QUESTION IMAGE
Question
- error analysis complete the following.
\\(\
\\) (with a red cross mark on the left of the equation)
Step1: Factor and Rewrite Denominator
First, factor \(x^2 - 25\) as \((x + 5)(x - 5)\) (difference of squares: \(a^2 - b^2=(a + b)(a - b)\) where \(a = x\), \(b = 5\)). Also, rewrite \(3 - x\) as \(-(x - 3)\) (since \(3 - x=-(x - 3)\)). So the original expression \(\frac{x^2 - 25}{3 - x}\cdot\frac{x - 3}{x + 5}\) becomes \(\frac{(x + 5)(x - 5)}{-(x - 3)}\cdot\frac{x - 3}{x + 5}\).
Step2: Cancel Common Factors
Now, cancel out the common factors. The \((x + 5)\) terms cancel, and the \((x - 3)\) terms cancel (note that \(x
eq3\) because \(x - 3\) is in the numerator and denominator, and \(x
eq - 5\) because \(x + 5\) is in the denominator). After canceling, we have \(\frac{(x - 5)}{-1}\) (since \(\frac{(x - 5)}{-(x - 3)}\cdot\frac{x - 3}{x + 5}\), the \((x - 3)\) and \((x + 5)\) cancel, leaving \(\frac{x - 5}{-1}\)).
Step3: Simplify the Result
Simplifying \(\frac{x - 5}{-1}\) gives \(-(x - 5)=-x + 5\) or \(5 - x\). Also, we need to check the domain restrictions. The original denominators are \(3 - x\) (which is \(-(x - 3)\)) and \(x + 5\), so \(x
eq3\) (because \(3 - x = 0\) when \(x = 3\)) and \(x
eq - 5\) (because \(x + 5 = 0\) when \(x=-5\)). The mistake in the original solution was not correctly handling the sign when rewriting \(3 - x\) as \(-(x - 3)\), leading to an incorrect simplification (they got \(x - 5\) instead of \(-x + 5\)).
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The error was in the sign when simplifying \(\frac{(x + 5)(x - 5)}{3 - x}\cdot\frac{x - 3}{x + 5}\). Correctly rewriting \(3 - x=-(x - 3)\) leads to \(\frac{(x + 5)(x - 5)}{-(x - 3)}\cdot\frac{x - 3}{x + 5}\), and after canceling, we get \(-(x - 5)=5 - x\) (or \(-x + 5\)) with \(x
eq3\) and \(x
eq - 5\). So the correct simplified form is \(5 - x\) (or \(-x + 5\)) where \(x
eq3\) and \(x
eq - 5\).