Question

Use the remainder theorem to determine the remainder when each polynomial is divided by \(x+2\) a) \(x^{3}+3 x^{2}-5 x+2\) b) \(2 x^{4}-2 x^{3}+5 x\) c) \(x^{4}+x^{3}-5 x^{2}+2 x-7\) d) \(8 x^{3}+4 x^{2}-19\) e) \(3 x^{3}-12 x-2\) f) \(2 x^{3}+3 x^{2}-5 x+2\)

Short answer

a) 16, b) 38, c) -23, d) -67, e) -2, f) 8

Step-by-step solution

- Review the Remainder Theorem

The Remainder Theorem states that if a polynomial f(x) is divided by x - c, the remainder is f(c).

- Identify the value for c

In this problem, the polynomial is divided by x + 2, so we need to find the remainder when x = -2.

- Evaluate each polynomial at x = -2

Substitute x = -2 into each polynomial and find the value of the polynomial at this point.

Step 4a - Evaluate the first polynomial

For the polynomial f(x) = x^3 + 3x^2 - 5x + 2, substitute x = -2:f(-2) = (-2)^3 + 3(-2)^2 - 5(-2) + 2f(-2) = -8 + 12 + 10 + 2 = 16

Step 4b - Evaluate the second polynomial

For the polynomial f(x) = 2x^4 - 2x^3 + 5x, substitute x = -2:f(-2) = 2(-2)^4 - 2(-2)^3 + 5(-2)f(-2) = 32 + 16 - 10 = 38

Step 4c - Evaluate the third polynomial

For the polynomial f(x) = x^4 + x^3 - 5x^2 + 2x - 7, substitute x = -2:f(-2) = (-2)^4 + (-2)^3 - 5(-2)^2 + 2(-2) - 7f(-2) = 16 - 8 - 20 - 4 - 7 = -23

Step 4d - Evaluate the fourth polynomial

For the polynomial f(x) = 8x^3 + 4x^2 - 19, substitute x = -2:f(-2) = 8(-2)^3 + 4(-2)^2 - 19f(-2) = -64 + 16 - 19 = -67

Step 4e - Evaluate the fifth polynomial

For the polynomial f(x) = 3x^3 - 12x - 2, substitute x = -2:f(-2) = 3(-2)^3 - 12(-2) - 2f(-2) = -24 + 24 - 2 = -2

Step 4f - Evaluate the sixth polynomial

For the polynomial f(x) = 2x^3 + 3x^2 - 5x + 2, substitute x = -2:f(-2) = 2(-2)^3 + 3(-2)^2 - 5(-2) + 2f(-2) = -16 + 12 + 10 + 2 = 8

Key concepts

Polynomial Division

Polynomial division is similar to traditional long division, but it involves dividing polynomials by other polynomials. Instead of numbers, you're working with variables and their coefficients. When dividing a polynomial, the goal is to determine the quotient and remainder. In simpler terms, you're breaking down a complex polynomial into smaller, more manageable parts. This helps in finding solutions to polynomial equations and understanding their behavior. Remember, a polynomial of degree n divided by a polynomial of degree m will have a quotient of degree n-m. If the divider is a linear polynomial (like x - c), the remainder will always be a constant value.

Evaluation of Polynomials

Evaluating polynomials is the process of finding the value of a polynomial function for a specific value of the variable. This often involves substituting a numerical value into the polynomial and simplifying. For example, if we have a polynomial f(x) = x^3 + 3x^2 - 5x + 2, and we want to evaluate it for x = -2, we replace x with -2:

f(-2) = (-2)^3 + 3(-2)^2 - 5(-2) + 2
f(-2) = -8 + 12 + 10 + 2
f(-2) = 16

By performing these calculations, we determine the value of the polynomial at x = -2. This method is often used in the Remainder Theorem.

Finding Remainders

The Remainder Theorem provides a quick way to find the remainder when a polynomial is divided by a linear divisor. According to the theorem, for a given polynomial f(x) and a linear divisor (x - c), the remainder of this division is simply f(c). This means you only need to substitute c into the polynomial and compute the result. Let's take f(x) = 3x^3 - 12x - 2 and divide it by (x + 2). According to the Remainder Theorem, we set x equal to -2:

f(-2) = 3(-2)^3 - 12(-2) - 2
f(-2) = -24 + 24 - 2
f(-2) = -2

Hence, -2 is the remainder when 3x^3 - 12x - 2 is divided by (x + 2).

Precalculus

Precalculus encompasses a range of mathematical concepts that prepare students for calculus. Understanding polynomials and their behavior is a fundamental part of this subject. Topics such as polynomial division, the Remainder Theorem, and the evaluation of polynomials are critical. They form the foundation for more complex calculus concepts like limits, derivatives, and integrals. In precalculus, students also work on mastering functions, trigonometry, and the analysis of mathematical models. These skills are essential as they lay the groundwork for higher-level math and various applications in science and engineering.

Mathematical Substitution

Mathematical substitution is an essential technique in algebra. It involves replacing a variable with a specific value or another expression. When evaluating polynomials or applying the Remainder Theorem, substitution helps determine the value of the polynomial for particular inputs. For example, in the polynomial f(x) = 2x^3 + 3x^2 - 5x + 2, if we want to find the remainder when divided by x + 2, we substitute x with -2:

f(-2) = 2(-2)^3 + 3(-2)^2 - 5(-2) + 2
f(-2) = -16 + 12 + 10 + 2
f(-2) = 8

This substitution reveals the remainder is 8. Substitution simplifies complex expressions, making them easier to solve and understand.