Problem 44 - Precalculus : Building Concepts and Connections

Question

For what value(s) of \(k\) do you get a remainder of 15 when you divide \(k x^{3}+2 x^{2}-10 x+3\) by \(x+2 ?\)

Step-by-Step Solution

Answer

The value of \(k\) that gives a remainder of 15 when \(kx^3+2x^2-10x+3\) is divided by \(x+2\) is \(k=-11\).

Step 1: Identify the Polynomial and the binomial

In our case, the polynomial \(P(x)\) is \(kx^{3}+2x^{2}-10x+3\) and the binomial is \(x+2\) .

Step 2: Use the Remainder Theorem

We plug in \(a=-2\) in the polynomial which is \(P(-2)\) which should equal to 15 as per the Remainder Theorem.

Step 3: Solve Equation

Solve the resulting equation from step 2 i.e., \(k*(-2)^{3}+2*(-2)^{2}-10*(-2) +3 = 15\). If you simplify it, you get \(k=-11\).

Key Concepts

Polynomial Division

When faced with an algebraic equation in the form of a polynomial, one way to find solutions is through polynomial division. This method is useful for breaking down complex polynomials into simpler parts or finding factors.

Polynomial division is similar to long division which is used with numbers. It involves dividing a polynomial (called the dividend) by another polynomial (called the divisor) to find another polynomial (the quotient) and often a remainder. In the given exercise, the dividend is the polynomial \( kx^{3}+2x^{2}-10x+3 \) and the divisor is the binomial \( x+2 \).

When performing polynomial division manually, align the terms in descending order of their powers, then divide the first term of the dividend by the first term of the divisor. This quotient term is then multiplied by the divisor, and the result is subtracted from the dividend to create a new, simpler polynomial. Continue this process until all terms of the dividend have been considered. The result is the quotient polynomial, and sometimes a remainder, similar to numerical division.

Synthetic Division

Another effective tool for dividing polynomials, especially when the divisor is a binomial of the form \( x-a \) is synthetic division. It's a shortcut method that simplifies the traditional long division steps and can quickly find polynomial roots or evaluate polynomials.

In synthetic division, you use only the coefficients of the polynomial and perform operations on these numbers, which represents an abbreviated form of the long division. To use synthetic division:

• Write down the coefficients of the dividend.
• Place the zero of the binomial divider (\(\ -a\)) to the left.
• Bring down the leading coefficient to the bottom row.
• Multiply this coefficient by \(\ -a\) and place the result under the next coefficient.
• Continue the process by adding the numbers in columns and then multiplying by \(\ -a\), working your way through all coefficients.

The result is the coefficients of the quotient polynomial. If there is any remainder, it will be the last number in your bottom row.

In the given exercise, the Remainder Theorem is an alternate way to find the remainder without performing the entire division, making synthetic division unnecessary for this particular problem.

Algebraic Equations

Algebraic equations are mathematical statements that show the equality between two expressions, typically including one or more variables. They are fundamental in finding unknowns in various mathematical and real-world problems. The provided exercise showcases the use of algebraic equations in determining the value of the variable \(k\) given a specific condition related to polynomial division.

To solve such algebraic equations, you'll often rearrange the terms, combine like terms, and use properties of operations such as the distributive property. The goal is to isolate the variable to find its value. In the example, \(k\) is the variable in the polynomial whose value we need to determine to achieve a remainder of 15 when the polynomial is divided by \(x+2\). By applying these principles and the Remainder Theorem, we set \(P(-2) = 15\) and solve for \(k\), obtaining \(k = -11\).

Understanding algebraic equations is crucial in various fields, such as physics, engineering, economics, and beyond, because they provide a means to model and solve real-world problems systematically.