Problem 39 - Algebra and Trigonometry

Question

\(39-51\) . Use synthetic division and the Remainder Theorem to evaluate \(P(c) .\) $$ P(x)=4 x^{2}+12 x+5, \quad c=-1 $$

Step-by-Step Solution

Answer

The value of \(P(-1)\) is \(-3\).

Step 1: Set Up Synthetic Division

Write down the coefficients of the polynomial \(P(x) = 4x^2 + 12x + 5\) in order: \(4, 12, 5\). Next, write \(c = -1\) to the left outside the synthetic division bracket.

Step 2: Begin Synthetic Division

Carry down the leading coefficient, which is \(4\), as the first number of the bottom row.

Step 3: Multiply and Add

Multiply \(c = -1\) by the carried down number \(4\) and write the result under the next coefficient, \(12\). Then, add this to \(12\) to get \(12 + (-4) = 8\). Write \(8\) below.

Step 4: Repeat Multiply and Add

Multiply \(-1\) (the divisor) by \(8\) (the new number in the bottom row), resulting in \(-8\). Add this to the next coefficient, \(5\), giving \(5 + (-8) = -3\). Write \(-3\) below.

Step 5: Interpret the Remainder

The final number, \(-3\), is the remainder. According to the Remainder Theorem, the remainder is equal to \(P(c)\), so \(P(-1) = -3\).

Key Concepts

Remainder Theorem

The Remainder Theorem is a useful tool when working with polynomials, especially in synthetic division. It states that when a polynomial \( P(x) \) is divided by a linear divisor of the form \( x - c \), the remainder of this division is simply \( P(c) \). This means that instead of performing the entire polynomial division, if we're only interested in finding \( P(c) \) for a specific value of \( c \), we can directly use this theorem.
When we apply the Remainder Theorem as shown in the original solution, we take the polynomial \( P(x) = 4x^2 + 12x + 5 \) and substitute the divisor \( c = -1 \). As synthetic division progresses, we reach the final step where the remainder from the division corresponds exactly to \( P(-1) \). Here, the remainder is -3, which tells us that the value of the polynomial at \( x = -1 \) is \(-3\). This greatly simplifies calculations and allows us to quickly evaluate polynomial expressions.

Evaluate Polynomial

Evaluating a polynomial at a particular value, say \( x = c \), means calculating the value of the polynomial when \( x \) is replaced by \( c \). For polynomial \( P(x) = 4x^2 + 12x + 5 \) and \( c = -1 \), to evaluate \( P(-1) \), we essentially follow the steps of the synthetic division.

• Start with the polynomial's coefficients: 4, 12, and 5.
• Use the divisor \( c = -1 \) in the synthetic division.
• Progress through the steps until the remainder is obtained.

The remainder from the synthetic division directly gives \( P(-1) \), meaning the process is simple and free from the operations of substituting \( x \) and calculating step-by-step. So when students want to find \( P(c) \) for any polynomial, they can use synthetic division to efficiently and accurately find the value. This is particularly handy for higher-degree polynomials where manual substitution could become cumbersome.

Polynomial Division

Polynomial division can often seem daunting, but it's made much simpler using synthetic division. This technique allows one to divide polynomials more quickly and with less error, especially when dealing with a linear factor such as \( x - c \). In general polynomial division, we try to determine the quotient and remainder when the polynomial is divided by a given factor.
Synthetic division simplifies this process by focusing on coefficients rather than the entire polynomial terms:

• Write down the coefficients of the polynomial in order.
• Use the given divisor outside the synthetic division setup.
• Proceed through the division by multiplying, adding, and writing results in sequence.

This process is similar to traditional long division but reduces the number of steps involved and the potential for errors in calculations. Moreover, synthetic division doesn't just find the remainder when dividing by \( x - c \); it provides a deeper understanding by highlighting the relationship between divisors and polynomial roots, bridging advanced mathematical concepts with practical computational techniques. This makes it a foundational skill necessary for anyone tackling polynomial problems effectively.