Problem 52 - A Graphical Approach to College Algebra

Question

Perform the indicated operations. $$(2 p-1)\left(3 p^{2}-4 p+5\right)$$

Step-by-Step Solution

Answer

The result of the operation is \(6p^3 - 11p^2 + 14p - 5\).

Step 1: Distribute Terms

To simplify the expression \((2p-1)(3p^2-4p+5)\), distribute each term in \(2p-1\) over each term in \(3p^2-4p+5\). This means we'll multiply \(2p\) with each term, and \(-1\) with each term.

Step 2: Multiply \(2p\) with Each Term

Multiply \(2p\) by each term in \(3p^2 - 4p + 5\):- \(2p \times 3p^2 = 6p^3\)- \(2p \times (-4p) = -8p^2\)- \(2p \times 5 = 10p\)The result is \(6p^3 - 8p^2 + 10p\).

Step 3: Multiply \(-1\) with Each Term

Multiply \(-1\) by each term in \(3p^2 - 4p + 5\):- \(-1 \times 3p^2 = -3p^2\)- \(-1 \times (-4p) = 4p\)- \(-1 \times 5 = -5\)The result is \(-3p^2 + 4p - 5\).

Step 4: Combine Like Terms

Combine the like terms from the distributed results\[6p^3 - 8p^2 + 10p - 3p^2 + 4p - 5\]:- Combine the \(p^2\) terms: \(-8p^2 - 3p^2 = -11p^2\)- Combine the \(p\) terms: \(10p + 4p = 14p\)The resulting expression is \[6p^3 - 11p^2 + 14p - 5\].

Key Concepts

Distributive Property

The Distributive Property is a fundamental tool in algebra that allows us to multiply a single term by each term in a polynomial. This property is essential for simplifying expressions, especially those involving polynomials.

• It states that for any three numbers or algebraic terms, say, a, b, and c, the expression \(a(b + c)\) is equivalent to \(ab + ac\).
• This property allows us to "distribute" the multiplication over addition or subtraction inside the parentheses, as seen in the expression \((2p-1)(3p^2-4p+5)\).

Let's see it in action:

• First, distribute each term of the binomial \((2p - 1)\) across each term of the trinomial \((3p^2 - 4p + 5)\).
• This involves multiplying \(2p\) with every term in \(3p^2 - 4p + 5\) and separately \(-1\) with every term in that polynomial.
• Through distributive property, both parts are expanded and simplified step by step.

Using this property makes complex multiplication straightforward and manageable by breaking it down into simpler, smaller parts.

Combining Like Terms

Combining like terms is a crucial step in simplifying algebraic expressions. This process involves merging terms that have the same variables raised to the same power. Let’s look at how we can perform this:

• Once you've distributed and expanded the terms using the distributive property, you might have several like terms that need combining.
• In our example, after distributing \((2p-1)(3p^2-4p+5)\), we have the expression \(6p^3 - 8p^2 + 10p - 3p^2 + 4p - 5\).
• To combine like terms: match the variables with similar powers. In our case:
• The \(p^2\) terms are \(-8p^2\) and \(-3p^2\). Adding them gives \(-11p^2\).
• The \(p\) terms are \(10p\) and \(4p\). Combined, they become \(14p\).
• The term \(6p^3\) and \(-5\) do not have like terms, so they remain as they are.

Through combining like terms, we compact our expression to its simplest form, making it easier to understand or further use in equations or inequalities.

Algebraic Expressions

Algebraic expressions consist of terms made up of constants, variables, and exponents. These components are combined through operations like addition, subtraction, multiplication, and division.

• An expression such as \((2p-1)(3p^2-4p+5)\) comprises multiple polynomials and requires an understanding of algebraic structures.
• Each part—\(2p-1\) and \(3p^2-4p+5\)—is a polynomial in itself. We identify these terms and their coefficients to use algebraic rules effectively.
• Polynomials are algebraic expressions that include variables raised to whole number powers and have more than one term. In this example, our polynomials are multiplied together to produce a new polynomial.
• These expressions need simplifying using distributive property and combining like terms, as previously described.
• Understanding how to manipulate algebraic expressions is crucial for more advanced mathematical operations, such as solving equations or working with functions.

Having a firm grasp of algebraic expressions and their elements equips you to tackle a wide range of problems, not just in algebra, but across all areas of mathematics.