Problem 45 - Introductory Algebra for College Students

Question

Divide as indicated. $$\left(y^{2}-16\right)+\frac{y^{2}+3 y-4}{y^{2}+4}$$

Step-by-Step Solution

Answer

The result of the given division and simplification is \(y^{2} + \frac{1}{4}y - 3\)

Step 1: Recognizing the Expressions

The given expression is \((y^{2}-16)+\frac{y^{2}+3 y-4}{y^{2}+4}\). Here, the first part is a simple quadratic \(y^{2}-16\). The second part is a division of two polynomials. Notably, the quadratic expression involved can be factored.

Step 2: Factoring Quadratic Terms

Start by factoring the quadratic terms into two binomial expressions. For the expression \(y^{2}-16\), apply the difference of squares rule, which states: if \(a^{2}-b^{2}\), it can be rewritten as \((a-b)(a+b)\). So, \(y^{2}-16\) can be factored as \((y-4)(y+4)\). The second term can be factored as \((y-1)(y+4)\), by searching for 2 numbers that multiply to -4 and add to 3.

Step 3: Substituting the Factored Terms

Substitute the factored quadratic expressions back into the original equation. This gives: \((y-4)(y+4)+\frac{(y-1)(y+4)}{y^{2}+4}\). Notice that \(y^{2}+4\) in the denominator can also be expressed as \(y^{2}-(-4)^{2}\), or \((y-2i)(y+2i)\), where \(i\) is the imaginary unit.

Step 4: Solving the Expressions

The expression can be further simplified by writing out the division as multiplication: \((y-4)(y+4)+(y-1)(y+4)\div(y^{2}+4)= (y-4)(y+4)+(y-1)(y+4) \cdot \frac{1}{(y+2i)(y-2i)}\)

Step 5: Simplifying the Expression

The final simplification involves expressing the complex conjugate in the denominator of the latter term and simplifying it. Doing so will lead to the final expression: After multiplying and simplifying: \((y-4)(y+4)+(y-1)(y+4) \cdot \frac{1}{4} = y^{2} + \frac{1}{4}y - 3\)

Key Concepts

Factoring Quadratics

Factoring quadratics is an essential algebraic technique that involves rewriting a quadratic expression as the product of two binomials. This technique is especially useful in simplifying expressions and solving quadratic equations. To factor a quadratic, typically given in the form \( ax^2 + bx + c \), one searches for two numbers that multiply to \( ac \) (the product of the coefficient of \( x^2 \) and the constant term) and add up to \( b \) (the coefficient of \( x \)).

Let's consider the quadratic term from our exercise \( y^2 + 3y - 4 \). We need to find two numbers that multiply to \( -4 \) and add to \( 3 \). These numbers are \( 4 \) and \( -1 \), which then gives us the factored form as \( (y + 4)(y - 1) \). This approach simplifies complex problems, making them easier to solve or further manipulate.

Difference of Squares

The difference of squares is a special product pattern in algebra, and recognizing it can greatly aid in factoring quadratic expressions. The rule states that if you have an expression in the form of \( a^2 - b^2 \), it can be broken down into the product of its conjugate pairs \( (a - b)(a + b) \). In our exercise, the term \( y^2 - 16 \) is a difference of squares since \( 16 \) can be written as \( 4^2 \).

Hence, this term factors neatly into \( (y - 4)(y + 4) \). Applying the difference of squares shortcut makes factoring instantaneous since it bypasses the need to guess-and-check for the two numbers in standard factoring methodology. To improve understanding, students should practice identifying such patterns as they significantly simplify polynomial division.

Complex Conjugate

A complex conjugate refers to a pair of complex numbers that have the same real part but opposite imaginary parts. In the context of polynomial division, you will occasionally encounter quadratics that can't be factored over the real numbers. The quadratic \( y^2 + 4 \) from our exercise is one such example, as it cannot be factored into real-numbered binomials. However, by embracing complex numbers, we rewrite \( y^2 + 4 \) as \( y^2 - (-2i)^2 \) and factor it as \( y + 2i \) and \( y - 2i \) — a pair of complex conjugates.

When dealing with polynomial division involving complex numbers, the conjugate allows us to multiply the numerator and denominator by the conjugate of the denominator to remove the complex number from the denominator. This process, known as 'rationalization', is crucial when we want our final expression to have real numbers only.