To solve each equation using the principle of zero products, we set each factor equal to zero and solve for \(x\).
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Equation: \((x - 5)(2x + 1) = 0\)
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Set each factor to zero:
- \(x - 5 = 0 \Rightarrow x = 5\)
- \(2x + 1 = 0 \Rightarrow 2x = -1 \Rightarrow x = -\frac{1}{2}\)
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Solutions: \(x = 5\) or \(x = -\frac{1}{2}\)
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Equation: \((x + 5)(2x - 1) = 0\)
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Set each factor to zero:
- \(x + 5 = 0 \Rightarrow x = -5\)
- \(2x - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}\)
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Solutions: \(x = -5\) or \(x = \frac{1}{2}\)
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Equation: \((5x + 1)(x - 2) = 0\)
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Set each factor to zero:
- \(5x + 1 = 0 \Rightarrow 5x = -1 \Rightarrow x = -\frac{1}{5}\)
- \(x - 2 = 0 \Rightarrow x = 2\)
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Solutions: \(x = 2\) or \(x = -\frac{1}{5}\)
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Equation: \((5x - 1)(x + 2) = 0\)
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Set each factor to zero:
- \(5x - 1 = 0 \Rightarrow 5x = 1 \Rightarrow x = \frac{1}{5}\)
- \(x + 2 = 0 \Rightarrow x = -2\)
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Solutions: \(x = -2\) or \(x = \frac{1}{5}\)
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Now let's match each set of solutions to the pairs given on the right:
- (x - 5)(2x + 1) = 0 gives \(x = 5\) or \(x = -\frac{1}{2}\).
- (x + 5)(2x - 1) = 0 gives \(x = -5\) or \(x = \frac{1}{2}\).
- (5x + 1)(x - 2) = 0 gives \(x = 2\) or \(x = -\frac{1}{5}\).
- (5x - 1)(x + 2) = 0 gives \(x = -2\) or \(x = \frac{1}{5}\).
Matching results:
- (x - 5)(2x + 1) = 0 → \(x = 5\) or \(x = -\frac{1}{2}\)
- (x + 5)(2x - 1) = 0 → \(x = -5\) or \(x = \frac{1}{2}\)
- (5x + 1)(x - 2) = 0 → \(x = 2\) or \(x = -\frac{1}{5}\)
- (5x - 1)(x + 2) = 0 → \(x = -2\) or \(x = \frac{1}{5}\)
In summary:
- (x - 5)(2x + 1) = 0 → x = 5 or x = -1/2
- (x + 5)(2x - 1) = 0 → x = -5 or x = 1/2
- (5x + 1)(x - 2) = 0 → x = 2 or x = -1/5
- (5x - 1)(x + 2) = 0 → x = -2 or x = 1/5
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