Question
Solve the polynomial equation.
3x3+2x2−27x−12=6
(1 point)
Responses
x = 3, x=−23
x = 3, x=−23
x = 3, x=−3, x=−23
x = 3, x=−3, x=−23
x = 3, x=−3, x=23
x = 3, x=−3, x=23
x=−3, x=−23
x=−3, x=−23
3x3+2x2−27x−12=6
(1 point)
Responses
x = 3, x=−23
x = 3, x=−23
x = 3, x=−3, x=−23
x = 3, x=−3, x=−23
x = 3, x=−3, x=23
x = 3, x=−3, x=23
x=−3, x=−23
x=−3, x=−23
Answers
Answered by
GPT-4o mini
To solve the polynomial equation \( 3x^3 + 2x^2 - 27x - 12 = 6 \), we first move all terms to one side of the equation:
\[
3x^3 + 2x^2 - 27x - 12 - 6 = 0
\]
This simplifies to:
\[
3x^3 + 2x^2 - 27x - 18 = 0
\]
Next, we can attempt to find rational roots using the Rational Root Theorem. We will test possible rational roots, which are factors of the constant term (-18) divided by factors of the leading coefficient (3).
The factors of -18 are: ±1, ±2, ±3, ±6, ±9, ±18.
The factors of 3 are: ±1, ±3.
The potential rational roots are therefore: ±1, ±2, ±3, ±6, ±9, ±18, ±1/3, ±2/3.
Let's test these values.
1. Testing \( x = 3 \):
\[
3(3^3) + 2(3^2) - 27(3) - 18 = 3(27) + 2(9) - 81 - 18
\]
\[
= 81 + 18 - 81 - 18 = 0
\]
So, \( x = 3 \) is a root.
Next, we can perform polynomial long division to divide \( 3x^3 + 2x^2 - 27x - 18 \) by \( x - 3 \):
1. Divide the leading term: \( 3x^3 / x = 3x^2 \).
2. Multiply \( 3x^2 \) by \( x - 3 \):
\[
3x^3 - 9x^2
\]
3. Subtract this from the original polynomial:
\[
(3x^3 + 2x^2 - 27x - 18) - (3x^3 - 9x^2) = 11x^2 - 27x - 18
\]
Now repeat the process with \( 11x^2 - 27x - 18 \):
1. Divide the leading term: \( 11x^2 / x = 11x \).
2. Multiply \( 11x \) by \( x - 3 \):
\[
11x^2 - 33x
\]
3. Subtract:
\[
(11x^2 - 27x - 18) - (11x^2 - 33x) = 6x - 18
\]
Now we’ve reduced it to \( 6x - 18 \).
1. Divide \( 6x - 18 \) by \( x - 3 \):
\[
6x / x = 6
\]
2. Multiply, then subtract:
\[
(6x - 18) - (6x - 18) = 0
\]
The complete factorization of the polynomial is:
\[
3(x - 3)(x^2 + 11x + 6) = 0
\]
Next, factor or use the quadratic formula on \( x^2 + 11x + 6 = 0 \):
Using the formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}
\]
\[
= \frac{-11 \pm \sqrt{121 - 24}}{2} = \frac{-11 \pm \sqrt{97}}{2}
\]
Thus, the equation has one rational root \( x = 3 \) and two irrational roots.
Finally, the choices provided correspond:
- \( x = 3 \) (rational root),
- \( x = \frac{-11 + \sqrt{97}}{2} \) (irrational root),
- \( x = \frac{-11 - \sqrt{97}}{2} \) (irrational root).
Since none of the options corresponds to the irrational roots found, the solution set is \( x = 3 \) plus the two irrational roots.
Thus the answer from the already listed options is that \( x = 3, x = -3, x = -23 \) is incorrect after reassessing, as only \( x = 3 \) is rational, and the roots derived are not akin to the other options.
The only valid answer based on \( x = 3 \) is the choices should conclude there, and the exact roots can be confirmed by checking if their values make the polynomial equal to 0, particularly validating \( x = 3 \):
Ultimately the equation has complex evaluations not producing simple values. Thus, I recommend you verify the surrounding polynomial provisions above if they affirmatively capture values \( -3 \), \( -23 \) which can roundabout add given differing estimations.
\[
3x^3 + 2x^2 - 27x - 12 - 6 = 0
\]
This simplifies to:
\[
3x^3 + 2x^2 - 27x - 18 = 0
\]
Next, we can attempt to find rational roots using the Rational Root Theorem. We will test possible rational roots, which are factors of the constant term (-18) divided by factors of the leading coefficient (3).
The factors of -18 are: ±1, ±2, ±3, ±6, ±9, ±18.
The factors of 3 are: ±1, ±3.
The potential rational roots are therefore: ±1, ±2, ±3, ±6, ±9, ±18, ±1/3, ±2/3.
Let's test these values.
1. Testing \( x = 3 \):
\[
3(3^3) + 2(3^2) - 27(3) - 18 = 3(27) + 2(9) - 81 - 18
\]
\[
= 81 + 18 - 81 - 18 = 0
\]
So, \( x = 3 \) is a root.
Next, we can perform polynomial long division to divide \( 3x^3 + 2x^2 - 27x - 18 \) by \( x - 3 \):
1. Divide the leading term: \( 3x^3 / x = 3x^2 \).
2. Multiply \( 3x^2 \) by \( x - 3 \):
\[
3x^3 - 9x^2
\]
3. Subtract this from the original polynomial:
\[
(3x^3 + 2x^2 - 27x - 18) - (3x^3 - 9x^2) = 11x^2 - 27x - 18
\]
Now repeat the process with \( 11x^2 - 27x - 18 \):
1. Divide the leading term: \( 11x^2 / x = 11x \).
2. Multiply \( 11x \) by \( x - 3 \):
\[
11x^2 - 33x
\]
3. Subtract:
\[
(11x^2 - 27x - 18) - (11x^2 - 33x) = 6x - 18
\]
Now we’ve reduced it to \( 6x - 18 \).
1. Divide \( 6x - 18 \) by \( x - 3 \):
\[
6x / x = 6
\]
2. Multiply, then subtract:
\[
(6x - 18) - (6x - 18) = 0
\]
The complete factorization of the polynomial is:
\[
3(x - 3)(x^2 + 11x + 6) = 0
\]
Next, factor or use the quadratic formula on \( x^2 + 11x + 6 = 0 \):
Using the formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}
\]
\[
= \frac{-11 \pm \sqrt{121 - 24}}{2} = \frac{-11 \pm \sqrt{97}}{2}
\]
Thus, the equation has one rational root \( x = 3 \) and two irrational roots.
Finally, the choices provided correspond:
- \( x = 3 \) (rational root),
- \( x = \frac{-11 + \sqrt{97}}{2} \) (irrational root),
- \( x = \frac{-11 - \sqrt{97}}{2} \) (irrational root).
Since none of the options corresponds to the irrational roots found, the solution set is \( x = 3 \) plus the two irrational roots.
Thus the answer from the already listed options is that \( x = 3, x = -3, x = -23 \) is incorrect after reassessing, as only \( x = 3 \) is rational, and the roots derived are not akin to the other options.
The only valid answer based on \( x = 3 \) is the choices should conclude there, and the exact roots can be confirmed by checking if their values make the polynomial equal to 0, particularly validating \( x = 3 \):
Ultimately the equation has complex evaluations not producing simple values. Thus, I recommend you verify the surrounding polynomial provisions above if they affirmatively capture values \( -3 \), \( -23 \) which can roundabout add given differing estimations.