Question

For each dividend, determine the value of \(k\) if the remainder is 3 a) \(\left(x^{3}+4 x^{2}-x+k\right) \div(x-1)\) b) \(\left(x^{3}+x^{2}+k x-15\right) \div(x-2)\) c) \(\left(x^{3}+k x^{2}+x+5\right) \div(x+2)\) d) \(\left(k x^{3}+3 x+1\right) \div(x+2)\)

Short answer

a) -1 b) 3 c) 2 d) -1

Step-by-step solution

Identify the divisor and apply the Remainder Theorem

The Remainder Theorem states that for a polynomial function \(f(x)\) and a divisor \(x - c\), the remainder of the division \(f(x)\) by \(x - c\) is \(f(c)\). Given that the remainder is 3, set \(f(c) = 3\).

Solve for Part (a)

Given \(f(x) = x^{3} + 4 x^{2} - x + k\) and \(x - 1\) as the divisor, apply the Remainder Theorem with \(c = 1\): \[ f(1) = 1^{3} + 4(1)^{2} - 1 + k = 3 \]This simplifies to: \[ 1 + 4 - 1 + k = 3 \]\[ 4 + k = 3 \]Solve for \(k\): \[ k = 3 - 4 \]\[ k = -1 \]

Solve for Part (b)

Given \(f(x) = x^{3} + x^{2} + kx - 15\) and \(x - 2\) as the divisor, apply the Remainder Theorem with \(c = 2\): \[ f(2) = 2^{3} + 2^{2} + k(2) - 15 = 3 \]This simplifies to: \[ 8 + 4 + 2k - 15 = 3 \]\[ -3 + 2k = 3 \]Solve for \(k\): \[ 2k = 6 \]\[ k = 3 \]

Solve for Part (c)

Given \(f(x) = x^{3} + kx^{2} + x + 5\) and \(x + 2\) as the divisor, apply the Remainder Theorem with \(c = -2\): \[ f(-2) = (-2)^{3} + k(-2)^{2} + (-2) + 5 = 3 \]This simplifies to: \[ -8 + 4k - 2 + 5 = 3 \]\[ -5 + 4k = 3 \]Solve for \(k\): \[ 4k = 8 \]\[ k = 2 \]

Solve for Part (d)

Given \(f(x) = kx^{3} + 3x + 1\) and \(x + 2\) as the divisor, apply the Remainder Theorem with \(c = -2\): \[ f(-2) = k(-2)^{3} + 3(-2) + 1 = 3 \]This simplifies to: \[ -8k - 6 + 1 = 3 \]\[ -8k - 5 = 3 \]Solve for \(k\): \[ -8k = 8 \]\[ k = -1 \]

Key concepts

polynomial division

Polynomial division is a process of dividing a polynomial by another polynomial of lesser degree. It is similar to long division of numbers. The objective is to determine how many times the divisor can be multiplied to get a polynomial that is either equal to or just less than the dividend. This method helps in simplifying polynomials, beyond just finding quotients and remainders. Here are the steps in polynomial division:
- Begin with the highest degree terms.
- Divide the leading term of the dividend by the leading term of the divisor to find the first term of the quotient.
- Multiply the entire divisor by this term and subtract the result from the dividend.
- Repeat with the new polynomial until you reach a remainder that is of a lower degree than the divisor.
Understanding polynomial division is crucial for grasping other algebraic concepts, like the Remainder Theorem.

divisors

Divisors refer to the polynomial by which another polynomial (the dividend) is being divided. In the problem provided, each polynomial is divided by a linear term. For example
- In part (a), the divisor is \( x - 1 \)
- In part (b), it is \( x - 2 \)
- In part (c), it is \( x + 2 \)
- and in part (d), it is \( x + 2 \)
These linear divisors play a key role when applying the Remainder Theorem. The Remainder Theorem simplifies finding the remainder, making it easier than performing long division. This step is crucial, as it turns a complex polynomial division into a much simpler algebraic substitution.

finding remainder

Finding the remainder when dividing a polynomial by a linear divisor can be done easily with the Remainder Theorem. The Remainder Theorem states that the remainder of the division of a polynomial \( f(x) \) by \( x - c \) is simply the value of \( f(c) \). This means that:
- Substitute the value of \( c \) into the polynomial \( f(x) \)
For instance, when dividing \( x^3 + 4x^2 - x + k \) by \( x - 1 \), substitute \( x = 1 \) into the polynomial:
\[ f(1) = 1^3 + 4(1)^2 - 1 + k = 3 \]
This simplifies to \[ 4 + k = 3 \], giving us \[ k = -1 \]
Using this method helps find the remainder quickly without performing detailed division.

solving for constants

Solving for constants involves substituting a specific value into a polynomial, as guided by the Remainder Theorem, and then solving the resulting equation. This is often used when given a condition for the remainder, like we have here where the remainder is always 3. Let's break it down:
1. **Identify \( c \)**: Determine the value of \( c \) from the divisor \( x - c \).
2. **Substitute \( c \)**: Plug this \( c \) into the polynomial \( f(x) \).
3. **Solve equation**: Set \( f(c) \) equal to the remainder (which is 3) and solve for the unknown constant \( k \).
For example, in part (b) \[ f(2) = 2^3 + 2^2 + 2k - 15 = 3 \] simplifies to \[ 4 + 2k - 3 = 3; 2k = 6; k = 3 \]. This process allows you to find the constant \( k \) required to meet the remainder condition.