Problem 58 - Elementary Algebra

Question

For the following problems, perform the multiplications and divisions. $$ \frac{b^{2}-5 b+6}{b^{2}-b-2} \cdot \frac{b^{2}-2 b-3}{b^{2}-9 b+20} $$

Step-by-Step Solution

Answer

Question: Simplify the following expression: $\frac{b^2 - 5b + 6}{b^2 - b - 2} \cdot \frac{b^2 - 2b - 3}{b^2 - 9b + 20}$. Answer: $\frac{(b - 3)^2}{(b - 5)(b - 4)}$

Step 1: Factor numerators and denominators

First, let's factor each of the numerators and denominators: $$b^2 - 5b + 6 = (b - 2)(b - 3)$$ $$b^2 - b - 2 = (b - 2)(b + 1)$$ $$b^2 - 2b - 3 = (b - 3)(b + 1)$$ $$b^2 - 9b + 20 = (b - 5)(b - 4)$$ Now let's rewrite the expression with these factors: $$\frac{(b - 2)(b - 3)}{(b - 2)(b + 1)} \cdot \frac{(b - 3)(b + 1)}{(b - 5)(b - 4)}$$

Step 2: Cancel out common factors

Now we can cancel out any common factors between the numerators and denominators: $$\require{cancel}\frac{\cancel{(b - 2)}(b - 3)}{\cancel{(b - 2)}(b + 1)} \cdot \frac{(b - 3)\cancel{(b + 1)}}{(b - 5)(b - 4)\cancel{(b + 1)}}$$

Step 3: Multiply simplified expressions

Finally, we multiply the simplified expressions: $$\frac{(b - 3)(b - 3)}{(b - 5)(b - 4)} = \frac{(b - 3)^2}{(b - 5)(b - 4)}$$ The simplified expression is: $$\frac{(b - 3)^2}{(b - 5)(b - 4)}$$

Key Concepts

Polynomial Factorization

Factoring polynomials is like finding the puzzle pieces that fit together to form a bigger picture. It is an essential skill in algebra that allows us to break down complex expressions into simpler parts. Polynomials, such as quadratics, are common and you'll typically start by finding two numbers that both add to the middle term and multiply to the last term.

• For example, let's factor the quadratic polynomial \(b^2 - 5b + 6\). We look for two numbers that multiply to 6 (the constant term) and add up to -5 (the coefficient of the middle term \(b\)). These numbers are -2 and -3, leading us to the factors: \((b - 2)(b - 3)\).
• Applying the same technique to \(b^2 - b - 2\), you find the factors are \((b - 2)(b + 1)\).
• With \(b^2 - 2b - 3\), the factors turn out to be \((b - 3)(b + 1)\).
• And for \(b^2 - 9b + 20\), they are \((b - 5)(b - 4)\).

When expressions are factored, they become much easier to work with, especially when simplifying fractions or solving equations. It's a powerful tool in the algebra toolbox.

Simplifying Expressions

Simplifying algebraic expressions involves cutting through the complexity to make them easier to handle. Once you've factored complex terms, look for common factors between numerators and denominators. This makes simplification possible, as it did in our example.

• Let's start by writing the expression with our factors: \[ \frac{(b - 2)(b - 3)}{(b - 2)(b + 1)} \cdot \frac{(b - 3)(b + 1)}{(b - 5)(b - 4)} \]
• Notice that \((b - 2)\) appears in both the numerator and the denominator of the first fraction. Thus, it can be canceled, as dividing by the same term in both parts equals one.
• Similarly, \((b + 1)\) can also be canceled between the second fraction's numerator and denominator.

After canceling these common factors, you're left with a simplified expression:\[ \frac{(b - 3)(b - 3)}{(b - 5)(b - 4)} \].
This process reduces complexity and makes multiplication or further operations more straightforward.

Multiplication and Division of Fractions

When it comes to multiplying and dividing algebraic fractions, the process mirrors that of numerical fractions. The numerators are multiplied together, as are the denominators. Let's look into this process step-by-step.

• Considering our simplified expression: \[ \frac{(b - 3)}{1} \cdot \frac{(b - 3)}{(b - 5)(b - 4)} \]
• The trick here is straight multiplication: multiply the tops (numerators) and multiply the bottoms (denominators): \[ (b - 3)(b - 3)/(b - 5)(b - 4) \]
• Note that division of fractions ideally is handled by flipping the second fraction (reciprocal) and then multiplying, but in this case, we directly multiplied since both expressions were already set for multiplication.

The result gives us: \[ \frac{(b - 3)^2}{(b - 5)(b - 4)} \].
By approach, this mirrors the same operation you perform in arithmetic, showing algebra maintains a consistent structure, just with variables!