Question

Perform the indicated operations and simplify the result. Leave your answer in factored form. $$ \frac{1}{h}\left[\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}\right] $$

Short answer

- \frac{(2x + h)}{(x+h)^2 x^2}

Step-by-step solution

Combine the fractions inside the brackets

Find a common denominator for the fractions inside the brackets: \[ \frac{1}{(x+h)^2} - \frac{1}{x^2} \] The common denominator will be \[(x+h)^2 x^2\]

Rewrite each fraction with the common denominator

Rewrite the fractions over the common denominator: \[ \frac{x^2 - (x+h)^2}{(x+h)^2 x^2} \]

Expand and simplify the numerator

Expand \((x+h)^2\) and simplify the numerator: \[ (x+h)^2 = x^2 + 2xh + h^2 \] \[ x^2 - (x^2 + 2xh + h^2) = x^2 - x^2 - 2xh - h^2 = -2xh - h^2 \] Therefore: \[ \frac{-2xh - h^2}{(x+h)^2 x^2} \]

Factor out h from the numerator

Factor out \h\ from the numerator: \[ \frac{-h(2x + h)}{(x+h)^2 x^2} \]

Divide by h

Now divide the entire expression by \h\: \[ \frac{1}{h} \times \frac{-h(2x + h)}{(x+h)^2 x^2} = \frac{- (2x + h)}{(x+h)^2 x^2} \]

Key concepts

common denominator

To effectively subtract two fractions, they must have the same denominator. This part of algebra is known as finding the 'common denominator'. When given fractions \(\frac{1}{(x+h)^2}\) and \(\frac{1}{x^2}\), the best common denominator includes factors from both denominators. In this case, \((x+h)^2 x^2\) works perfectly because it contains both \(x^2\) and \((x+h)^2\). This step is crucial as it ensures both fractions are expressed in terms of the same base, allowing subtraction to occur seamlessly.

fraction subtraction

Once both fractions share a common denominator, the next step is straightforward: subtracting the numerators while keeping the common denominator the same. For our example: We thus rewrite the fractions: \(\frac{x^2 - (x+h)^2}{(x+h)^2 x^2}\). Now, it’s all about cleaning up the expression in the numerator, while the denominator remains continual. This unified platform simplifies the operation and aids in the further simplification of complex terms.

factoring

Factoring plays a pivotal role in algebra. In our solution, we expanded \((x+h)^2\) which equals \(x^2 + 2xh + h^2\). When we subtract this from \(x^2\), the result is \(x^2 - x^2 - 2xh - h^2 = -2xh - h^2\). Factoring out the common factor \(h\), we end up with \(-h(2x+h)\), a more manageable form. Factoring helps isolate significant variables and constants, simplifying expressions, and preparing for the final steps of simplification.

simplification

Simplification generally means reducing an expression to its simplest form. For our algebraic fraction, this was achieved step-by-step: from identifying the common denominator, subtracting fractions, expanding and combining like terms, to eventually factoring. Dividing by \(h\), our expression \(\frac{-h(2x + h)}{(x+h)^2 x^2}\) simplifies to \(\frac{-(2x + h)}{(x+h)^2 x^2}\) upon canceling \(h\) from the numerator and the multiplying factor outside the fraction. Each step unravels the complexity of the problem, providing clarity and a final, simplified solution.