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Chapter 0: Problem 11

\(7-20=\) Simplify the rational expression. $$ \frac{3(x+2)(x-1)}{6(x-1)^{2}} $$

Short Answer

Expert verified
\(\frac{x+2}{2(x-1)}\)

Step by step solution

01

Identify and Cancel Common Factors

Look for common factors in the numerator and the denominator. In the expression \(\frac{3(x+2)(x-1)}{6(x-1)^{2}}\), the common factor is \((x-1)\). You can cancel \((x-1)\) from both the numerator and the denominator.
02

Simplify the Remaining Expression

After canceling \((x-1)\), the expression simplifies to \(\frac{3(x+2)}{6(x-1)}\). Now, look for any remaining factors in the numerator and the denominator that can be simplified.
03

Further Simplify the Coefficients

Simplify the coefficients of the fraction. In \(\frac{3(x+2)}{6(x-1)}\), you can simplify the coefficient \(\frac{3}{6}\) to \(\frac{1}{2}\). Thus, the expression becomes \(\frac{1(x+2)}{2(x-1)}\).
04

Final Simplified Expression

With all possible simplifications completed, the final simplified form of the expression is \(\frac{x+2}{2(x-1)}\).

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Key Concepts

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

Rational Expressions
Rational expressions are similar to fractions, but instead of just numbers, they include polynomials. These are expressions of the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \). They are used frequently in algebra to represent quantities that vary based on another variable.
  • Just like regular fractions, rational expressions can be simplified, multiplied, divided, added, and subtracted.
  • The main goal in simplifying rational expressions is to make them as simple as possible by reducing them to their lowest terms.
By understanding rational expressions, you can solve complex algebraic equations with ease and elegance. They are fundamental to mastering broader algebraic concepts.
Canceling Common Factors
Canceling common factors is the process of simplifying an expression by removing terms that appear in both the numerator and the denominator. The goal is to reduce the expression to its simplest form.
  • The key is to factor both the numerator and the denominator completely.
  • Once factored, identify any terms that appear in both and cancel them out.
For example, in the expression \( \frac{3(x+2)(x-1)}{6(x-1)^2} \), the term \((x-1)\) is present in both the numerator and the denominator.Cancelling these leaves you with \( \frac{3(x+2)}{6(x-1)} \). Remember, canceling is only possible for factors, not terms added or subtracted.
Simplifying Coefficients
Simplifying coefficients in a rational expression involves dealing with the numerical part, very much like reducing fractions. Coefficients are the numbers placed in front of the variables.
  • Once common factors are canceled, you often need to simplify these coefficients to ensure the expression is fully reduced.
  • Consider the example \( \frac{3(x+2)}{6(x-1)} \). The coefficients \( 3 \) and \( 6 \) share a factor of \( 3 \).
Simplifying \( \frac{3}{6} \) gives you \( \frac{1}{2} \). Thus, the expression simplifies further to \( \frac{1(x+2)}{2(x-1)} \). Reducing coefficients is an essential step in making the expression as concise as possible.
Algebraic Fractions
Algebraic fractions are fractions where the numerator, denominator, or both contain algebraic expressions. Simplifying these requires a combination of the principles of arithmetic fractions and algebra.
  • Start by factoring each part of the expression completely.
  • Use the skills of canceling common factors and simplifying coefficients to reduce them.
For instance, an algebraic fraction like \( \frac{x^2 - 9}{x^2 - 1} \) would need you to factor as \( \frac{(x+3)(x-3)}{(x+1)(x-1)} \).After factoring, check if any factors can be canceled to simplify the expression to its lowest terms. Understanding how to navigate and simplify algebraic fractions will aid significantly in progressing through algebra studies.

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Most popular questions from this chapter

\(65-70\) m Simplify the fractional expression. (Expressions like these arise in calculus.) $$ \frac{\frac{1}{a+h}-\frac{1}{a}}{h} $$

Factoring \(x^{4}+a x^{2}+b\) A trinomial of the form \(x^{4}+a x^{2}+b\) can sometimes be factored easily. For example \(x^{4}+3 x^{2}-4=\left(x^{2}+4\right)\left(x^{2}-1\right) .\) But \(x^{4}+3 x^{2}+4\) cannot be factored in this way. Instead, we can use the following method. \(x^{4}+3 x^{2}+4=\left(x^{4}+4 x^{2}+4\right)-x^{2} \quad\) Add and subtract \(x^{2}\) \(=\left(x^{2}+2\right)^{2}-x^{2} \quad\) Factor perfect square \(=\left[\left(x^{2}+2\right)-x\right]\left[\left(x^{2}+2\right)+x\right] \quad\) Difference of squares \(=\left(x^{2}-x+2\right)\left(x^{2}+x+2\right)\) Factor the following using whichever method is appropriate. (a) \(x^{4}+x^{2}-2\) (b) \(x^{4}+2 x^{2}+9\) (c) \(x^{4}+4 x^{2}+16\) (d) \(x^{4}+2 x^{2}+1\)

Easy Powers That Look Hard Calculate these expressions in your head. Use the Laws of Exponents to help you. $$ \begin{array}{ll}{\text { (a) } \frac{18^{5}}{9^{5}}} & {\text { (b) } 20^{6} \cdot(0.5)^{6}}\end{array} $$

The Form of an Algebraic Expression An algebraic expression may look complicated, but its 'form" is always simple; it must be a sum, a product, a quotient, or a power. For example, consider the following expressions: $$ \begin{array}{ll}{\left(1+x^{2}\right)^{2}+\left(\frac{x+2}{x+1}\right)^{3}} & {(1+x)\left(1+\frac{x+5}{1+x^{4}}\right)} \\\ {\frac{\left(5-x^{3}\right)}{1+\sqrt{1+x^{2}}}} & {\sqrt{\frac{1+x}{1-x}}}\end{array} $$ With appropriate choices for \(A\) and \(B\) , the first has the form \(A+B,\) the second \(A B\) , the third \(A / B,\) and the fourth \(A^{1 / 2}\) . Recognizing the form of an expression helps us expand, simplify, or factor it correctly. Find the form of the following algebraic expressions. $$ \begin{array}{ll}{\text { (a) } x+\sqrt{1+\frac{1}{x}}} & {\text { (b) }\left(1+x^{2}\right)(1+x)^{3}} \\ {\text { (c) } \sqrt[3]{x^{4}\left(4 x^{2}+1\right)}} & {\text { (d) } \frac{1-2 \sqrt{1+x}}{1+\sqrt{1+x^{2}}}}\end{array} $$

\(89-96\) m State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) $$ 2\left(\frac{a}{b}\right)=\frac{2 a}{2 b} $$
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