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Factor completely. \( 36 s^{3}+9 s^{2}-20 s-5 \)

Ask by Burton May. in the United States
Mar 11,2025

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Tutor-Verified Answer

Answer

The expression \(36s^3+9s^2-20s-5\) factors to \((4s+1)(9s^{2}-5)\).

Solution

Factor the expression \( 36s^3+9s^2-20s-5 \). Factor the expression by following steps: - step0: Factor: \(36s^{3}+9s^{2}-20s-5\) - step1: Calculate: \(36s^{3}-20s+9s^{2}-5\) - step2: Rewrite the expression: \(4s\times 9s^{2}-4s\times 5+9s^{2}-5\) - step3: Factor the expression: \(4s\left(9s^{2}-5\right)+9s^{2}-5\) - step4: Factor the expression: \(\left(4s+1\right)\left(9s^{2}-5\right)\) The factored form of the expression \(36s^3+9s^2-20s-5\) is \((4s+1)(9s^{2}-5)\).

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The Deep Dive

To factor the polynomial \( 36 s^{3}+9 s^{2}-20 s-5 \) completely, we first look for common factors. We can notice that the polynomial can be grouped: 1. Group terms: \( (36 s^{3} + 9 s^{2}) + (-20 s - 5) \) 2. Factor out the greatest common factor (GCF) from each group: - From the first group: \( 9s^{2}(4s + 1) \) - From the second group: \( -5(4s + 1) \) This can be written as: \[ 9s^{2}(4s + 1) - 5(4s + 1) \] Now, we notice that \( (4s + 1) \) is a common factor, so we can factor that out: \[ (4s + 1)(9s^{2} - 5) \] Next, \( 9s^{2} - 5 \) is a difference of squares: \[ 9s^{2} - 5 = (3s)^{2} - (\sqrt{5})^{2} = (3s - \sqrt{5})(3s + \sqrt{5}) \] Combining everything, we end up with: \[ (4s + 1)(3s - \sqrt{5})(3s + \sqrt{5}) \] So the complete factorization of \( 36 s^{3}+9 s^{2}-20 s-5 \) is: \[ (4s + 1)(3s - \sqrt{5})(3s + \sqrt{5}) \]

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