Problem 71 - Elementary Algebra

Question

For the following problems, perform the indicated operations. $$ \frac{y^{2}+4 y+4}{y^{2}+10 y+21} \div(y+2) $$

Step-by-Step Solution

Answer

Answer: The simplified form of the expression is $\frac{y+2}{(y+3)(y+7)}$.

Step 1: Factor the quadratic expressions

We will factor both the numerator and denominator where possible: $$ \frac{y^{2}+4 y+4}{y^{2}+10 y+21} = \frac{(y+2)(y+2)}{(y+3)(y+7)} $$

Step 2: Rewrite the division as multiplication

We will rewrite the division with the divisor (y+2) as multiplication by its reciprocal: $$ \frac{(y+2)(y+2)}{(y+3)(y+7)} \div (y+2) = \frac{(y+2)(y+2)}{(y+3)(y+7)} \times \frac{1}{y+2} $$

Step 3: Simplify the expression

Now, we will cancel out the common factors (y+2) from the numerator and denominator: $$ \frac{(y+2)(y+2)}{(y+3)(y+7)} \times \frac{1}{y+2} = \frac{y+2}{(y+3)(y+7)} $$ The simplified expression is: $$ \frac{y+2}{(y+3)(y+7)} $$

Key Concepts

Factoring Quadratics

To simplify rational expressions, understanding how to factor quadratics is essential. Factoring a quadratic means expressing it as the product of two linear factors. The standard form of a quadratic equation is \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. Factoring involves finding two numbers that add up to \(b\) and multiply to \(ac\).

In our exercise, we factored \(y^2+4y+4\) into \(y+2)(y+2)\) and \(y^2+10y+21\) into \(y+3)(y+7)\). Notice how the factors correspond to numbers that multiply to the constant term and add to the coefficient of \(y\). Factoring makes it possible to simplify rational expressions by dividing out common factors, which is precisely why it's so important in algebra.

Reciprocal of a Number

The reciprocal of a number is simply one divided by that number. In other words, if you have a number \(n\), its reciprocal is \(\frac{1}{n}\). When dealing with algebraic expressions, the concept remains the same. For any non-zero polynomial \(P(x)\), the reciprocal is \(\frac{1}{P(x)}\).

In our initial problem, to manage the division by \(y+2\), we multiplied by the reciprocal, which is \(\frac{1}{y+2}\). Multiplying by the reciprocal of a number (or a polynomial) is a fundamental operation in algebra because it allows you to turn division into multiplication. This simplifies the process of manipulating expressions, especially when dividing polynomials.

Division of Polynomials

Division of polynomials can seem daunting, but by using the reciprocal, we can greatly simplify the process. When one polynomial is divided by another, instead of performing long division, we can multiply by the reciprocal of the divisor. This converts a complex fraction into a simpler one or even into a product of factors.

In the solution to our exercise, the operation \(\frac{y^2+4y+4}{y^2+10y+21} \div (y+2)\) was transformed into a multiplication problem \(\frac{(y+2)(y+2)}{(y+3)(y+7)} \times \frac{1}{y+2}\). Approaching division this way makes it easier to identify and cancel common factors, which is the next step toward simplification.

Canceling Common Factors

Canceling common factors is a crucial skill in algebra because it allows us to simplify complex rational expressions. When a factor appears both in the numerator and denominator of a fraction, we can cancel it out. This is based on the fundamental property that \(\frac{a}{a} = 1\), whenever \(a \eq 0\).

Using this method, we canceled the common factor \(y+2\) in our exercise, which appeared in the transformed multiplication problem \(\frac{(y+2)(y+2)}{(y+3)(y+7)} \times \frac{1}{y+2}\) to obtain \(\frac{y+2}{(y+3)(y+7)}\). It is essential to remember only to cancel out factors that are exactly the same, and only if they are not zero, because division by zero is undefined.