Problem 45 - Precalculus with Calculus Previews

Question

Find a value of \(k\) such that the remainder in the division of \(f(x)=3 x^{2}-4 k x+1\) by \(d(x)=x+3\) is \(r\) \(=-20\)

Step-by-Step Solution

Answer

The value of \(k\) is \(-4\).

Step 1: Identify the Remainder Theorem

According to the Remainder Theorem, if a polynomial \(f(x)\) is divided by \(x-a\), then the remainder is \(f(a)\). In your case, the divisor is \(x+3\), which can be rewritten as \(x-(-3)\), so you need to find \(f(-3)\) to solve the problem.

Step 2: Set Up the Equation

Since the problem states that the remainder \(r = -20\), you need to set up the equation \(f(-3) = -20\). The function \(f(x) = 3x^2 - 4kx + 1\), so substitute \(-3\) for \(x\) to get \(3(-3)^2 - 4k(-3) + 1 = -20\).

Step 3: Simplify and Solve for \(k\)

First, calculate \((-3)^2 = 9\). Therefore, the equation becomes \(3(9) + 12k + 1 = -20\), which simplifies to \(27 + 12k + 1 = -20\). Combine like terms to get \(28 + 12k = -20\).

Step 4: Isolate \(k\)

Subtract 28 from both sides of the equation: \(12k = -20 - 28\), which simplifies to \(12k = -48\).

Step 5: Divide to Find \(k\)

Finally, divide both sides by 12 to solve for \(k\): \(k = \frac{-48}{12} = -4\).

Key Concepts

Polynomial Division

Polynomial division is a process similar to long division but applied to polynomials. Understanding it is crucial for tackling many algebraic problems. In simple terms, it involves dividing one polynomial—called the dividend—by another, called the divisor, to obtain a quotient and possibly a remainder. This is often written in the form of: \[ f(x) = d(x) \cdot q(x) + r(x) \] where:

• \( f(x) \) is the dividend (the polynomial we are dividing),
• \( d(x) \) is the divisor,
• \( q(x) \) is the quotient, and
• \( r(x) \) is the remainder.

The main goal is to find values for \( q(x) \) and \( r(x) \). In our exercise, the function \( f(x) = 3x^2 - 4kx + 1 \) was divided by a simple linear function \( d(x) = x+3 \). Understanding polynomial division makes it easier to manipulate and solve equations, especially when the remainder needs to be determined or used, as seen in our example.

Remainder Theorem Application

The Remainder Theorem is a neat trick in algebra that simplifies finding the remainder of polynomial division. It states that if a polynomial \( f(x) \) is divided by \( x-a \), the remainder of this division is simply \( f(a) \).
For our problem, we are dividing \( f(x) = 3x^2 - 4kx + 1 \) by \( d(x) = x+3 \), identified by rewriting the divisor as \( x-(-3) \).
To apply the theorem, we substitute the value \( x = -3 \) into \( f(x) \). This yields the equation:

• \( f(-3) = 3(-3)^2 - 4k(-3) + 1 = -20 \)

The theorem provides a straightforward way to find the remainder without needing to do full polynomial division, making it quicker and more efficient. Thus, once the polynomial is evaluated, setting it equal to the given remainder (\(-20\) in this case) allows us to solve for the unknown \( k \).

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to find unknown variables. In mathematics, it's a crucial skill that allows us to solve equations more efficiently. Within the context of this problem, after applying the Remainder Theorem, algebraic manipulation was used to find the value of \( k \).
The transformation of the equation was as follows:

• Substitute \( -3 \) into \( f(x) \): \( 3(9) + 12k + 1 = -20 \)
• Simplify to: \( 27 + 12k + 1 = -20 \)
• Combine like terms: \( 28 + 12k = -20 \)
• Isolate \( k \): \( 12k = -48 \)
• Solve for \( k \): \( k = \frac{-48}{12} = -4 \)

Each step in the manipulation brought us closer to the solution by isolating \( k \) on one side of the equation. This demonstrates the power of algebraic manipulation in solving polynomial-related problems, facilitating a deeper understanding of the relationships between the elements of the polynomial.