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Factor the trinomial completely. If the trinomial cannot be factored, say it is prime. \[ r^{2}+r s-6 s^{2} \]

Ask by Rojas Kirk. in the United States
Jan 24,2025

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Answer

The trinomial factors to \((r + 3s)(r - 2s)\).

Solution

The given trinomial is: \[ r^{2} + r s - 6 s^{2} \] To factor this, we look for two binomials of the form \((r + a s)(r + b s)\) such that: 1. The coefficients of \(r\) add up to 1 (from the \(rs\) term): \(a + b = 1\). 2. The product of the constants from both binomials equals \(-6\) (from the \(-6s^2\) term): \(a \cdot b = -6\). Let's find the values of \(a\) and \(b\): - **Possible pairs for \(a\) and \(b\):** - \(3\) and \(-2\): \(3 + (-2) = 1\) and \(3 \cdot (-2) = -6\). So, the factors are: \[ (r + 3s)(r - 2s) \] **Final Factored Form:** \[ (r + 3s)(r - 2s) \]

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To factor the trinomial \( r^{2}+rs-6s^{2} \), we look for two numbers that multiply to give the product of the leading coefficient (1) and the constant term (-6s²), which is -6s², and add up to the coefficient of the middle term (1). The numbers that meet these requirements are 3s and -2s. Thus, we can rewrite the middle term: \[ r^{2} + 3rs - 2rs - 6s^{2} \] Next, we group the terms: \[ (r^{2} + 3rs) + (-2rs - 6s^{2}) \] Now we can factor by grouping: \[ r(r + 3s) - 2s(r + 3s) \] Factoring out the common binomial \((r + 3s)\): \[ (r + 3s)(r - 2s) \] So the factored form of the trinomial \( r^{2}+rs-6s^{2} \) is \((r + 3s)(r - 2s)\).

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