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Divide: (2x^4+5x^3+6x-1)/(x+3) A. 2x^3+11x^2+39x+117+(350)/(x+3) B 2x^3-x^2+9x-27+(80)/(x+3) C 2x^3+11x^2+33x+105+(314)/(x+3) D. 2x^3-x^2+3x-3+(8)/(x+3)

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Divide: (2x^4+5x^3+6x-1)/(x+3) A. 2x^3+11x^2+39x+117+(350)/(x+3) B 2x^3-x^2+9x-27+(80)/(x+3) C 2x^3+11x^2+33x+105+(314)/(x+3) D. 2x^3-x^2+3x-3+(8)/(x+3)

Divide: (2x^4+5x^3+6x-1)/(x+3) A. 2x^3+11x^2+39x+117+(350)/(x+3) B 2x^3-x^2+9x-27+(80)/(x+3) C 2x^3+11x^2+33x+105+(314)/(x+3) D. 2x^3-x^2+3x-3+(8)/(x+3)

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DylanElite · Tutor for 8 years

Answer

The provided expression is a polynomial long division problem. We start by dividing the highest term of the numerator by the highest term of the denominator.<br />Dividing $2x^4$ by $x$ results into $2x^3$. Then multiply $2x^3$ by $x+3$ to get $2x^4+6x^3$, and subtract this from the first two terms in the numerator. <br /><br />This is shown as:<br />$2x^{4}+5x^{3}-(2x^{4}+6x^{3})=-x^{3}$.<br /><br />Next, include down the next term from the numerator and repeat the process. <br />Divide $-x^3$ by $x$ to get $-x^2$, multiply $-x^{2}$ by $x+3$ to get $-x^{3}-3x^{2}$ and subtract it from $-x^{3}+6x$ to get $3x^2+6x$.<br /><br />For the next step, we bring down the last term from the numerator and divide $3x^2$ by $x$ to get $3x$, multiply $3x$ onto $x+3$ to become $3x^{2}+9x$. Subtract $3x^{2}+9x$ from $3x^{2}+6x$ leaving us $-3x-1$.<br /><br />Finally, divide $-3x$ by $x$ to yield $-3$, calculate $-3(x+3)= -3x-9$ and subtract $-3x-9$ from $-3x-1$ to get the rest as $8$. This rest is then written as a fraction over the denominator $(x+3)$. As a result, the quotient obtained is $2x^{3}-x^{2}+3x-3+\frac {8}{x+3}$, which is answer D.
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