(8t ^ { 3 } +3tr ^ { 2 } -10t ^ { 2 } r) \div (4t ^ { 2 } -3tr)

asked by guest
on Mar 26, 2025 at 8:06 pm



You asked:

Evaluate the expression: 8t3+3tr2−10t2r4t2−3tr\frac{8 {t}^{3} + 3 t {r}^{2} - 10 {t}^{2} r}{4 {t}^{2} - 3 t r}

MathBot Answer:

Evaluated



8t3+3tr2−10t2r4t2−3tr=3r2t−10t2r+8t3−3rt+4t2\displaystyle \frac{8 {t}^{3} + 3 t {r}^{2} - 10 {t}^{2} r}{4 {t}^{2} - 3 t r} = \frac{3 r^{2} t - 10 t^{2} r + 8 t^{3}}{- 3 r t + 4 t^{2}}

Expanded

8t3+3tr2−10t2r4t2−3tr=3r2t−3rt+4t2−10t2r−3rt+4t2+8t3−3rt+4t2\frac{8 {t}^{3} + 3 t {r}^{2} - 10 {t}^{2} r}{4 {t}^{2} - 3 t r} = \frac{3 r^{2} t}{- 3 r t + 4 t^{2}} - \frac{10 t^{2} r}{- 3 r t + 4 t^{2}} + \frac{8 t^{3}}{- 3 r t + 4 t^{2}}

Factored

8t3+3tr2−10t2r4t2−3tr=−r+2t\frac{8 {t}^{3} + 3 t {r}^{2} - 10 {t}^{2} r}{4 {t}^{2} - 3 t r} = - r + 2 t

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