Problem 26 - Algebra Form and Function

Question

If \(p(x)=x^{4}-2 x^{2}+1,\) find (a) \(p(0)\) (b) \(p(2)\) (c) \(p\left(t^{2}\right)\). (d) The values of \(x\) such that \(p(x)=0\)

Step-by-Step Solution

Answer

Answer: (a) p(0) = 1 (b) p(2) = 9 (c) p(t^2) = t^8 - 2t^4 + 1 (d) The values of x that make p(x) = 0 are x = ±1.

Step 1: Evaluate p(0)

Plug in \(x = 0\) into the polynomial function: $$p(0) = 0^4 - 2\cdot0^2 + 1 = 1$$

Step 2: Evaluate p(2)

Plug in \(x = 2\) into the polynomial function: $$p(2) = 2^4 - 2\cdot2^2 + 1 = 16 - 8 + 1 = 9$$

Step 3: Evaluate p(t^2)

Replace \(x\) with \(t^2\) in the polynomial function: $$p(t^2) = (t^2)^4 - 2(t^2)^2 + 1 = t^8 - 2t^4 + 1$$

Step 4: Find x such that p(x) = 0

We need to find the roots of the polynomial function: \begin{align*} x^4 - 2x^2 + 1 &= 0 \\ (x^2 - 1)^2 &= 0 \\ x^2 - 1 &= 0 \\ x^2 &= 1 \end{align*} The solutions for the equation are \(x = \pm 1\). So the answers are: (a) \(p(0) = 1\) (b) \(p(2) = 9\) (c) \(p(t^2) = t^8 - 2t^4 + 1\) (d) The values of \(x\) such that \(p(x) = 0\) are \(x = \pm 1\).

Key Concepts

Plug-in Method

The plug-in method is a straightforward approach to evaluate polynomial functions. It involves substituting a specific value into the polynomial in place of the variable. This method helps in quickly determining the result of the polynomial for particular values, such as finding \(p(0)\) or \(p(2)\) in the original exercise.

To use the plug-in method effectively:

• Identify the variable in the polynomial, often denoted as \(x\).
• Choose the value you want to substitute for \(x\), such as 0 or 2 in the exercise.
• Replace \(x\) in the polynomial with the chosen value.
• Simplify the expression to find the result.

For example, plugging in \(x = 2\) into the polynomial \(p(x) = x^4 - 2x^2 + 1\) means substituting every occurrence of \(x\) with 2, leading to \(p(2) = 2^4 - 2 \cdot 2^2 + 1\). Simplifying this gives \(16 - 8 + 1 = 9\).

Using the plug-in method can also be extended to expressions like \(p(t^2)\) where you substitute \(t^2\) for \(x\). This flexibility makes it a powerful tool for polynomial evaluations.

Roots of a Polynomial

The roots of a polynomial are the values of \(x\) that make the polynomial equal to zero. Finding these roots is a fundamental problem in algebra because they represent where the polynomial graph crosses the x-axis. In the context of our exercise, we need to find the roots for the equation \(p(x) = x^4 - 2x^2 + 1 = 0\).

To find the roots of a polynomial:

• Set the polynomial equation equal to zero.
• Simplify the equation, if possible, by factoring or other algebraic methods.
• Solve the resulting equations to find the values of \(x\).

In our example, we rewrite \(x^4 - 2x^2 + 1\) as \((x^2 - 1)^2 = 0\). Solving \((x^2 - 1) = 0\) gives \(x^2 = 1\), which provides the roots \(x = \pm 1\). These are the values where the polynomial equals zero, indicating the points where the polynomial graph intersects the x-axis.

Evaluation of Polynomials

Evaluating polynomials is the process of calculating the value of a polynomial at specific points. This can be for numerical values, as seen when finding \(p(0)\) or \(p(2)\), or for algebraic expressions, such as \(p(t^2)\). Understanding how to evaluate polynomials is essential for many mathematical applications, including solving equations and analyzing functions.

To evaluate a polynomial:

• Identify the polynomial function, which is usually in the form \(p(x)\).
• Substitute the given value or expression into the polynomial, replacing \(x\).
• Perform all arithmetic operations to simplify the expression and arrive at the final value.

In the exercise, evaluating \(p(t^2)\) involves substituting \(x\) with \(t^2\), resulting in the transformed polynomial \(p(t^2) = t^8 - 2t^4 + 1\). This demonstrates that polynomial evaluation isn't limited to just numbers but can involve other polynomials or algebraic expressions.

Proper evaluation allows students to verify solutions, understand polynomial behaviors, and apply them in various contexts seamlessly.