(t ^ { 5 } -3t ^ { 3 } +4t-6) \div (t-2)
asked by guest on Mar 26, 2025 at 8:59 pm
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MathBot Answer:
t5−3t3+4t−6t−2=t5−3t3+4t−6t−2\displaystyle \frac{{t}^{5} - 3 {t}^{3} + 4 t - 6}{t - 2} = \frac{t^{5} - 3 t^{3} + 4 t - 6}{t - 2} t−2t5−3t3+4t−6=t−2t5−3t3+4t−6
t5−3t3+4t−6t−2=t5t−2−3t3t−2+4tt−2−6t−2\frac{{t}^{5} - 3 {t}^{3} + 4 t - 6}{t - 2} = \frac{t^{5}}{t - 2} - \frac{3 t^{3}}{t - 2} + \frac{4 t}{t - 2} - \frac{6}{t - 2}t−2t5−3t3+4t−6=t−2t5−t−23t3+t−24t−t−26
