Problem 27 - Algebra for College Students

Question

Find each indicated product. Remember the shortcut for multiplying binomials and the other special patterns we discussed in this section. $$(x-3)(x+3)(x-1)$$

Step-by-Step Solution

Answer

The expanded form is \(x^3 - x^2 - 9x + 9\).

Step 1: Recognize Special Patterns

Notice that \((x-3)(x+3)\) represents a difference of squares pattern. The difference of squares formula is: \[(a-b)(a+b) = a^2 - b^2\].In this case, \(a = x\) and \(b = 3\).

Step 2: Apply the Difference of Squares Formula

Using the formula for the difference of squares, expand \((x-3)(x+3)\):\[(x-3)(x+3) = x^2 - 3^2 = x^2 - 9\].

Step 3: Multiply the Result with the Remaining Binomial

Next, multiply the result \((x^2 - 9)\) by the remaining binomial \((x-1)\):\[(x^2 - 9)(x-1) = x^2(x-1) - 9(x-1)\].

Step 4: Distribute Each Term

Distribute \(x^2\) and \(-9\) into \((x-1)\):\[x^2(x-1) = x^3 - x^2\]\[-9(x-1) = -9x + 9\].

Step 5: Combine Like Terms

Combine all the terms from the previous step:\[x^3 - x^2 - 9x + 9\].This is the expanded form of the original expression.

Key Concepts

Difference of Squares

The concept of the difference of squares is incredibly useful when multiplying binomials. This occurs when you have two terms in the form

• \((a-b)(a+b)\), where \(a\) and \(b\) each occur twice but with opposite signs relative to each other.

This pattern helps because it simplifies the multiplication process to a single step. The formula for simplifying the difference of squares is: \[ (a-b)(a+b) = a^2 - b^2 \] In our original exercise,

• both binomials \((x-3)\) and \((x+3)\) fit this pattern.
• Set \(a = x\) and \(b = 3\).
• So, using the formula, \((x-3)(x+3)\) becomes \(x^2 - 9\) because \(9\) is \(3^2\).

This simplification makes future steps more manageable by directly reducing a component of the multiplication process.

Polynomial Multiplication

When multiplying polynomials, approach each term one at a time. Initial organization is key. Our task is to multiply \((x^2 - 9)\) by \((x-1)\). Start by distributing each element of the first polynomial across each element of the second one.

• You'll distribute \(x^2\) to both \(x\) and \(-1\) to get \(x^3\) and \(-x^2\).
• Then, distribute \(-9\) to \(x\) and \(-1\) to get \(-9x\) and \(+9\).

The result is an expansion comprising multiple terms: \(x^3 - x^2 - 9x + 9\). Be careful to maintain proper sign alignment, as each term's sign is crucial when you add or combine like terms in the next step.

Binomial Distribution

In polynomial multiplication, a core principle includes the binomial distribution, often referred to in practices like the Distributive Property.

• Here, you need to ensure every term in one binomial \((x^2 - 9)\) is distributed to each term in the other factor \((x-1)\).

This creates two rounds of multiplication for each individual term:

• First, multiple \(x^2\) involves: \(x^2\times x = x^3\) and \(x^2\times -1 = -x^2\).
• Subsequently, handle constant \(-9\), multiplying denoting separate interactions:\(-9\times x = -9x\) and \(-9\times -1 = +9\).

In polynomial multiplication, take note of each component's interaction; properly managing these aspects ensures the combination of like terms in the final step results in a clean, simplified polynomial form.
The binomial distribution process is integral to ensuring that no term is missed when expanding, which could affect the overall result.