+0  
 
0
206
2
avatar

What are the roots of the polynomial equation?

 0=x4-3x3+10x2-24x+16

 Jan 27, 2021
 #1
avatar+120023 
+1

x^4  - 3x^3  + 10x^2 - 24  +   16     =  0

 

Note  that if we  can   add   the coefficients  and the   constant and  we  get 0 ,  then 1  is a root

1 + 10  + 16    - 3   - 24   =

27  - 27   = 0

So 1 is a root

 

Using  some synthetic division   to find the  reduced polynomial, we  have

 

1  [     1       - 3       10          -24         16    ]

                     1         -2            8         -16

        ________________________________

          1        -2        8            -16            0

 

So....the reduced polynomial is

 

x^2  - 2x^2  +  8x    -  16       =     0        this factors  as

 

x (x -2)  +  8 ( x -2)     = 0

 

(x + 8) ( x -2)     =   0

 

Setting each factor to 0  and  solving for  x  gives us

 

x = -8     and   x  = 2

 

So   the roots are

 

x =  -8       x  = 1      and     x  =    2

 

 

cool cool cool

 Jan 27, 2021
 #2
avatar
0

I'm not sure but after the synthetic division, wouldn't the equation be x^3 instead of x^2? Because if you look at the original equation, it starts with x^4, not x^3. So after recollecting the numbers, the equation would be

x^3 - 2x^2 + 8x -16 = 0

From here, it would factor out as,

x^2(x-2) +8(x-2) = 0

(x-2) (x^2 +8) = 0

With the zero product property,

x - 2 = 0

x = 2

 

x^2 +8 = 0

x^2 = -8

After sqaure rooting each side,

x = 2 sqrt 2i, x = -2 sqrt 2i

So, in total, all the roots would be x = 1, 2, 2 sqrt 2i and -2 sqrt 2i

Guest Feb 5, 2021
edited by Guest  Feb 5, 2021
edited by Guest  Feb 5, 2021

28 Online Users

avatar