How do you divide $(- 2 v^{3} + 10 v^{2} - 27) \div (v - 4)$ $(- 2v ^ { 3} + 10v ^ { 2} - 27) \div ( v - 4)$ ?

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Answer 1 · 2018-07-24T21:13:26

$\frac{- 2 v^{3} + 10 v^{2} - 27}{v - 4} = - 2 v^{2} + 2 v + 8 + \frac{5}{v - 4}$ $(-2v^3+10v^2-27)/(v-4)=-2v^2+2v+8+5/(v-4)$

$\frac{- 2 v^{3} + 10 v^{2} - 27}{v - 4}$ $(-2v^3+10v^2-27)/(v-4)$

$= \frac{- 2 v^{2} (v - 4) + 2 v^{2} - 27}{v - 4}$ $=(-2v^2(v-4)+2v^2-27)/(v-4)$

$= \frac{- 2 v^{2} (v - 4) + 2 v (v - 4) + 8 v - 27}{v - 4}$ $=(-2v^2(v-4)+2v(v-4)+8v-27)/(v-4)$

$= \frac{- 2 v^{2} (v - 4) + 2 v (v - 4) + 8 (v - 4) + 5}{v - 4}$ $=(-2v^2(v-4)+2v(v-4)+8(v-4)+5)/(v-4)$

$= - 2 v^{2} + 2 v + 8 + \frac{5}{v - 4}$ $=-2v^2+2v+8+5/(v-4)$

\0/ here's our answer !

How do you divide (- 2v ^ { 3} + 10v ^ { 2} - 27) \div ( v - 4)(−2v3+10v2−27)÷(v−4)(- 2v ^ { 3} + 10v ^ { 2} - 27) \div ( v - 4)?