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Synthetic division is a streamlined way to divide polynomials when you know one of its roots. Here, we used it knowing that the polynomial root is given as \(k = -4\). Since \(k = -4\) is a root, the factor \(x + 4\) should correctly divide the polynomial.For synthetic division, follow these steps:

• Set up by writing down the coefficients of the polynomial, which for the given problem are 6, 25, 3, and -4.
• Use the root \(-4\), the opposite of \(4\) from the factor \(x + 4\).

Start division by bringing down the leading coefficient, which is 6. Multiply 6 by \(-4\) and add this result to the next coefficient (25) to secure the new number in the quotient. This step creates a new sequence of coefficients from which the quadratic factor can be extracted once the remainder is zero, indicating a complete division.In our exercise, after performing synthetic division, the polynomial divides into \(6x^2 + x - 1\) with no remainder.

The next step in factoring involves using the quadratic formula to find the roots of the remaining quadratic, \(6x^2 + x - 1\), derived from synthetic division. The quadratic formula is crucial when factoring quadratics that do not factor neatly using common numbers or factoring techniques.Recall the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, identify \(a = 6\), \(b = 1\), and \(c = -1\). Plug these into the formula, beginning by calculating the discriminant \(b^2 - 4ac\), which helps determine the nature of the roots:

• The discriminant is \(25\), indicating two distinct real roots as it's positive.
• Compute the square root, \(\sqrt{25} = 5\).
• Find the roots: \(x = \frac{-1 \pm 5}{12}\). After completing the calculations, you get the roots \(x = \frac{2}{3}\) and \(x = -1\).

These roots help convert the quadratic into its linear factors.

The last phase involves writing the polynomial in its linear factor form using the roots and division results. Initially, with a known root of \(k = -4\), we already have a factor \(x + 4\).After achieving the quadratic roots \(\frac{2}{3}\) and \(-1\) from the quadratic formula, these are converted into factors:

• For the root \(\frac{2}{3}\), the associated factor is \(3x - 2\), derived from setting \(3x - 2 = 0\).
• For the root \(-1\), the factor is \(x + 1\).

Initially, from synthetic division, the expression was \((x + 4)(6x - 2)(x + 1)\). Simplifying the middle factor gives \((x + 4)(3x - 1)(x + 1)\), which is the simplest form of the factored polynomial. This completes the full factorization of the polynomial into linear factors, making solving polynomial equations or analyzing properties of the function more accessible.