Question

Which of the values shown are potential roots of f(x) = 3x3 â€"" 13x2 â€"" 3x 45? Select all that apply.

Answer 1

The potential roots of the function are, $\pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45,\ \pm \dfrac{1}{3},\ \pm \dfrac{5}{3}$

And the accurate root is 3 it can be determined by using rules of the rational root equation.

Given that,

Function; $\rm f(x) = 3x^3 - 13x^2 -3x + 45$

We have to determine,

Which of the values shown are potential roots of the given equation?

According to the question,

Potential roots of the polynomial are all possible roots of f(x).

$\rm f(x) = 3x^3 - 13x^2 -3x + 45$

Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

$\rm p=\dfrac{All\ the \ positive}{Negative\ factors \ of\ 45}$

$\rm q=\dfrac{All\ the \ positive}{Negative\ factors \ of\ 3}$

The factor of the term 45 are,

$\pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45$

And The factor of 3 are,

$\pm1, \ \pm3$

All the possible roots are,

$\dfrac{p}{q} = \pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45,\ \pm \dfrac{1}{3},\ \pm \dfrac{5}{3}$

Now check for all the rational roots which are possible for the given function,

$\rm f(x) = 3x^3 - 13x^2 -3x + 45\\\\ f(1) = 3(1)^3 - 13(1)^2 -3(1) + 45 = 3-13-3+45 = 32\neq 0\\\\ f(-1) = 3(-1)^3 - 13(-1)^2 -3(-1) + 45 =- 3-13+3+45 = 32\neq 0\\\\ f(3) = 3(3)^3 - 13(3)^2 -3(3) + 45 = 81-117-9+45 =0\\\\ f(-3) = 3(-3)^3 - 13(-3)^2 -3(-3) + 45 = -81+117+9+45 =-144\neq 0$

Therefore, x = 3 is the potential root of the given function.

Hence, The potential roots of the function are, $\pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45,\ \pm \dfrac{1}{3},\ \pm \dfrac{5}{3}$.

For more details about Potential roots refer to the link given below.

brainly.com/question/25873992