Problem 108 - Algebra and Trigonometry

Question

State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) $$ \frac{1+x+x^{2}}{x}=\frac{1}{x}+1+x $$

Step-by-Step Solution

Answer

The equation is true for all values of \( x \) except \( x = 0 \).

Step 1: Analyze the Given Equation

The original equation is given as \( \frac{1+x+x^2}{x} = \frac{1}{x} + 1 + x \). We need to determine if this equation holds true for all values of \( x \), except when the denominator is zero, which would be when \( x = 0 \).

Step 2: Simplify the Left Side

Start simplifying the left side of the equation, \( \frac{1+x+x^2}{x} \). Split the fraction:\[ \frac{1+x+x^2}{x} = \frac{1}{x} + \frac{x}{x} + \frac{x^2}{x} = \frac{1}{x} + 1 + x \].

Step 3: Verify Simplification

Compare the simplified form of the left side \( \frac{1}{x} + 1 + x \) to the right side of the equation \( \frac{1}{x} + 1 + x \). Since both sides are equal, the equation is verified.

Step 4: Conclusion

Since we have shown that the left side simplifies exactly to the right side, the given equation is true for all values of \( x \), except when \( x = 0 \), as we must not divide by zero.

Key Concepts

Simplifying Expressions

Simplifying expressions in algebra is like tidying up your room; it makes everything easier to understand and work with. When simplifying an expression, we break down complex fractions or terms into simpler forms that are equivalent to the original expression.
To simplify the expression \( \frac{1+x+x^2}{x} \), notice how we can separate this single fraction into three distinct parts:

• \( \frac{1}{x} \)
• \( \frac{x}{x} \), which simplifies to 1
• \( \frac{x^2}{x} \), which simplifies to \( x \)

This separation allows us to rewrite the expression as \( \frac{1}{x} + 1 + x \), matching the simpler form on the right side of the equation.
By doing this, we confirm that both sides of the equation remain equal, aiding in the verification of the equation's truth for all allowed values of \( x \).

Dividing by Zero

Dividing by zero is one of the big no-no's in mathematics. It's like trying to share zero candies with your friends; it simply doesn't work!
Any fraction that has zero in the denominator is undefined, meaning we cannot calculate or resolve this expression. That's why, in this problem, we state that the equation is true for all values of \( x \) except \( x = 0 \).
Zero in the denominator would make the fraction \( \frac{1}{x} \) undefined, as there is no number you can multiply by zero to give you something other than zero. Always remember to avoid dividing by zero to keep your algebraic concepts sound! That's why we exclude \( x = 0 \) from possible values in this exercise.

Polynomial Division

Polynomial division is a key skill in algebra that allows us to divide polynomials, which are expressions featuring variables and coefficients, like \( x^2 + x + 1 \).
When dividing a polynomial by a monomial, which is a single term, you handle each term of the polynomial individually.
In our case, we take the polynomial \( 1 + x + x^2 \) and divide each term by \( x \). Here's how:

• \( \frac{1}{x} \): remains as is
• \( \frac{x}{x} \): simplifies to 1
• \( \frac{x^2}{x} \): simplifies to \( x \)

This results in the expression \( \frac{1}{x} + 1 + x \).
The polynomial division helps determine equivalency on both sides of the equation, proving the equation's truthfulness for allowed values of \( x \). Polynomial division is a valuable tool, often used in simplifying complex algebraic expressions like the one here.