$16 \sqrt{3}$ 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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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 $x^3$ $0$ .

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.