Problem 5 - Basic Technical Mathematics with Calculus

Question

Find the remainder by long division. $$\left(x^{3}+2 x-8\right) \div(x-2)$$

Step-by-Step Solution

Answer

The remainder is 4.

Step 1: Set Up the Long Division

Write down the dividend \(x^3 + 0x^2 + 2x - 8\) and the divisor \(x - 2\). Ensure all terms of the polynomial are represented, filling in any gaps with zero coefficients. Then, write it in a long division format.

Step 2: Divide the Leading Terms

Divide the leading term of the dividend \(x^3\) by the leading term of the divisor \(x\). This gives \(x^2\). Write \(x^2\) on top of the division bracket. This represents the first term of the quotient.

Step 3: Multiply and Subtract

Multiply \(x^2\) by \(x - 2\) yielding \(x^3 - 2x^2\). Write this beneath the dividend, and subtract it from the original dividend to get the new polynomial \(2x^2 + 2x - 8\).

Step 4: Repeat Division

Divide \(2x^2\) by \(x\) to get \(2x\). Write it above the division line next to \(x^2\). Multiply \(2x\) by \(x - 2\) to get \(2x^2 - 4x\). Subtract this result from the current polynomial, giving \(6x - 8\).

Step 5: Final Division and Remainder

Divide \(6x\) by \(x\) to get \(6\). Write \(6\) on top next to \(2x\). Multiply \(6\) by \(x - 2\), resulting in \(6x - 12\). Subtract this from \(6x - 8\) to find the remainder which is \(4\).

Step 6: Conclusion

The division is now complete. The result of the division is \(x^2 + 2x + 6\) with a remainder of \(4\).

Key Concepts

Remainder Theorem

The Remainder Theorem offers a way to find the remainder of a polynomial when divided by a linear factor. It states that if you divide a polynomial \( f(x) \) by \( x - c \), the remainder is \( f(c) \). This means you can find the remainder without performing the full division process.
For example, in our original exercise, the polynomial \( x^3 + 2x - 8 \) is divided by \( x - 2 \). According to the Remainder Theorem, we substitute \( x = 2 \) into the polynomial to find \( f(2) \).

Compute each term separately:

• First, \( 2^3 = 8 \).
• Then, \( 2 \times 2 = 4 \).
• Since there's no \( x^2 \) term, you might feel tempted to skip it, but remember the placeholder \( 0 \times 2^2 = 0 \) plays no role here.
• Finally, subtract 8 to reflect the constant term.

This arithmetic results in \( 8 + 0 + 4 - 8 = 4 \). Hence, the remainder is \( 4 \), which aligns perfectly with our division results.

Polynomial Division

Polynomial Division is similar to long division with numbers, but includes handling of variables and exponents. It involves dividing a polynomial by another, simpler polynomial, usually of lower degree. The goal is to express the dividend as the divisor times a quotient plus a remainder.

To begin, arrange both the dividend and divisor according to descending powers of \( x \). The division process starts by dividing the leading term of the dividend by the leading term of the divisor, determining the first term of the quotient.

• For example, in our problem: Divide \( x^3 \) by \( x \), resulting in \( x^2 \).
• Multiply \( x^2 \) by the divisor \( x - 2 \), resulting in \( x^3 - 2x^2 \).
• Subtract these from the original dividend, repeat with the new polynomial.

Continue this cycle until you've dealt with all terms of the dividend. The remainder is simply what's left after the last subtraction.
In our case, the remainder was \( 4 \). This method systematically breaks down complex polynomials into simpler components.

Algebraic Operations

Algebraic operations, like those used in polynomial division, rely on the basic principles of multiplying, dividing, adding, and subtracting both constants and variables. These operations allow the simplification and manipulation of expressions to solve equations. The key to solving polynomials is properly applying these techniques.

In dividing polynomials, careful attention must be paid to:

• Order of Operations: Ensure you follow the correct sequence, starting with dividing leading terms, to prevent mistakes.
• Sign Changes: When subtracting terms, change the signs accurately to avoid computational errors.
• Handling Zeros: Insert zero coefficients for missing terms to maintain alignment in subtraction steps. This ensures every power of \( x \) is represented, even if its coefficient is zero.

These algebraic operations provide the foundational skills needed to solve polynomial problems reliably. Understanding and executing these steps flawlessly can greatly decrease the intimidation factor associated with complex algebraic expressions.