16sqrt3 is a root of f(x)=x^(3)+x^(2)- 768 x-768. Find the other roots of f(x). Write your answer as a list of simplified values separated by commas, if there is more than one value.

Set Up Synthetic Division: Since 16√3 is a root of the polynomial f(x), we can use synthetic division or polynomial division to divide f(x) by (x - 16√3) to find the other factors of the polynomial.Find Quadratic Factor: First, let's set up the synthetic division with 16√3 as the divisor and the coefficients of f(x) as the dividend: 1 (for x^3), 1 (for x^2), -768 (for x), and -768 (for the constant term).Solve System of Equations: Performing the synthetic division, we bring down the leading coefficient (1) and multiply it by 16√3, then add this to the next coefficient (1), and continue this process. However, since this is a complex process and requires careful calculation, we will use the fact that the polynomial is of degree 3 and we already have one real root to find the other roots by factoring or using the quadratic formula.Apply Quadratic Formula: To find the quadratic factor, we can use the fact that the sum of the roots of the polynomial, given by -b/a for a cubic polynomial ax^3 + bx^2 + cx + d, should equal the negative of the coefficient of x^2 divided by the coefficient of x^3. In this case, the sum of the roots is -1/1 = -1.Calculate Discriminant: Since we know one root is 16√3, the sum of the other two roots must be -1 - 16√3. Let's call these two roots r1 and r2. So, r1 + r2 = -1 - 16√3.Find Roots: We also know that the product of the roots of the polynomial, given by -d/a for a cubic polynomial ax^3 + bx^2 + cx + d, should equal the negative of the constant term divided by the coefficient of x^3. In this case, the product of the roots is -(-768)/1 = 768.Since we know one root is 16√3, the product of the other two roots must be 768 / (16√3). Let's calculate this value: 768 / (16√3) = 48 / √3 = 48√3 / 3 = 16√3.Now we have a system of equations for r1 and r2: r1 + r2 = -1 - 16√3 r1 * r2 = 16√3 We can solve this system by using the quadratic formula, where the sum of the roots is the negative coefficient of x, and the product of the roots is the constant term. The quadratic equation will be x^2 + (1 + 16√3)x + 16√3 = 0.Applying the quadratic formula, x = [-b ± √(b^2 - 4ac)] / (2a), with a = 1, b = 1 + 16√3, and c = 16√3, we find the other two roots.Calculating the discriminant, Δ = b^2 - 4ac = (1 + 16√3)^2 - 4(1)(16√3).Simplifying the discriminant, Δ = 1 + 32√3 + 768 - 64√3 = 769 - 32√3.Now we can find the roots using the quadratic formula: x = [-(1 + 16√3) ± √(769 - 32√3)] / 2This gives us two roots, which we can simplify further. However, we made a mistake in the calculation of the discriminant. The correct calculation should be Δ = (1 + 16√3)^2 - 4 * 1 * 16√3 = 1 + 32√3 + 256 * 3 - 64√3 = 1 - 32√3 + 768 + 64√3 = 769.

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16sqrt3 is a root of
f(x)=x^(3)+x^(2)-
768 x-768. Find the other roots of
f(x).
Write your answer as a list of simplified values separated by commas, if there is more than one value.

$16 \sqrt{3}$ is a root of $f(x)=x^{3}+x^{2}-$ $768 x-768$ . Find the other roots of $f(x)$ . $\newline$ Write your answer as a list of simplified values separated by commas, if there is more than one value.

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Conjugate root theorems

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16 \sqrt{3}

is a root of

f(x)=x^{3}+x^{2}-

768 x-768

. Find the other roots of

f(x)

\newline

Write your answer as a list of simplified values separated by commas, if there is more than one value.

Set Up Synthetic Division: Since $16\sqrt{3}$ is a root of the polynomial $f(x)$ , we can use synthetic division or polynomial division to divide $f(x)$ by $(x - 16\sqrt{3})$ to find the other factors of the polynomial.
Find Quadratic Factor: First, let's set up the synthetic division with $16\sqrt{3}$ as the divisor and the coefficients of $f(x)$ as the dividend: $1$ (for $x^3$ ), $1$ (for $x^2$ ), $-768$ (for $x$ ), and $-768$ (for the constant term).
Solve System of Equations: Performing the synthetic division, we bring down the leading coefficient $1$ and multiply it by $16\sqrt{3}$ , then add this to the next coefficient $1$ , and continue this process. However, since this is a complex process and requires careful calculation, we will use the fact that the polynomial is of degree $3$ and we already have one real root to find the other roots by factoring or using the quadratic formula.
Apply Quadratic Formula: To find the quadratic factor, we can use the fact that the sum of the roots of the polynomial, given by $-b/a$ for a cubic polynomial $ax^3 + bx^2 + cx + d$ , should equal the negative of the coefficient of $x^2$ divided by the coefficient of $x^3$ . In this case, the sum of the roots is $-1/1 = -1$ .
Calculate Discriminant: Since we know one root is $16\sqrt{3}$ , the sum of the other two roots must be $-1 - 16\sqrt{3}$ . Let's call these two roots $r_1$ and $r_2$ . So, $r_1 + r_2 = -1 - 16\sqrt{3}$ .
Find Roots: We also know that the product of the roots of the polynomial, given by $-\frac{d}{a}$ for a cubic polynomial $ax^3 + bx^2 + cx + d$ , should equal the negative of the constant term divided by the coefficient of $x^3$ . In this case, the product of the roots is $-(-768)/1 = 768$ .
Find Roots: We also know that the product of the roots of the polynomial, given by $-d/a$ for a cubic polynomial $ax^3 + bx^2 + cx + d$ , should equal the negative of the constant term divided by the coefficient of $x^3$ . In this case, the product of the roots is $-(-768)/1 = 768$ .Since we know one root is $16\sqrt{3}$ , the product of the other two roots must be $768 / (16\sqrt{3})$ . Let's calculate this value: $768 / (16\sqrt{3}) = 48 / \sqrt{3} = 48\sqrt{3} / 3 = 16\sqrt{3}$ .
Find Roots: We also know that the product of the roots of the polynomial, given by $-\frac{d}{a}$ for a cubic polynomial $ax^3 + bx^2 + cx + d$ , should equal the negative of the constant term divided by the coefficient of $x^3$ . In this case, the product of the roots is $-(-768)/1 = 768$ .Since we know one root is $16\sqrt{3}$ , the product of the other two roots must be $768 / (16\sqrt{3})$ . Let's calculate this value: $768 / (16\sqrt{3}) = 48 / \sqrt{3} = 48\sqrt{3} / 3 = 16\sqrt{3}$ .Now we have a system of equations for $r1$ and $r2$ : $\newline$ $r1 + r2 = -1 - 16\sqrt{3}$ $\newline$ $ax^3 + bx^2 + cx + d$ $0$ $\newline$ We can solve this system by using the quadratic formula, where the sum of the roots is the negative coefficient of $ax^3 + bx^2 + cx + d$ $1$ , and the product of the roots is the constant term. The quadratic equation will be $ax^3 + bx^2 + cx + d$ $2$ .
Find Roots: We also know that the product of the roots of the polynomial, given by $-\frac{d}{a}$ for a cubic polynomial $ax^3 + bx^2 + cx + d$ , should equal the negative of the constant term divided by the coefficient of $x^3$ . In this case, the product of the roots is $-(-768)/1 = 768$ .Since we know one root is $16\sqrt{3}$ , the product of the other two roots must be $768 / (16\sqrt{3})$ . Let's calculate this value: $768 / (16\sqrt{3}) = 48 / \sqrt{3} = 48\sqrt{3} / 3 = 16\sqrt{3}$ .Now we have a system of equations for $r_1$ and $r_2$ : $r_1 + r_2 = -1 - 16\sqrt{3}$ $r_1 \cdot r_2 = 16\sqrt{3}$ We can solve this system by using the quadratic formula, where the sum of the roots is the negative coefficient of $x$ , and the product of the roots is the constant term. The quadratic equation will be $ax^3 + bx^2 + cx + d$ $0$ .Applying the quadratic formula, $ax^3 + bx^2 + cx + d$ $1$ , with $ax^3 + bx^2 + cx + d$ $2$ , $ax^3 + bx^2 + cx + d$ $3$ , and $ax^3 + bx^2 + cx + d$ $4$ , we find the other two roots.
Find Roots: We also know that the product of the roots of the polynomial, given by $-\frac{d}{a}$ for a cubic polynomial $ax^3 + bx^2 + cx + d$ , should equal the negative of the constant term divided by the coefficient of $x^3$ . In this case, the product of the roots is $-(-768)/1 = 768$ .Since we know one root is $16\sqrt{3}$ , the product of the other two roots must be $768 / (16\sqrt{3})$ . Let's calculate this value: $768 / (16\sqrt{3}) = 48 / \sqrt{3} = 48\sqrt{3} / 3 = 16\sqrt{3}$ .Now we have a system of equations for $r_1$ and $r_2$ : $r_1 + r_2 = -1 - 16\sqrt{3}$ $r_1 \cdot r_2 = 16\sqrt{3}$ We can solve this system by using the quadratic formula, where the sum of the roots is the negative coefficient of $x$ , and the product of the roots is the constant term. The quadratic equation will be $ax^3 + bx^2 + cx + d$ $0$ .Applying the quadratic formula, $ax^3 + bx^2 + cx + d$ $1$ , with $ax^3 + bx^2 + cx + d$ $2$ , $ax^3 + bx^2 + cx + d$ $3$ , and $ax^3 + bx^2 + cx + d$ $4$ , we find the other two roots.Calculating the discriminant, $ax^3 + bx^2 + cx + d$ $5$ .
Find Roots: We also know that the product of the roots of the polynomial, given by $-\frac{d}{a}$ for a cubic polynomial $ax^3 + bx^2 + cx + d$ , should equal the negative of the constant term divided by the coefficient of $x^3$ . In this case, the product of the roots is $-(-768)/1 = 768$ .Since we know one root is $16\sqrt{3}$ , the product of the other two roots must be $768 / (16\sqrt{3})$ . Let's calculate this value: $768 / (16\sqrt{3}) = 48 / \sqrt{3} = 48\sqrt{3} / 3 = 16\sqrt{3}$ .Now we have a system of equations for $r_1$ and $r_2$ : $r_1 + r_2 = -1 - 16\sqrt{3}$ $r_1 \cdot r_2 = 16\sqrt{3}$ We can solve this system by using the quadratic formula, where the sum of the roots is the negative coefficient of $x$ , and the product of the roots is the constant term. The quadratic equation will be $ax^3 + bx^2 + cx + d$ $0$ .Applying the quadratic formula, $ax^3 + bx^2 + cx + d$ $1$ , with $ax^3 + bx^2 + cx + d$ $2$ , $ax^3 + bx^2 + cx + d$ $3$ , and $ax^3 + bx^2 + cx + d$ $4$ , we find the other two roots.Calculating the discriminant, $ax^3 + bx^2 + cx + d$ $5$ .Simplifying the discriminant, $ax^3 + bx^2 + cx + d$ $6$ .
Find Roots: We also know that the product of the roots of the polynomial, given by $-\frac{d}{a}$ for a cubic polynomial $ax^3 + bx^2 + cx + d$ , should equal the negative of the constant term divided by the coefficient of $x^3$ . In this case, the product of the roots is $-(-768)/1 = 768$ .Since we know one root is $16\sqrt{3}$ , the product of the other two roots must be $768 / (16\sqrt{3})$ . Let's calculate this value: $768 / (16\sqrt{3}) = 48 / \sqrt{3} = 48\sqrt{3} / 3 = 16\sqrt{3}$ .Now we have a system of equations for $r_1$ and $r_2$ : $r_1 + r_2 = -1 - 16\sqrt{3}$ $r_1 \cdot r_2 = 16\sqrt{3}$ We can solve this system by using the quadratic formula, where the sum of the roots is the negative coefficient of $x$ , and the product of the roots is the constant term. The quadratic equation will be $ax^3 + bx^2 + cx + d$ $0$ .Applying the quadratic formula, $ax^3 + bx^2 + cx + d$ $1$ , with $ax^3 + bx^2 + cx + d$ $2$ , $ax^3 + bx^2 + cx + d$ $3$ , and $ax^3 + bx^2 + cx + d$ $4$ , we find the other two roots.Calculating the discriminant, $ax^3 + bx^2 + cx + d$ $5$ .Simplifying the discriminant, $ax^3 + bx^2 + cx + d$ $6$ .Now we can find the roots using the quadratic formula: $x = \frac{-(1 + 16\sqrt{3}) \pm \sqrt{769 - 32\sqrt{3}}}{2}$
Find Roots: We also know that the product of the roots of the polynomial, given by $-\frac{d}{a}$ for a cubic polynomial $ax^3 + bx^2 + cx + d$ , should equal the negative of the constant term divided by the coefficient of $x^3$ . In this case, the product of the roots is $-(-768)/1 = 768$ .Since we know one root is $16\sqrt{3}$ , the product of the other two roots must be $768 / (16\sqrt{3})$ . Let's calculate this value: $768 / (16\sqrt{3}) = 48 / \sqrt{3} = 48\sqrt{3} / 3 = 16\sqrt{3}$ .Now we have a system of equations for $r_1$ and $r_2$ : $\newline$ $r_1 + r_2 = -1 - 16\sqrt{3}$ $\newline$ $ax^3 + bx^2 + cx + d$ $0$ $\newline$ We can solve this system by using the quadratic formula, where the sum of the roots is the negative coefficient of $ax^3 + bx^2 + cx + d$ $1$ , and the product of the roots is the constant term. The quadratic equation will be $ax^3 + bx^2 + cx + d$ $2$ .Applying the quadratic formula, $ax^3 + bx^2 + cx + d$ $3$ , with $ax^3 + bx^2 + cx + d$ $4$ , $ax^3 + bx^2 + cx + d$ $5$ , and $ax^3 + bx^2 + cx + d$ $6$ , we find the other two roots.Calculating the discriminant, $ax^3 + bx^2 + cx + d$ $7$ .Simplifying the discriminant, $ax^3 + bx^2 + cx + d$ $8$ .Now we can find the roots using the quadratic formula: $\newline$ $ax^3 + bx^2 + cx + d$ $9$ This gives us two roots, which we can simplify further. However, we made a mistake in the calculation of the discriminant. The correct calculation should be $x^3$ $0$ .

$16 \sqrt{3}$ is a root of $f(x)=x^{3}+x^{2}-$ $768 x-768$ . Find the other roots of $f(x)$ . $\newline$ Write your answer as a list of simplified values separated by commas, if there is more than one value.

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163 16 \sqrt{3} 163​ is a root of f(x)=x3+x2− f(x)=x^{3}+x^{2}- f(x)=x3+x2− 768x−768 768 x-768 768x−768. Find the other roots of f(x) f(x) f(x).\newlineWrite your answer as a list of simplified values separated by commas, if there is more than one value.

$16 \sqrt{3}$ is a root of $f(x)=x^{3}+x^{2}-$ $768 x-768$ . Find the other roots of $f(x)$ . $\newline$ Write your answer as a list of simplified values separated by commas, if there is more than one value.