Problem 29 - Algebra 2

Question

Multiply or divide. State any restrictions on the variable. $$ \frac{6 x^{3}-6 x^{2}}{x^{4}+5 x^{3}} \div \frac{3 x^{2}-15 x+12}{2 x^{2}+2 x-40} $$

Step-by-Step Solution

Answer

The simplified form of the given expression is \( \frac{4}{3}\). The restrictions are x ≠ -5, 0, 4.

Step 1: Factorize the Numerator and Denominator of Each Fraction

Rewrite every term as a product of its factors. \[\frac{6 x^{3}-6 x^{2}}{x^{4}+5 x^{3}} \div \frac{3 x^{2}-15 x+12}{2 x^{2}+2 x-40} \]becomes\[\frac{6x^{2}(x-1)}{x^{3}(x+5)} \div \frac{3(x-4)(x-1)}{2(x-4)(x+5)} \]

Step 2: Dividing Fractions

You will need to remember that dividing fractions is equivalent to multiplying by the reciprocals. Swap the terms in the second fraction and change the division operation to a multiplication. \[\frac{6x^{2}(x-1)}{x^{3}(x+5)} \times \frac{2(x-4)(x+5)}{3(x-4)(x-1)} \]

Step 3: Cancel out Similar Terms

You will notice some terms are present on both the numerator and denominator of the expression. Cancel out these identical terms.\[\frac{4x^{2}}{3x^{2}}\]

Step 4: Simplify the Result

Simplify the resulting polynomial.\[\frac{4}{3}\]

Step 5: Identify the Restrictions on the Variable

The original problem included terms that were in the denominator of a fraction, since division by zero is undefined, we need to find values of x that make the denominator zero and exclude them. The second fraction had an \(x-4\) term in its denominator, therefore x cannot be 4. Similarly, \(x+5\) in the denominator of both fractions means x cannot be -5. Also, the first fraction's denominator included an \(x^{3}\) term and therefore x cannot be 0.

Key Concepts

Polynomial Factorization

Polynomial factorization is breaking down a polynomial into simpler "factor" polynomials, similar to how numbers are factorized into prime numbers.
These factors can then be easily manipulated, compared, or simplified in subsequent calculations. In the given exercise, the expression \(6x^3 - 6x^2\) is factorized as \(6x^2(x-1)\). Similarly, the expression \(3x^2 - 15x + 12\) is factorized as \(3(x-4)(x-1)\).
This factorization simplifies the process of multiplication and division of rational expressions. You can perform factorization by checking for common factors or using techniques like the quadratic formula or synthetic division, if applicable. It is a fundamental skill in algebra that makes complex problems more manageable.

Rational Expressions

Rational expressions are fractions involving polynomials in the numerator and the denominator. In mathematical terms, they are expressions of the form \(\frac{P(x)}{Q(x)}\) where both \(P(x)\) and \(Q(x)\) are polynomials.To correctly work with rational expressions, understanding how to simplify, multiply, divide, add, or subtract them is crucial.
In the exercise, after the factorization, each fraction is an example of a rational expression:

• \(\frac{6x^2(x-1)}{x^3(x+5)}\)
• \(\frac{3(x-4)(x-1)}{2(x-4)(x+5)}\)

These kinds of expressions are common in algebra and it is important to manipulate them correctly for simplification or solving equations.

Division of Fractions

Dividing fractions in algebra can initially seem challenging, but it simplifies greatly once you realize that it is the same as multiplying by the reciprocal.
This means to divide by a fraction, simply flip the numerator and denominator of the second fraction and change the division to multiplication.For example, in the given problem, the division \(\frac{6x^2(x-1)}{x^3(x+5)} \div \frac{3(x-4)(x-1)}{2(x-4)(x+5)}\) is converted to multiplication by rearranging:\[\frac{6x^2(x-1)}{x^3(x+5)} \times \frac{2(x-4)(x+5)}{3(x-4)(x-1)}\]By understanding this method, you can effectively handle more complicated expressions and avoid common pitfalls.

Variable Restrictions

Variable restrictions refer to the values that a variable cannot take in an algebraic expression, particularly when they can cause the denominator to be zero.
In mathematics, division by zero is undefined, hence identifying these restrictions is crucial.In the exercise, variable restrictions are deduced from the denominators:\(x^3(x+5)\), \(2(x-4)(x+5)\), and \(3(x-4)(x-1)\).
The values that make these expressions zero need to be excluded:

• \(x^3\) suggests \(x eq 0\)
• \(x+5\) indicates \(x eq -5\)
• \(x-4\) denotes \(x eq 4\)

Understanding and applying these restrictions ensure the expression remains valid and calculable.