Answer :
To determine all possible rational solutions of the polynomial equation [tex]\( x^3 - 16x^2 - 6x + 8 = 0 \)[/tex], we can use the Rational Root Theorem. The Rational Root Theorem states that any possible rational root, [tex]\( \frac{p}{q} \)[/tex], of the polynomial [tex]\( p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \)[/tex] must be a factor of the constant term [tex]\( a_0 \)[/tex] divided by a factor of the leading coefficient [tex]\( a_n \)[/tex].
Here, the polynomial is [tex]\( x^3 - 16x^2 - 6x + 8 \)[/tex].
1. Identify the constant term ([tex]\(a_0\)[/tex]) and the leading coefficient [tex]\(a_n\)[/tex]:
- The constant term [tex]\(a_0\)[/tex] is 8.
- The leading coefficient [tex]\(a_n\)[/tex] is 1.
2. List the factors of the constant term 8:
- The factors of 8 are [tex]\( \pm 1, \pm 2, \pm 4, \pm 8 \)[/tex].
3. List the factors of the leading coefficient 1:
- The factors of 1 are [tex]\( \pm 1 \)[/tex].
4. According to the Rational Root Theorem, the possible rational roots are the factors of the constant term divided by the factors of the leading coefficient. Thus, the possible rational solutions are:
[tex]$ \pm 1, \pm 2, \pm 4, \pm 8 $[/tex]
Having determined the possible rational roots, we should verify which, if any, are actual roots of the polynomial by substituting them into the original polynomial equation and checking if the equation equals zero.
Let's test each possible root:
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ 1^3 - 16(1)^2 - 6(1) + 8 = 1 - 16 - 6 + 8 = -13 \neq 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is not a root.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ (-1)^3 - 16(-1)^2 - 6(-1) + 8 = -1 - 16 + 6 + 8 = -3 \neq 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a root.
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ 2^3 - 16(2)^2 - 6(2) + 8 = 8 - 64 - 12 + 8 = -60 \neq 0 \][/tex]
So, [tex]\( x = 2 \)[/tex] is not a root.
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ (-2)^3 - 16(-2)^2 - 6(-2) + 8 = -8 - 64 + 12 + 8 = -52 \neq 0 \][/tex]
So, [tex]\( x = -2 \)[/tex] is not a root.
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ 4^3 - 16(4)^2 - 6(4) + 8 = 64 - 256 - 24 + 8 = -208 \neq 0 \][/tex]
So, [tex]\( x = 4 \)[/tex] is not a root.
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ (-4)^3 - 16(-4)^2 - 6(-4) + 8 = -64 - 256 + 24 + 8 = -288 \neq 0 \][/tex]
So, [tex]\( x = -4 \)[/tex] is not a root.
- For [tex]\( x = 8 \)[/tex]:
[tex]\[ 8^3 - 16(8)^2 - 6(8) + 8 = 512 - 1024 - 48 + 8 = -552 \neq 0 \][/tex]
So, [tex]\( x = 8 \)[/tex] is not a root.
- For [tex]\( x = -8 \)[/tex]:
[tex]\[ (-8)^3 - 16(-8)^2 - 6(-8) + 8 = -512 - 1024 + 48 + 8 = -1480 \neq 0 \][/tex]
So, [tex]\( x = -8 \)[/tex] is not a root.
Since none of the possible rational roots satisfy the polynomial equation, there are no rational solutions for the polynomial [tex]\( x^3 - 16x^2 - 6x + 8 = 0 \)[/tex].
Thus, the list of all possible rational solutions is:
[tex]\(\boxed{\text{none}}\)[/tex]
Here, the polynomial is [tex]\( x^3 - 16x^2 - 6x + 8 \)[/tex].
1. Identify the constant term ([tex]\(a_0\)[/tex]) and the leading coefficient [tex]\(a_n\)[/tex]:
- The constant term [tex]\(a_0\)[/tex] is 8.
- The leading coefficient [tex]\(a_n\)[/tex] is 1.
2. List the factors of the constant term 8:
- The factors of 8 are [tex]\( \pm 1, \pm 2, \pm 4, \pm 8 \)[/tex].
3. List the factors of the leading coefficient 1:
- The factors of 1 are [tex]\( \pm 1 \)[/tex].
4. According to the Rational Root Theorem, the possible rational roots are the factors of the constant term divided by the factors of the leading coefficient. Thus, the possible rational solutions are:
[tex]$ \pm 1, \pm 2, \pm 4, \pm 8 $[/tex]
Having determined the possible rational roots, we should verify which, if any, are actual roots of the polynomial by substituting them into the original polynomial equation and checking if the equation equals zero.
Let's test each possible root:
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ 1^3 - 16(1)^2 - 6(1) + 8 = 1 - 16 - 6 + 8 = -13 \neq 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is not a root.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ (-1)^3 - 16(-1)^2 - 6(-1) + 8 = -1 - 16 + 6 + 8 = -3 \neq 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a root.
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ 2^3 - 16(2)^2 - 6(2) + 8 = 8 - 64 - 12 + 8 = -60 \neq 0 \][/tex]
So, [tex]\( x = 2 \)[/tex] is not a root.
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ (-2)^3 - 16(-2)^2 - 6(-2) + 8 = -8 - 64 + 12 + 8 = -52 \neq 0 \][/tex]
So, [tex]\( x = -2 \)[/tex] is not a root.
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ 4^3 - 16(4)^2 - 6(4) + 8 = 64 - 256 - 24 + 8 = -208 \neq 0 \][/tex]
So, [tex]\( x = 4 \)[/tex] is not a root.
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ (-4)^3 - 16(-4)^2 - 6(-4) + 8 = -64 - 256 + 24 + 8 = -288 \neq 0 \][/tex]
So, [tex]\( x = -4 \)[/tex] is not a root.
- For [tex]\( x = 8 \)[/tex]:
[tex]\[ 8^3 - 16(8)^2 - 6(8) + 8 = 512 - 1024 - 48 + 8 = -552 \neq 0 \][/tex]
So, [tex]\( x = 8 \)[/tex] is not a root.
- For [tex]\( x = -8 \)[/tex]:
[tex]\[ (-8)^3 - 16(-8)^2 - 6(-8) + 8 = -512 - 1024 + 48 + 8 = -1480 \neq 0 \][/tex]
So, [tex]\( x = -8 \)[/tex] is not a root.
Since none of the possible rational roots satisfy the polynomial equation, there are no rational solutions for the polynomial [tex]\( x^3 - 16x^2 - 6x + 8 = 0 \)[/tex].
Thus, the list of all possible rational solutions is:
[tex]\(\boxed{\text{none}}\)[/tex]
