Respuesta :
Answer:
The fourth root is 3
If the 4th root is not a real therefore it must be a complex number (a+ib),and its conjugate will be also a root ,therefore there would be 5 roots instead of 4 roots.
Therefore the fourth root is real.
The roots are -1 with multiplicity 2 and 3 with multiplicity 2
Therefore it has four roots
Step-by-step explanation:
Given polynomial equation is [tex]X^4-4X^3-2X^2+12X+9=0[/tex]
And also given that 3,-1 and -1 are the roots of the given polynomial equation
To find the fourth root of the polynomial equation and to solve the fourth root is real :
By synthetic division
_3| 1 -4 -2 12 9
0 3 -3 -15 -9
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_-1| 1 -1 -5 -3 0
0 -1 2 3
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1 -2 -3 0
Therefore x-3 and x+1 is a factor
Therefore 3 and -1 are roots
Now we have the quadratic equation [tex]x^2-2x-3=0[/tex]
[tex](x+1)(x-3)=0[/tex]
Therefore x=-1,3 are the roots
Therefore the fourth root is 3
If the 4th root is not a real therefore it must be a complex number (a+ib),and its conjugate will be also a root ,therefore there would be 5 roots instead of 4 roots.
Therefore the fourth root is real.
The roots are -1 with multiplicity 2 and 3 with multiplicity 2
Therefore it has four roots.
