Problem 24 - Algebra for College Students

Question

Perform the indicated divisions. $$ \left(6 x^{3}-2 x^{2}+4 x-3\right) \div(x+1) $$

Step-by-Step Solution

Answer

Quotient is \(6x^2 - 8x + 12\) with remainder \(-15\).

Step 1: Set Up the Synthetic Division

To divide the polynomial \( 6x^3 - 2x^2 + 4x - 3 \) by \( x+1 \), we first set up synthetic division. Note that \( x+1 \) can be rewritten as \( x - (-1) \), so we use -1 as the divisor. Write the coefficients of the dividend polynomial, which are 6, -2, 4, and -3.

Step 2: Start Synthetic Division

Write the -1 outside the synthetic division box and bring down the first coefficient, 6, as it is. This is the beginning of a new row.

Step 3: Multiply and Add

Multiply -1 (the divisor) by the number just brought down (6) to get -6. Write this result under the next coefficient (-2). Add -2 and -6 to get -8.

Step 4: Repeat Multiply and Add

Multiply -1 by -8 (the result from the last addition) to get 8. Write this under the next coefficient (4). Add 4 and 8 to get 12.

Step 5: Final Multiply and Add

Multiply -1 by 12 (the result from the last addition) to get -12. Write this result under the last coefficient (-3). Add -3 and -12 to get -15, which is the remainder.

Step 6: Write the Quotient and Remainder

The resulting polynomial from synthetic division is written using the coefficients 6, -8, and 12. Therefore, the quotient is \(6x^2 - 8x + 12\) and the remainder is \(-15\).

Key Concepts

Polynomial Division

Polynomial division is a method used to divide one polynomial by another. The process is similar to long division but involves algebraic expressions instead of numbers. In order to perform polynomial division, especially in cases like dividing \(6x^3 - 2x^2 + 4x - 3\) by \(x+1\), synthetic division is often used because it simplifies the process significantly.Here are some key points about polynomial division:

• The dividend is the polynomial which you are dividing, for example, \(6x^3 - 2x^2 + 4x - 3\).
• The divisor is the polynomial you are dividing by, such as \(x+1\).
• If the divisor is in the form of \(x \pm a\), synthetic division becomes a viable option.

Synthetic division is particularly efficient because it reduces the complexity of polynomial division by transforming it into a series of simple arithmetic operations.

Remainder Theorem

The Remainder Theorem is a principle in algebra that provides a quick way to find the remainder of a polynomial division. Specifically, when a polynomial \(f(x)\) is divided by \(x-a\), the remainder of this division will be \(f(a)\). This theorem is extremely useful as it allows us to evaluate the polynomial without actually performing the entire division.In the example division problem given, we are dividing by \(x + 1\). Therefore, you would substitute \(-1\) into the polynomial \(f(x) = 6x^3 - 2x^2 + 4x - 3\) to find the remainder. The Remainder Theorem ensures that the remainder calculated through synthetic division, which is \(-15\), matches \(f(-1)\).Knowing the remainder helps verify your work or determine a polynomial's specific characteristics when the divisor doesn't perfectly divide into the dividend.

Algebraic Expressions

Algebraic expressions consist of variables, coefficients, and constants. They form the basis of polynomial equations and are used extensively in algebra for solving problems, modeling relationships, and describing various mathematical phenomena.Algebraic expressions may resemble \(6x^3 - 2x^2 + 4x - 3\), comprising different terms with coefficients and exponents. Understanding these expressions is essential:

• **Variables**: Symbols that represent unknown values (e.g., \(x\)).
• **Coefficients**: Numbers multiplying the variables (e.g., 6 in \(6x^3\)).
• **Constants**: Numbers without variables (e.g., -3 in the given polynomial).

When working with polynomial division through synthetic division, grasping algebraic expressions helps you see how coefficients are manipulated to arrive at the quotient and remainder. By fully understanding these basic components, tasks like dividing complex polynomials become more attainable, giving you confidence in tackling further algebraic challenges.