Problem 57 - Algebra A Combined Function

Question

Solve. The area of the parallelogram shown is \(\left(10 x^{2}+31 x+15\right)\) square meters. If its base is \((5 x+3)\) meters, find its height.

Step-by-Step Solution

Answer

The height is \(2x + 5\) meters.

Step 1 - Recall the Area Formula for a Parallelogram

The area of a parallelogram can be calculated using the formula: \( \text{Area} = \text{base} \times \text{height} \). In this problem, we know the area \((10x^2 + 31x + 15)\) and the base \((5x + 3)\). We are asked to find the height.

Step 2 - Set Up the Equation

Using the area formula \( \text{Area} = \text{base} \times \text{height} \), we set up the equation: \( 10x^2 + 31x + 15 = (5x + 3) \times \text{height} \).

Step 3 - Solve for the Height

To find the height, solve for height by isolating it on one side of the equation. Divide both sides of the equation by the base: \( \text{height} = \frac{10x^2 + 31x + 15}{5x + 3} \).

Step 4 - Perform the Division

Perform polynomial division on \( \frac{10x^2 + 31x + 15}{5x + 3} \):- The first term is \( \frac{10x^2}{5x} = 2x \).- Multiply \( 2x \times (5x + 3) = 10x^2 + 6x \).- Subtract from the original polynomial: \((10x^2 + 31x + 15) - (10x^2 + 6x) = 25x + 15 \).- The next term is \( \frac{25x}{5x} = 5 \).- Multiply \( 5 \times (5x + 3) = 25x + 15 \).- Subtract: \((25x + 15) - (25x + 15) = 0 \).The quotient is \( 2x + 5 \), which is the height.

Key Concepts

Parallelogram Area Calculation

When dealing with the area of a parallelogram, it's usually about finding how much space is inside the shape. A key formula here is: \( \text{Area} = \text{base} \times \text{height} \). For any parallelogram, if you know two of these elements, you can find the third. In our example, we know:

• Area: \(10x^2 + 31x + 15\) square meters.
• Base: \(5x + 3\) meters.

The main goal is to solve for the height by rearranging the formula. Make sure you always plug in what you know, then solve for the unknown.This step often requires knowledge of polynomial manipulation, especially when the area and base themselves are algebraic expressions.It’s important here to be comfortable working with variables and expressions, which we will discuss further.

Polynomial Division

Polynomial division is essential in solving equations with polynomials, especially when dealing with expressions like fractions involving polynomials. It’s similar to long division with numbers.Here's a quick breakdown of how polynomial division works using the problem's expression:

• Divide the first term of the numerator \(10x^2\) by the first term of the divisor \(5x\), which gives \(2x\).
• Multiply the entire divisor \((5x + 3)\) by this quotient \(2x\) to get \(10x^2 + 6x\).
• Subtract this result from the original polynomial to find the new dividend: \((10x^2 + 31x + 15) - (10x^2 + 6x) = 25x + 15\).
• Repeat the process with the new dividend. Divide \(25x\) by \(5x\) to get \(5\). Then multiply and subtract again to eliminate the remaining terms.

This process will give you the quotient, which is \(2x + 5\), hence solving for height in our problem. Understanding this division process is key because it helps simplify complex algebraic expressions into something more manageable.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations. In this exercise, we have expressions like \(10x^2 + 31x + 15\) and \(5x + 3\).A good grasp of how to manipulate and simplify these expressions is crucial. Each part (terms) of the expressions has a role:

• The **constants** (like 15 and 3) are terms known without variables.
• The **coefficients** are numbers multiplying the variables (like 10 and 5 in front of \(x^2\) and \(x\)).
• The **degree** of the polynomial is determined by the highest exponent (\(x^2\) in this case).

Manipulating these requires understanding operations such as addition, subtraction, and division of like terms. In solving the exercise, the aim was to find the height, which involved dealing with these expressions systematically.Mastering the manipulation of algebraic expressions is a foundational skill in solving equations across many math problems.