if f(x)=6x^4-17x^3-57x^2+153x+27, and the graphing calculator tells you that x=3 and x=-3 are integer roots, algebraically deter
mine when f(x)>0
2 answers:
Answer:
The roots are x = 3, x = - 3 and x = - 1/6
Step-by-step explanation:
The given polynomial is expressed as
f(x) = 6x^4 - 17x^3 - 57x^2 + 153x + 27
Since x = 3 and x = - 3 are roots of the polynomial, it means that the factors are
(x - 3) and (x + 3)
We would divide the polynomial by either of the factors. Let us divide it by x - 3 by using the long division method. The steps are shown below:
The quotient from the division is
6x^3 + x^2 - 54x - 9
We would divide 6x^3 + x^2 - 54x - 9 by (x - 3) again. The steps are shown below:
We would factor 6x^2 + 19x + 3
We have
6x^2 + 18x + x + 3
6x(x + 3) + 1(x + 3) = 0
(6x + 1)(x + 3) = 0
6x + 1 = 0 or x + 3 = 0
6x = - 1 or x = - 3
x = - 1/6 or x = 3
The roots are x = 3, x = - 3 and x = - 1/6
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