Answer:
6,-4,3
Step-by-step explanation:
We are given that a polynomial
[tex]x^3+72=5x^+18x[/tex]
[tex]x^3-5x^2-18x+72=0[/tex]
We have to find the original roots of polynomial equation
Substitute -6 in the polynomial
Then , we get
[tex](-6)^3-5(-6)^2-18(-6)+72[/tex]
[tex]-216-180+108+72\neq 0[/tex]
Therefore, it is not roots of the given polynomial.
Substitute x=0 then we get
[tex]72\neq 0[/tex]
Hence, o is not a root of given polynomial.
Substitute x=6 then we get
[tex](6)^3-5(6)^2-18(6)+72[/tex]
[tex]216-180-108+72=288-288=0[/tex]
Hence, 6 is a root of given polynomial because it satisfied the given polynomial.
Substitute x=-4 then we get
[tex](-4)^3-5(-4)^2-18(-4)+72[/tex]
[tex]-64-80+72+72=-144+144=0[/tex]
Hence, -4 is a root of given polynomial .
Substitute x=3 then we get
[tex](3)^3-5(3)^2-18(3)+72[/tex]
[tex]27-45-54+72[/tex]
[tex]99-99=0[/tex]
Hence, 3 is a root of given polynomial.
Substitute x=8 then we get
[tex](8)^3-5(8)^2-18(8)+72[/tex]
[tex]512-320-144+72\neq 0 [/tex]
Hence, 8 is not a roots of given polynomial.
