91Ó°ÊÓ

Cubic polynomials are polynomials of degree three. This means the highest power of the variable (usually denoted as \(z\) or \(x\)) is three. A general form of a cubic polynomial looks like this: \(az^3 + bz^2 + cz + d\), where \(a\), \(b\), \(c\), and \(d\) are constants, and \(a eq 0\). Cubic polynomials can have up to three real roots, and they often show a wavy curve when plotted on a graph. When solving cubic polynomials, you usually start by rearranging the equation to bring all terms to one side, setting it to equal zero. This lays the groundwork for factoring or applying other methods to find the roots. In our exercise, the original equation was transformed to \(3z^3 + 5z^2 - 12z - 20 = 0\).

Factoring is the process of breaking down an equation into simpler components (factors) that when multiplied together give the original equation. This is particularly useful in solving polynomial equations. There are several factoring techniques, one of the most useful being factoring by grouping. For cubic polynomials, you initially may try to separate terms to identify common factors. In the example, the equation \(3z^3 + 5z^2 - 12z - 20\) was factored by grouping into two parts: \((3z^3 + 5z^2)\) and \((-12z - 20)\). Then common factors were pulled out: \(z^2(3z + 5)\) from the first part and \(-4(3z + 5)\) from the second. This reveals the common factor \((3z + 5)\). Finally, factoring by grouping allowed us to express the cubic polynomial as \((3z + 5)(z^2 - 4)\).

Real solutions are the values of the variable that make the equation true when substituted back into it. They are the roots that lie on the real number line. Once a polynomial is factored, finding the real solutions involves setting each factor equal to zero and solving for the variable. For the factored cubic polynomial \((3z + 5)(z^2 - 4) = 0\), we split it into \(3z + 5 = 0\) and \(z^2 - 4 = 0\). Solving these equations, we find:

• \(3z + 5 = 0 \rightarrow z = -\frac{5}{3}\)
• \(z^2 - 4 = 0 \rightarrow (z - 2)(z + 2) \rightarrow z = 2\) and \(z = -2\)

Therefore, the real solutions are \(z = -\frac{5}{3}\), \(z = 2\), and \(z = -2\). These are the values where the original equation equals zero.

The difference of squares is a specific factoring technique used for expressions of the form \(a^2 - b^2\). This can be factored into \((a - b)(a + b)\). The reason it's called the 'difference of squares' is that it expresses one square subtracted from another. In the context of our exercise, after factoring the cubic polynomial into \((3z + 5)(z^2 - 4)\), we applied this method to the quadratic factor \(z^2 - 4\). Here \(a\) is \(z\) and \(b\) is \(2\), so \(z^2 - 4\) can be rewritten as \((z - 2)(z + 2)\). This reveals that \(z^2 - 4\) is a difference of squares, which further simplifies our equation, facilitating the identification of real solutions.