Problem 20 - College Algebra with Corequisite Support

Question

For the following exercises, find the product. \((3 d-5)(2 d+9)\)

Step-by-Step Solution

Answer

The product is \(6d^2 + 17d - 45\).

Step 1: Distribute First Term

Take the first term in the first binomial, which is \(3d\), and distribute it across each term in the second binomial, \((2d + 9)\). This gives us two terms: \((3d) \times (2d)\) and \((3d) \times 9\).

Step 2: Simplify First Distribution

Multiply the terms from Step 1. For \((3d) \times (2d)\), compute the product: \(3 \times 2 = 6\) and \(d \times d = d^2\). This results in \(6d^2\). For \((3d) \times 9\), compute the product: \(3 \times 9 = 27\) and keep the \(d\), resulting in \(27d\).

Step 3: Distribute Second Term

Now take the second term in the first binomial, which is \(-5\), and distribute it across each term in the second binomial, \((2d + 9)\). This gives us two terms: \((-5) \times (2d)\) and \((-5) \times 9\).

Step 4: Simplify Second Distribution

Multiply the terms from Step 3. For \((-5) \times (2d)\), compute the product: \(-5 \times 2 = -10\) and keep the \(d\), resulting in \(-10d\). For \((-5) \times 9\), compute the product: \(-5 \times 9 = -45\).

Step 5: Combine All Products

Combine all the terms obtained from the previous steps: \(6d^2, 27d, -10d,\) and \(-45\). Combine the like terms \(27d\) and \(-10d\), which results in \(17d\).

Step 6: Write Final Expression

The final product of \((3d-5)(2d+9)\) is obtained by combining all the terms into a single quadratic expression: \(6d^2 + 17d - 45\).

Key Concepts

Quadratic Expression

A quadratic expression is a polynomial of degree 2, typically written in the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Quadratic expressions are foundational in algebra and appear frequently in various mathematical contexts like geometry, physics, and economics. They can be used to represent parabolas, and depending on the signs and values of the coefficients, they can describe various behaviors such as opening upwards or downwards, and positioning on a graph.
The quadratic expression derived in our example, \(6d^2 + 17d - 45\), includes:

• A quadratic term: \(6d^2\), indicating the parabolic shape.
• A linear term: \(17d\), which affects the slope and orientation.
• A constant term: \(-45\), determining the y-intercept on a graph.

Understanding each component is crucial because they each play a role in solving quadratic equations through various methods such as factoring, completing the square, or using the quadratic formula.

Binomial Product

A binomial product involves two binomials, which are polynomials with two terms. In our example, these are \((3d - 5)\) and \((2d + 9)\). Finding the product of these binomials means multiplying them together to form a larger, often quadratic, expression.
This process is a fundamental aspect of algebra because:

• It reinforces understanding of polynomial operations.
• It lays the groundwork for more complex algebraic manipulations such as expanding polynomials or solving equations.

In practical situations, knowing how to find binomial products can be useful in areas such as physics, where equations often model real-world situations, or finance, where different scenarios need to be calculated.
Recognizing patterns in binomial multiplication, such as (a + b)(c + d), helps in mentally predicting the type of terms that should appear in the result, thanks to consistent algebraic structures.

Distributive Property

The distributive property is a key principle in algebra that dictates how multiplication over addition or subtraction works. Written as \(a(b + c) = ab + ac\), it helps expand expressions and simplify calculations.
The distributive property is essential in understanding how to multiply binomials effectively. In our cased exercise, this property is applied when we multiplied each term in \((3d - 5)\) by each term in \((2d + 9)\).
Here’s why the distributive property is critical:

• It allows the breaking down of complex multiplication into simpler tasks.
• It enables the expansion of expressions in steps, making it easier to identify errors and comprehend each part.
• It supports the foundational transition from arithmetic to algebraic manipulation.

This property not only simplifies computations but also aids in a deeper understanding of algebraic expressions and equations. Employing it correctly ensures that each term is correctly accounted for in polynomial operations.