Find all the roots of the polynomial

Question

The polynomial 

f(x)= 7x³-3x²+252x-108

has 6i as a root. Give all of the roots of f in a list, including the given one. 

Answer 1

f(x) = 7x^3 - 3x^2 + 252x - 108 has 6i as a root.

Give all of the roots of f in a list, including the given one.

 

For polynomial functions with real coefficients any complex roots occur in conjugate pairs. Since 6i is a root so is -6i.

 

(x - 6i)(x + 6i) = x^2 + 36

 

                 7x     - 3                          

x^2 + 36 | 7x^3 - 3x^2 + 252x - 108

                 7x^3            + 252x

                          - 3x^2

                          - 3x^2            - 108

 

 

f(x) = (7x - 3)(x - 6i)(x + 6i) with roots 3/7 and ±6i.

Answer 2

Hi Dalia,

as 6i is one of root so its conjugate -6i is another root.

( x-6i) & (x+6i) are factors.

 

f(x)= 7x³-3x²+252x-108

     

    (x-6i)(x+6i) = x²-36i²= x²+36  ........using i²=-1

Dividing polynomial f(x) by x²+36

 

 

             7x-3

x²+36 |¯¯¯¯¯¯¯¯¯¯¯

          | 7x³-3x²+252x-108

            7x³       +252x

           - _______-__________

                 -3x²-108

                 -3x²-108

                 +    +

_____________________

               o

Ans: roots are 6i, -6i, 3/7.

 

Meena from Strongsville, OH

Answer 3

Hi Dalia;

f(x)= 7x³-3x²+252x-108

0=(...)(...)

FIRST must be (7x)(x²) or (7x²)(x)

0=(7x....)(x²....)

0=(7x²...)(x....)

The x² must combine with -3 in either the OUTER or INNER...

0=(7x-3)(x²....)

OUTER must be(7x)(252/7)...(252/7)=36

0=(7x-3)(x²+36)

LAST must be (-3)(36)=108

0=(7x-3)(x²+36)

Either or both parenthetical equation(s) must equal zero...

0=7x-3...........0=x²+36

3=7x.............-36=x²

3/7=x............6i=x