Question

Which of the following are roots of the polynomial function below?

Check all that apply.
F(x)=x²-3x²+2
A. 3-√17
4
B. 2+√/12
C. 1
D. 3+√17
4
□ E. 2-12
SUBMIT

Answer 1

`\textsf{B.} \quad (2 +√(12))/(2)`

`\textsf{C.} \quad 1`

`\textsf{E.} \quad (2-√(12))/(2)`

Explanation:

Given polynomial function:

`f(x)=x^3-3x^2+2`

The roots of a polynomial function are the values of the variable x that make the function equal to zero.

To find the roots of the given polynomial function, we can apply the factor theorem.

Factor Theorem

If f(x) is a polynomial, and f(a) = 0, then (x - a) is a factor of f(x).

If the coefficients of a polynomial add up to 0, then (x - 1) is a factor.

Sum the coefficients of f(x):

`1-3+2=0`

As the sum of the coefficients equals zero, then (x - 1) is a factor of the polynomial.

Find the other factor by dividing the polynomial by (x - 1):

`\large \begin{array}{r}x^2-2x-2\phantom{)}\\x-1{\overline{\smash{\big)}\,x^3-3x^2+2\phantom{)}}\\{-~\phantom{(}\underline{(x^3-x^2)\phantom{-)..)}}\\-2x^2+2\phantom{)}\\-~\phantom{()}\underline{(-2x^2+2x)}\\-2x+2\phantom{)}\\\phantom{)}-~\phantom{()}\underline{(-2x+2)}\\0\phantom{)}\end{array}`

Therefore, the factored form of the polynomial is:

`f(x)=(x-1)(x^2-2x-2)`

To find the roots, set each factor to zero and solve for x.

Set the linear factor to zero and solve for x:

`(x-1)=0 \implies x=1`

Set the quadratic factor to zero and solve for x using the quadratic formula:

`x=(-b \pm √(b^2-4ac))/(2a)`

`x=(-(-2) \pm √((-2)^2-4(1)(-2)))/(2(1))`

`x=(2 \pm √(12))/(2)`

Therefore, the roots of the given polynomial function are:

`x=1, \quad x=(2 -√(12))/(2),\quad x=(2 +√(12))/(2)`

`\hrulefill`

Additional notes

The non-integer roots can be simplified further, as follows:

`x=(2 \pm √(2^2\cdot 3))/(2)`

`x=(2 \pm √(2^2)√(3))/(2)`

`x=(2 \pm 2√(3))/(2)`

`x=1\pm √(3)`