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3x^4+6x^3-2x^2-10x-5 Because the constant of the degree of the polynomial is not 1 but 3 so we have to use the method of rational roots theorem to find the roots of the polynomial. According to this theorem we first have to find the factors of the numerical constant of the degree of the polynomial and of the constant. Then Factors of 3 are 1,-1,3,-3 Factors of 5 are1,-1,5,-5 To find the roots of the polynomial we have to by hit and trial method put the factors:1,-1,3,-3,5,-5,1/3,-1/3,1/5,-1/5,3/5,-3/5 After trial and error method it comes out that -1 satisfies the x. So x=-1 x+1=0 3x^3(x+1)+3x^2(x+1)-5x(x+1)-5(x+1) (x+1)(3x^3+3x^2-5x-5) Again by using the same method we can solve the cubic equation. On hit and trial we find that again (x+1) is a factor of the polynomial. (x+1)(3x^2(x+1)+0x(x + 1)-5(x + 1) (x + 1)(3x^2+ 0 x -5)(x + 1) as we know that the quadratic equation cannot be solved sothe zeros of the polynomial are -1 and (5/3)^1/2
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