-t3t^{3}t3 - 3t23t^{2}3t2 +t+3
asked by guest on Mar 22, 2025 at 10:15 am
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MathBot Answer:
−t3−3t2+t+3=−t3−3t2+t+3\displaystyle -{t}^{3} - 3 {t}^{2} + t + 3 = - t^{3} - 3 t^{2} + t + 3 −t3−3t2+t+3=−t3−3t2+t+3
−t3−3t2+t+3=−(t−1)(t+1)(t+3)-{t}^{3} - 3 {t}^{2} + t + 3 = - \left(t - 1\right) \left(t + 1\right) \left(t + 3\right)−t3−3t2+t+3=−(t−1)(t+1)(t+3)
