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Factoring a quadratic expression like \( 3p^2 - 2p - 5 \) involves finding two binomials that multiply together to give the original expression. Here’s one way to approach it: 1. **Identify coefficients**: For the quadratic expression \( 3p^2 - 2p - 5 \), the coefficients are: - \( a = 3 \) (coefficient of \( p^2 \)) - \( b = -2 \) (coefficient of \( p \)) - \( c = -5 \) (constant term) 2. **Multiply \( a \) and \( c \)**: - \( ac = 3 \times -5 = -15 \) 3. **Find two numbers that multiply to \( ac \) and add to \( b \)**: - We need to find two numbers that multiply to \(-15\) and add to \(-2\). - Those numbers are \( -5 \) and \( 3 \). 4. **Rewrite the middle term using these two numbers**: - Rewrite \( -2p \) as \( -5p + 3p \): \[ 3p^2 - 5p + 3p - 5 \] 5. **Factor by grouping**: - Group the terms into pairs and factor out the common factors in each pair: \[ = (3p^2 - 5p) + (3p - 5) \] We can see that factoring by grouping does not directly work here, so we may need another approach. Since direct factoring seems complicated due to prime coefficients, we use the quadratic formula to solve for roots, then express as binomials: \[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plug in \( a = 3 \), \( b = -2 \), and \( c = -5 \) into the quadratic formula: \[ p = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} \] \[ p = \frac{2 \pm \sqrt{4 + 60}}{6} \] \[ p = \frac{2 \pm \sqrt{64}}{6} \] \[ p = \frac{2 \pm 8}{6} \] So, we get the roots: \[ p = \frac{2 + 8}{6} = \frac{10}{6} = \frac{5}{3} \] \[ p = \frac{2 - 8}{6} = \frac{-6}{6} = -1 \] The factors of the quadratic expression are thus: \[ 3p^2 - 2p - 5 = 3(p - \frac{5}{3})(p + 1) \] \[ = (3p - 5)(p + 1) \] Therefore, \( 3p^2 - 2p - 5 \) factors to \( (3p - 5)(p + 1) \).

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