Example Question - simplify rational expressions
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Simplifying Rational Expression
To simplify the given expression, you should recognize that the numerator can be factored using the difference of squares rule, and the denominator appears to be already factored. Here is the step-by-step simplification: The given expression is \(\dfrac{x^2 - 16}{x^2 - 3x - 28}\). Factor the numerator which is a difference of squares: \(\dfrac{(x + 4)(x - 4)}{x^2 - 3x - 28}\). Now let's factor the denominator. To factor a quadratic trinomial, we look for two numbers that multiply to the constant term (-28) and add to the linear coefficient (-3). Those two numbers are -7 and +4. \(\dfrac{(x + 4)(x - 4)}{(x - 7)(x + 4)}\). Now, the \((x + 4)\) term is both in the numerator and the denominator, so they can be canceled out: \(\dfrac{x - 4}{x - 7}\). The expression is now simplified to: \(x - 4 \over x - 7\), which cannot be simplified further.
Simplifying a Rational Expression with Exponents
To simplify the given expression, we will use properties of exponents and combine like terms. Here's the expression, with each term separated for clarity: \[ \frac{10x^{5n+4} + 4.125x^{5n+2}}{3x^{5n+3} - 20x^{5n+1}} \] First, let's look at the exponents in the numerator and the denominator to see if we can simplify by factoring out common factors. We want to identify the term with the lowest exponent for \( x \) that is common to all terms. In this case, the term with the lowest exponent for \( x \) is \( x^{5n+1} \), so let's factor that out from each term: Numerator: \[ 10x^{5n+4} = 10x^{5n+1} \cdot x^3 \] \[ 4.125x^{5n+2} = 4.125x^{5n+1} \cdot x \] Denominator: \[ 3x^{5n+3} = 3x^{5n+1} \cdot x^2 \] \[ 20x^{5n+1} \] We can now rewrite the expression: \[ \frac{x^{5n+1} (10x^3 + 4.125x)}{x^{5n+1} (3x^2 - 20)} \] We simplify by canceling the common factor \( x^{5n+1} \) from both the numerator and the denominator: \[ \frac{10x^3 + 4.125x}{3x^2 - 20} \] The expression is now simplified to the extent that we can without further information about \( x \). We cannot simplify it further algebraically, as there are no common factors left to cancel between the numerator and the denominator. The result is a rational expression in terms of \( x \).
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