  X = -5
                             Doing the long division you should come up with the same answer.