Problem 35
Question
Find values of \(k\) so that each remainder is \(3 .\) $$ \left(x^{2}+k x-17\right) \div(x-2) $$
Step-by-Step Solution
Verified Answer
The value of \( k \) is 8.
1Step 1: Understand the Remainder Theorem
According to the Remainder Theorem, the remainder of the division of a polynomial \( f(x) \) by \( x-a \) is \( f(a) \). So in this case, we need to substitute \( x = 2 \) into \( x^2 + kx - 17 \) and set it equal to the given remainder, 3.
2Step 2: Set Up the Equation from the Remainder Theorem
The given polynomial is \( f(x) = x^2 + kx - 17 \). According to the Remainder Theorem, we have:\[f(2) = 2^2 + k(2) - 17 = 3\]Simplifying, this becomes:\[4 + 2k - 17 = 3\]
3Step 3: Simplify the Equation
Simplify the equation obtained in Step 2:\[2k + 4 - 17 = 3\]\[2k - 13 = 3\]
4Step 4: Solve for \(k\)
Add 13 to both sides of the equation:\[2k = 16\]Now, divide both sides by 2 to solve for \( k \):\[k = 8\]
Key Concepts
Polynomial DivisionSolving EquationsSubstitution Method
Polynomial Division
Polynomial division is a method used to divide polynomials, similar to how we divide numbers. It's often employed when trying to simplify complex polynomials or when solving polynomial equations. In polynomial division, you divide the terms of the dividend by the highest degree term of the divisor, subtract, and then bring down the next term, much like long division with numbers. This process is repeated until you reach a remainder or other stopping point.
In our exercise, we're determining what makes the remainder of a polynomial division equal a specific value (3, in this case). This often involves using techniques like the Remainder Theorem for efficiency, rather than performing polynomial long division directly.
- Start by focusing on the highest power terms.
- Work your way down to constants.
- Record each step to track division progress.
In our exercise, we're determining what makes the remainder of a polynomial division equal a specific value (3, in this case). This often involves using techniques like the Remainder Theorem for efficiency, rather than performing polynomial long division directly.
Solving Equations
Solving equations involves finding the values of variables that make the equations true, which is a fundamental skill in mathematics. Often, especially in polynomial contexts, equations are formed and solved using the properties of polynomial functions and theorems like the Remainder Theorem, as illustrated in the given exercise.
In this exercise, the equation we solved was formed by substituting a value for \(x\) to make the polynomial expression equal to the remainder, as guided by the Remainder Theorem.
- Identify the equation you need to solve.
- Isolate variables of interest.
- Apply appropriate mathematical operations to unveil the variable values.
In this exercise, the equation we solved was formed by substituting a value for \(x\) to make the polynomial expression equal to the remainder, as guided by the Remainder Theorem.
Substitution Method
The substitution method is a key approach used in algebra to find the value of unknowns. It's also particularly useful in polynomial division, especially when linked with the Remainder Theorem. Substitution involves replacing variables with specific numbers to simplify the solving process or to verify the solutions of equations.
In our example problem, substituting \( x = 2 \) into the polynomial \( x^2 + kx - 17 \) allowed us to find \( k \), ensuring the remainder was consistent with the desired value (3). This transformation is crucial as it directly connects each term of the polynomial to its divisor, offering a straightforward path to the solution.
- Select the variable or expression to be substituted.
- Substitute the chosen variable/expression with its known value.
- Simplify the equation to solve for any remaining unknowns.
In our example problem, substituting \( x = 2 \) into the polynomial \( x^2 + kx - 17 \) allowed us to find \( k \), ensuring the remainder was consistent with the desired value (3). This transformation is crucial as it directly connects each term of the polynomial to its divisor, offering a straightforward path to the solution.
Other exercises in this chapter
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