The roots of the given function are [tex]x = 3[/tex], [tex]x = \frac{3+\sqrt{29} }{2}[/tex] and [tex]x = \frac{3-\sqrt{29} }{2}[/tex]. The correct options are A, E and F
Determining roots of a polynomial function
From the question, we are to determine which of the given options are roots of the polynomial function
The given polynomial function is
[tex]f(x) = x^{3} -6x^{2}+4x+15[/tex]
First, we will determine one of the zeros of the given function.
Testing for some integers by putting them into the function, we determine that
[tex]f(3) =0[/tex]
As shown here,
[tex]f(3) = 3^{3} -6(3)^{2}+4(3)+15[/tex]
[tex]f(3) = 27 -54+12+15[/tex]
[tex]f(3) = 0[/tex]
Then, one of the roots of the function is 3
∴ x = 3
If x = 3, then x - 3 must be a factor of the equation.
By factoring,
[tex]x^{3} -6x^{2}+4x+15=(x-3)(x^{2} -3x-5)[/tex]
To determine the remaining roots of the function, we will solve x² -3x -5 quadratically
x² -3x -5 = 0
Using the general formula for quadratic equation,
[tex]x = \frac{-b \pm\sqrt{b^{2}-4ac } }{2a}[/tex]
a = 1, b = -3, and c = -5
Putting the values into the formula,
[tex]x = \frac{-(-3) \pm\sqrt{(-3)^{2}-4(1)(-5)} }{2(1)}[/tex]
[tex]x = \frac{3\pm\sqrt{9+20} }{2}[/tex]
[tex]x = \frac{3\pm\sqrt{29} }{2}[/tex]
∴ [tex]x = \frac{3+\sqrt{29} }{2}[/tex] OR [tex]x = \frac{3-\sqrt{29} }{2}[/tex]
Hence, the roots of the given function are [tex]x = 3[/tex], [tex]x = \frac{3+\sqrt{29} }{2}[/tex] and [tex]x = \frac{3-\sqrt{29} }{2}[/tex]. The correct options are A, E and F.
Learn more on Determining the roots of a function here: https://brainly.com/question/14193054
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