Problem 46 - Precalculus with Calculus Previews

Question

When \(f(x)=x^{2}-3 x-1\) is divided by \(x-c\), the remainder is \(r=3 .\) Determine \(c .\)

Step-by-Step Solution

Answer

The values of \(c\) that satisfy the condition are \(c = 4\) and \(c = -1\).

Step 1: Understand the Remainder Theorem

The Remainder Theorem states that if a polynomial \(f(x)\) is divided by \(x-c\), the remainder is \(f(c)\). In this case, we are given that the remainder is 3, so \(f(c) = 3\).

Step 2: Substitute into the Polynomial

Substitute \(c\) into the given polynomial \(f(x) = x^2 - 3x - 1\) to form the equation \(f(c) = c^2 - 3c - 1\).

Step 3: Set Equal to Remainder

Since we know \(f(c) = 3\), set up the equation \(c^2 - 3c - 1 = 3\).

Step 4: Simplify the Equation

Subtract 3 from both sides to simplify the equation to \(c^2 - 3c - 4 = 0\).

Step 5: Factor the Quadratic Equation

Factor the quadratic equation to \((c - 4)(c + 1) = 0\).

Step 6: Solve for \(c\)

Set each factor equal to zero to find \(c\):- For \(c - 4 = 0\), \(c = 4\).- For \(c + 1 = 0\), \(c = -1\).

Step 7: Verify the Solutions

Since we need the remainder to be 3, substitute each solution back into \(f(x)\) and verify which value satisfies \(f(c) = 3\):- Check \(f(4): 16 - 12 - 1 = 3\). Thus, \(c = 4\) works.- Check \(f(-1): 1 + 3 - 1 = 3\). Thus, \(c = -1\) also works.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial that takes the form of \[ ax^2 + bx + c = 0 \]where \(a \), \(b\), and \(c\) are constants with \(a eq 0\). These equations are fundamental in algebra and have numerous applications in fields like physics, engineering, and medicine. Quadratic equations focus on finding values of \(x\) that make the entire expression equal to zero.

Quadratic equations can be solved through various methods including:

• Factoring: Breaking down the equation into two binomials that, when multiplied, give the original quadratic.
• The Quadratic Formula: Using \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) to find solutions.
• Completing the Square: Rewriting the equation in the form of a perfect square trinomial.

In this exercise's solution, factoring was used to solve the quadratic equation \(c^2 - 3c - 4 = 0\), which revealed two potential solutions for \(c\), namely 4 and -1.

Polynomial Division

Polynomial division is the process of dividing one polynomial by another, much like long division in numbers. When dividing a polynomial \(f(x)\) by a linear divisor \(x-c\), the result includes a quotient and possibly a remainder.

There are several steps to perform polynomial long division:

• Set up the division similar to regular long division, with the polynomial under the division bar.
• Divide the leading term of the numerator by the leading term of the divisor.
• Multiply the entire divisor by this result and subtract it from the original polynomial.
• Repeat the process with the new polynomial formed after subtraction.

The purpose of polynomial division is often to simplify expressions, find roots, and understand the behavior of functions. In the context of the problem, polynomial division led us to apply the Remainder Theorem, which connects division to the remainder in a straightforward fashion.

Polynomial Remainders

Polynomial remainders can arise naturally when dividing polynomials, especially when the division is not complete. The Remainder Theorem helps us understand these remainders better.

The Remainder Theorem states:

• If a polynomial \(f(x)\) is divided by \(x-c\), the remainder of this division is simply the value of the polynomial evaluated at \(c\): \(f(c)\).
• This means you can find the remainder without performing complete polynomial division, just by evaluating \(f(c)\).

In this exercise, when \(f(x) = x^2 - 3x - 1\) is divided by \(x-c\), the remainder theorem tells us that \(f(c) = 3\). Solving for \(c\) involves setting up the equation \(c^2 - 3c - 1 = 3\) and simplifying it to find the exact values, confirming with \(f(4)\) and \(f(-1)\), both equal 3, validating our remainder deductions.