Problem 41 - College Algebra

Question

For the following exercises, multiply the polynomials. $$ (y-2)\left(y^{2}-4 y-9\right) $$

Step-by-Step Solution

Answer

The product of the polynomials is \(y^3 - 6y^2 - y + 18\).

Step 1: Distribute the First Term

The given expression is \((y-2)(y^2 - 4y - 9)\). Start by distributing the first term \(y\) across the second polynomial. This involves multiplying \(y\) by each term in \(y^2 - 4y - 9\):\(y \cdot y^2 = y^3\), \(y \cdot (-4y) = -4y^2\), and \(y \cdot (-9) = -9y\). The resulting expression is \(y^3 - 4y^2 - 9y\).

Step 2: Distribute the Second Term

Now distribute the second term \(-2\) across the second polynomial \(y^2 - 4y - 9\): \(-2 \cdot y^2 = -2y^2\), \(-2 \cdot (-4y) = 8y\), and \(-2 \cdot (-9) = 18\). This gives the expression \(-2y^2 + 8y + 18\).

Step 3: Combine Like Terms

Combine the expressions from Steps 1 and 2: \(y^3 - 4y^2 - 9y\) and \(-2y^2 + 8y + 18\). Group the same kinds of terms together: \(y^3\), \(-4y^2 - 2y^2 = -6y^2\), \(-9y + 8y = -y\), and the constant \(+ 18\). The combined expression is \(y^3 - 6y^2 - y + 18\).

Key Concepts

Distributive property

When multiplying polynomials, the distributive property plays a crucial role. The distributive property states that for any numbers or algebraic expressions, the equation \(a(b + c) = ab + ac\) holds true. In simpler terms, this means that you distribute, or apply, the term outside the parentheses to each term inside the parentheses.
The original problem, \((y-2)(y^2 - 4y - 9)\), uses the distributive property to multiply two polynomials. First, you distribute the \(y\) across the binomial \((y^2 - 4y - 9)\), multiplying \(y\) by each term inside:

• \(y \cdot y^2 = y^3\)
• \(y \cdot (-4y) = -4y^2\)
• \(y \cdot (-9) = -9y\)

This results in the expression \(y^3 - 4y^2 - 9y\).
Next, you distribute the \(-2\) and repeat the process:

• \(-2 \cdot y^2 = -2y^2\)
• \(-2 \cdot (-4y) = 8y\)
• \(-2 \cdot (-9) = 18\)

The result is \(-2y^2 + 8y + 18\). Using this property effectively allows you to simplify the multiplication of polynomials.

Combining like terms

After distributing and obtaining separate expressions, combining like terms is the next step for simplifying polynomial expressions. Like terms are those that have the exact same variable parts raised to the same power. The coefficients of these terms can then be added or subtracted.
In the problem solution provided, once you have \(y^3 - 4y^2 - 9y\) and \(-2y^2 + 8y + 18\) from using the distributive property, it's time to combine these like terms:

• The \(y^3\) term appears only once, so you leave it as \(y^3\).
• For the \(y^2\) terms, combine \(-4y^2\) and \(-2y^2\) to get \(-6y^2\).
• For the \(y\) terms, combine \(-9y\) and \(8y\) to end up with \(-y\).
• The constant terms, \(+ 18\), remain unchanged.

So, the final expression is \(y^3 - 6y^2 - y + 18\). By grouping and simplifying these terms, you've effectively reduced the polynomial to its simplest form.

Algebraic expressions

Algebraic expressions are mathematical phrases involving numbers, variables, and operational symbols. They do not have an equals sign, which differentiates them from algebraic equations. When working with polynomial multiplication, like in the given exercise, you're dealing primarily with expressions.
The expression \((y-2)(y^2 - 4y - 9)\) is a product of two polynomials. A polynomial is an algebraic expression with more than one term, and it's characterized by variables raised to whole number exponents and their respective coefficients.
Handling algebraic expressions involves understanding:

• **Terms:** The individual parts of the expression separated by plus or minus signs.
• **Coefficients:** The numerical factor in front of a term's variable.
• **Exponents:** The power to which a variable is raised.
• **Operations:** How you simplify, factor, and manipulate these expressions.

These tools are essential for multiplying polynomials, simplifying expressions, and solving algebraic equations. Working step-by-step through distributing and combining like terms helps to evaluate these expressions thoroughly. By practicing this method, students can enhance their skills in dealing with various algebraic operations.