91Ó°ÊÓ

Synthetic division is a simplified way to divide polynomials, particularly when you need to divide by a linear expression of the form \(x - c\). In this problem, since we know that \(x = 8\) is a zero of the polynomial \(-x^3 + 8x^2 + 3x - 24\), we can use synthetic division to factor \((x - 8)\) out of the polynomial.

To begin synthetic division, list the coefficients of the polynomial in descending order. For the given polynomial, these are \(-1\), \(8\), \(3\), and \(-24\). Next, write the zero \(8\) to the left and initiate the division process:

• Bring down the leading coefficient, \(-1\), to the bottom row.
• Multiply it by \(8\) and place the result, \(-8\), under the next coefficient.
• Add these numbers up to get the new bottom row entry, repeat the multiplication, continue until the end.
• The final line represents the coefficients of the quotient polynomial.

By executing these steps, we've discovered that the quotient polynomial is \(-x^2 + 3\). The lack of a remainder confirms that our division was precise.

Factoring a polynomial refers to breaking it down into a product of simpler polynomials. This is crucial for finding the roots or zeros of the polynomial. With our problem, after using synthetic division, the polynomial \(-x^3 + 8x^2 + 3x - 24\) was reduced to \(-x^2 + 3\).

To factor \(-x^2 + 3\), we first consider it as \(-1\cdot(x^2 - 3)\). We focus on \(x^2 - 3\) since it's a difference of squares type, which takes the form \((x - a)(x + a)\) where \(a = \sqrt{3}\).

• This means \(x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})\).

The factoring process simplifies the task of finding zeros, making it clear that our zeros are \(x = \sqrt{3}\) and \(x = -\sqrt{3}\). Combined with the zero we already know \(x = 8\), we have found all zeros of the original polynomial.

A quadratic equation is a second-degree polynomial equation in the form \(ax^2 + bx + c = 0\). In this scenario, the polynomial \(-x^2 + 3 = 0\) represents a type of quadratic where the linear term \(b\) is absent. This means it's simpler to solve using factoring or by rearranging terms.

Here, since \(-x^2 + 3 = 0\) can be seen through the lens of rearranging terms, we get \(x^2 = 3\). Solving for \(x\), we take the square root of both sides, giving us two solutions:

• \(x = \sqrt{3}\)
• \(x = -\sqrt{3}\)

Understanding these solutions helps complete the process of identifying all zeros of our original cubic polynomial, along with the already known zero \(x = 8\). This reinforces the concept that once a quadratic is factored, finding solutions becomes straightforward.