Problem 41 - Intermediate Algebra

Question

Let \(P(x)=2 x^{3}-4 x^{2}+2 x-1 .\) Evaluate \(P(x)\) by substituting the given value of \(x\) into the polynomial and simplifying. Then evaluate the polynomial by using the remainder theorem and synthetic division. See Example 4. $$ P(1) $$

Step-by-Step Solution

Answer

The value of \(P(1)\) is \(-1\).

Step 1: Substitute x and Evaluate

First, substitute the value of \(x = 1\) into the polynomial \(P(x) = 2x^3 - 4x^2 + 2x - 1\). This gives us: \[P(1) = 2(1)^3 - 4(1)^2 + 2(1) - 1\] Simplify the expression: \[= 2(1) - 4(1) + 2 - 1\] \[= 2 - 4 + 2 - 1\] \[= -1\]

Step 2: Apply the Remainder Theorem

The Remainder Theorem states that for a polynomial \(P(x)\), the remainder of the division of \(P(x)\) by \(x - a\) is \(P(a)\). Here, to find \(P(1)\), we can use synthetic division to divide \(P(x)\) by \(x - 1\).

Step 3: Set Up Synthetic Division

Write the coefficients of the polynomial \(P(x)\): \[2, -4, 2, -1\]. Write \(1\) to the left (the divisor \(x - 1\) gives us 1 because \(x - a = x - 1\)).

Step 4: Perform Synthetic Division

1. Bring down the leading coefficient (2).2. Multiply 1 by this leading coefficient: \(1 \times 2 = 2\).3. Add this result to the next coefficient: \(-4 + 2 = -2\).4. Multiply 1 by the result \(-2\): \(1 \times (-2) = -2\).5. Add this to the next coefficient: \(2 + (-2) = 0\).6. Multiply 1 by the result \(0\): \(1 \times 0 = 0\).7. Add this to the last coefficient: \(-1 + 0 = -1\). The remainder is \(-1\).

Step 5: Interpret the Result

Seing that the remainder of the synthetic division is \(-1\), this confirms that \(P(1) = -1\) through both methods: substitution and the Remainder Theorem.

Key Concepts

Remainder Theorem

The Remainder Theorem offers a quick way to evaluate polynomials at specific values. It tells us that if a polynomial \(P(x)\) is divided by \(x - a\), the remainder of that division is exactly \(P(a)\). This means you don't have to go through the entire long division process each time you need to evaluate \(P(x)\) at a given \(x\). For example, if we want to find out the value of \(P(1)\) for the polynomial \(P(x) = 2x^3 - 4x^2 + 2x - 1\), we substitute the value into the polynomial to direct find the remainder: - Substitute \(x = 1\).- Calculate the simplified expression: \(P(1) = 2(1) - 4(1) + 2 - 1 = -1\). Thus, by the Remainder Theorem, the remainder when \(P(x)\) is divided by \(x-1\) is \(-1\). This highlights the theorem's utility in evaluating polys without full division.

Synthetic Division

Synthetic division is a streamlined method to divide polynomials when the divisor is of the form \(x - a\). It's faster and less cumbersome than traditional polynomial long division. Let's walk through how it works, using \(x - 1\) as the divisor for our polynomial \(P(x) = 2x^3 - 4x^2 + 2x - 1\).

• Represent the polynomial by its coefficients: \(2, -4, 2, -1\).
• Write down \(a = 1\) derived from the divisor \(x - 1\).
• Bring down the first coefficient (2).
• Multiply \(1\) by this leading coefficient. Add the result to the next coefficient: \(-4 + 2 = -2\).
• Continue: \(1\) times \(-2\) results in another interim value \(-2\), affecting the next coefficient: \(2 + (-2) = 0\).
• Repeat one more time for the last part: \(1 \times 0 = 0\). Finally add this to the last: \(-1 + 0 = -1\).

The final remainder, \(-1\), matches \(P(1)\). This concise division confirms calculations easily!

Substitution Method

The substitution method in polynomial evaluation is the most straightforward. Simply replace \(x\) with the value provided and calculate. This involves several operations based on the polynomial's structure. Let's see it in action for the polynomial \(P(x) = 2x^3 - 4x^2 + 2x - 1\), specifically for \(x = 1\):

• Substitute \(x = 1\) into \(P(x)\).
• Solve step-by-step: \(P(1) = 2(1)^3 - 4(1)^2 + 2(1) - 1\).
• Simplify the expression: \(= 2(1) - 4(1) + 2 - 1\).
• Perform the arithmetic: \(= 2 - 4 + 2 - 1\).

The result is \(-1\), which confirms the reliability of simple arithmetic for polynomial evaluation. It's all about accurately substituting and simplifying.