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# 1. What are the roots of the polynomial equation?

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What are the roots of the polynomial equation?

0=x4-3x3+10x2-24x+16

Jan 27, 2021

#1
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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

Jan 27, 2021
#2
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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
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