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.

Question

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.  f(x)=x^3-5x^2-61x-55???

Answer 1

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

 

 

Answer 2

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)

 

Answer 3

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 + bX^{n -1} + cX^n-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 + X_{3 + ....}+ X_n = -b/a

                                          X₁ . X₂ . X₃ .....X_{n = X}⁰ /a

 

            Like in the case of  f(x) = X³ - 5X² - 61X -55

                

                                                 X₁ + X₂ + X_{3 = 5}

                                                  X_{1 .} X_{2  .} X _{3 = 55}

 

                                        X_{1 = -}1        X₂ + X_{3 = 6}

                                                           X₂ . X_{3 =-55}

 

                              Quadratic  with roots of X₂ , X_{2 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.