Problem 56 - Algebra and Trigonometry

Question

\(53-56\) . Use the Factor Theorem to show that \(X-c\) is a factor of \(P(x)\) for the given value(s) of \(c .\) $$ P(x)=x^{4}+3 x^{3}-16 x^{2}-27 x+63, \quad c=3,-3 $$

Step-by-Step Solution

Answer

Both \(x-3\) and \(x+3\) are factors of \(P(x)\).

Step 1: Recall Factor Theorem

The Factor Theorem states that \(x-c\) is a factor of \(P(x)\) if and only if \(P(c) = 0\). Therefore, to determine if \(x-c\) is a factor, we need to evaluate \(P(c)\) for the given values of \(c\).

Step 2: Evaluate P(x) for c = 3

Substitute \(c = 3\) into \(P(x)\):\[P(3) = 3^4 + 3 \cdot 3^3 - 16 \cdot 3^2 - 27 \cdot 3 + 63.\]Calculate each term separately:- \(3^4 = 81\)- \(3 \cdot 3^3 = 3 \cdot 27 = 81\)- \(-16 \cdot 3^2 = -16 \cdot 9 = -144\)- \(-27 \cdot 3 = -81\)- Constant term \(+63\)Combine these results:\[P(3) = 81 + 81 - 144 - 81 + 63 = 0.\] Thus, \(x - 3\) is a factor of \(P(x)\).

Step 3: Evaluate P(x) for c = -3

Substitute \(c = -3\) into \(P(x)\):\[P(-3) = (-3)^4 + 3 \cdot (-3)^3 - 16 \cdot (-3)^2 - 27 \cdot (-3) + 63.\]Calculate each term separately:- \((-3)^4 = 81\)- \(3 \cdot (-3)^3 = 3 \cdot (-27) = -81\)- \(-16 \cdot (-3)^2 = -16 \cdot 9 = -144\)- \(-27 \cdot (-3) = 81\)- Constant term \(+63\)Combine these results:\[P(-3) = 81 - 81 - 144 + 81 + 63 = 0.\] Thus, \(x + 3\) is a factor of \(P(x)\).

Step 4: Conclusion

Based on the evaluations, both \(x - 3\) and \(x + 3\) (equivalent to \(x - (-3)\)) are factors of \(P(x)\). This confirms the Factor Theorem.

Key Concepts

Polynomial Functions

Polynomial functions are a class of functions that are essential in both algebra and calculus. They are expressions that involve variables raised to whole number powers. A polynomial function can be described in the form:

• \( P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\),
• where each term consists of a coefficient \(a_i\) and a non-negative integer exponent \(n\).

The highest power of the variable is called the degree of the polynomial. For example, in the polynomial \(x^4 + 3x^3 - 16x^2 - 27x + 63\), which is degree 4, the highest power is 4.
Polynomials can have various numbers of terms, but what unites them is this structure of coefficients and powers. Polynomials are smooth, continuous functions and are defined for all values of \(x\). This makes them vital for graphing and analyzing real-life phenomena, from predicting population growth to designing roller coasters.

Factoring Polynomials

Factoring polynomials is a critical process in algebra, allowing us to express a polynomial as a product of its factors. This technique is used to simplify expressions, solve polynomial equations, and evaluate limits.By applying the Factor Theorem, which states that \(x - c\) is a factor of \(P(x)\) if and only if \(P(c) = 0\), we can confirm whether a given polynomial can be divided evenly by a certain binomial.
For example, consider \(P(x) = x^4 + 3x^3 - 16x^2 - 27x + 63\). To factor this polynomial, we evaluate at specific values, like \(c = 3\) and \(c = -3\). If substituting these values into \(P(x)\) renders a zero result, it means exactly that \(x-c\) (or \(x-(-c)\)) is indeed a factor.
Factoring reduces complex problems into simpler, more manageable expressions. It promotes accuracy and efficiency, especially when solving larger algebraic equations.

Roots of Equations

Roots of equations are the solutions where the polynomial equals zero. Finding these roots is pivotal because they reveal the values of \(x\) where \(P(x) = 0\). Each root is essentially an intercept on the x-axis of a graphical representation and corresponds to a factor of the polynomial.Using the factor theorem to evaluate potential roots, as demonstrated in the exercise, helps identify these solutions efficiently. In the solved example, both \(c = 3\) and \(c = -3\) yield roots when substituted into the polynomial \(P(x)\). This confirms that \(x = 3\) and \(x = -3\) make the function equal zero, thus they are the roots.
Once the roots are found, they can help further understand the behavior of the polynomial function. This includes where the function intersects the x-axis, and what factors the polynomial can be broken down into. Understanding roots is essential for students to master anything from simple equations to complex calculus problems.