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Chapter 6: Problem 88

Add or subtract as indicated. $$\frac{2}{m^{2}-4 m+4}+\frac{3}{m^{2}+m-6}$$

Short Answer

Expert verified
\(\frac{2}{m^{2}-4m+4} + \frac{3}{m^{2}+m-6} = \frac{5m}{(m-2)^{2}(m+3)}\).

Step by step solution

01

Factor the Denominators

First, factor the denominators of each fraction. For \(\frac{2}{m^{2}-4m+4}\): Let鈥檚 write \(m^{2}-4m+4\) as \((m-2)(m-2)=(m-2)^{2}\). For \(\frac{3}{m^{2}+m-6}\): Factor \(m^{2}+m-6\) as \((m+3)(m-2)\).
02

Find the Least Common Denominator (LCD)

Determine the least common denominator (LCD) for the fractions. The denominators are \((m-2)^{2}\) and \((m+3)(m-2)\). The LCD will be \((m-2)^{2}(m+3)\).
03

Rewrite Each Fraction with the LCD

Rewrite each fraction so that they have the same denominator, which is the LCD \((m-2)^{2}(m+3)\). For \(\frac{2}{(m-2)^{2}}\), multiply numerator and denominator by \(m+3\): \(\frac{2}{(m-2)^{2}} \times \frac{m+3}{m+3} = \frac{2(m+3)}{(m-2)^{2}(m+3)} = \frac{2m+6}{(m-2)^{2}(m+3)}\). For \(\frac{3}{(m+3)(m-2)}\), multiply numerator and denominator by \(m-2\): \(\frac{3}{(m+3)(m-2)} \times \frac{m-2}{m-2} = \frac{3(m-2)}{(m+3)(m-2)^{2}} = \frac{3m-6}{(m-2)^{2}(m+3)}\).
04

Add the Fractions

Now that the fractions have a common denominator, add the numerators: \(\frac{2m+6}{(m-2)^{2}(m+3)} + \frac{3m-6}{(m-2)^{2}(m+3)} = \frac{(2m+6) + (3m-6)}{(m-2)^{2}(m+3)} = \frac{5m}{(m-2)^{2}(m+3)}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

factoring polynomials
Factoring polynomials is an essential skill in algebra that simplifies complex expressions. When you factor a polynomial, you break it down into simpler factors that, when multiplied together, give the original polynomial. For example, the expression \(m^{2}-4m+4\) can be factored as \( (m-2)(m-2) \).... Let's look step-by-step:
  • Identify the polynomial.
  • See if it fits any common factoring patterns, like the difference of squares, perfect square trinomials, or simple trinomials.
In our case, \(m^{2}-4m+4 \) is factored into \((m-2)^{2}\) because \(4m \) is split into \(-2m \) and \(-2m \). This pattern matches a perfect square trinomial.
least common denominator (LCD)
Finding the Least Common Denominator (LCD) is crucial when adding or subtracting fractions. The LCD is the smallest number, or in our case, the polynomial expression, that both denominators divide evenly into. Here鈥檚 a simple way to determine the LCD:
  • Factorize both denominators.
  • Identify the distinct factors in both denominators.
  • The LCD is composed of all distinct factors, each raised to the power of their highest occurrence in any denominator.

In our problem, we have the denominators \( (m-2)^{2} \) and \( (m+3)(m-2) \). The LCD is then \((m-2)^{2}(m+3) \) because we take the factor \( (m-2) \) to its highest power (2) and also include the distinct factor \((m+3) \).
adding and subtracting fractions
Adding and subtracting fractions with different denominators involves several steps. Here is a straightforward approach:
  • Find the LCD of the denominators.
  • Rewrite each fraction with the LCD as the new denominator.
  • Combine the numerators.
  • Simplify the result if possible.

In our example, we found the LCD was \((m-2)^{2}(m+3)\). We rewrote each fraction: \[ \frac{2}{(m-2)^{2}} \times \frac{m+3}{m+3} = \frac{2(m+3)}{(m-2)^{2}(m+3)} = \frac{2m+6}{(m-2)^{2}(m+3)} \] and \[ \frac{3}{(m+3)(m-2)} \times \frac{m-2}{m-2} = \frac{3(m-2)}{(m+3)(m-2)^{2}} = \frac{3m-6}{(m-2)^{2}(m+3)} \]. Finally, we added the numerators and simplified to \[ \frac{5m}{(m-2)^{2}(m+3)} \].

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Most popular questions from this chapter

Multiply or divide as indicated. $$ \frac{(-3 m n)^{2} \cdot 64\left(m^{2} n\right)^{3}}{16 m^{2} n^{4}\left(m n^{2}\right)^{3}} \div \frac{24\left(m^{2} n^{2}\right)^{4}}{\left(3 m^{2} n^{3}\right)^{2}} $$

Multiply or divide as indicated. $$ \frac{12 x-10 y}{3 x+2 y} \cdot \frac{6 x+4 y}{10 y-12 x} $$

Write each rational expression in lowest terms. $$ \frac{2 c^{2}+2 c d-60 d^{2}}{2 c^{2}-12 c d+10 d^{2}} $$

Give the domain of each rational function using (a) set-builder notation and (b) interval notation. $$ f(x)=\frac{9 x^{2}-8 x+3}{4 x^{2}+1} $$

Write each rational expression in lowest terms. $$ \frac{12 x^{2}-4 x-5}{8 x^{2}-6 x-5} $$
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