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Answer:

[tex]\frac{9x^3\:+\:18x\:^2\:-\:13x\:+\:5}{3x-1}=9x^3+18x^2-13x+5[/tex]

Step-by-step explanation:

DIVISION ALGORITHM: If [tex]p(x)[/tex] and [tex]d(x)\neq 0[/tex]  are polynomials, and the degree of [tex]d(x)[/tex] is less than or equal to the degree of [tex]f(x)[/tex],  then there exist unique polynomials [tex]q(x)[/tex] and [tex]r(x)[/tex], so that

                                               [tex]\frac{p(x)}{d(x)} =q(x)+\frac{r(x)}{d(x)}[/tex]

and so that the degree of [tex]r(x)[/tex]  is less than the degree of [tex]d(x)[/tex].

To find [tex]\frac{9 x^{3} + 18 x^{2} - 13 x + 5}{3 x - 1}[/tex] you must:

[tex]\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}9x^3+18x^2-13x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{9x^3}{3x}=3x^2[/tex]

[tex]\mathrm{Quotient}=3x^2[/tex]

[tex]\mathrm{Multiply\:}3x-1\mathrm{\:by\:}3x^2:\:9x^3-3x^2\\\mathrm{Subtract\:}9x^3-3x^2\mathrm{\:from\:}9x^3+18x^2-13x+5\mathrm{\:to\:get\:new\:remainder}\\[/tex]

[tex]\mathrm{Remainder}=21x^2-13x+5[/tex]

Therefore,

[tex]\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+\frac{21x^2-13x+5}{3x-1}[/tex]

[tex]\mathrm{Divide}\:\frac{21x^2-13x+5}{3x-1}[/tex]

[tex]\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}21x^2-13x+54\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{21x^2}{3x}=7x\\\\\mathrm{Quotient}=7x[/tex]

[tex]\mathrm{Multiply\:}3x-1\mathrm{\:by\:}7x:\:21x^2-7x\\\mathrm{Subtract\:}21x^2-7x\mathrm{\:from\:}21x^2-13x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=-6x+5[/tex]

Therefore,

[tex]\frac{21x^2-13x+5}{3x-1}=7x+\frac{-6x+5}{3x-1}\\\\\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+7x+\frac{-6x+5}{3x-1}[/tex]

[tex]\mathrm{Divide}\:\frac{-6x+5}{3x-1}[/tex]

[tex]\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{-6x}{3x}=-2\\\\\mathrm{Quotient}=-2[/tex]

[tex]\mathrm{Multiply\:}3x-1\mathrm{\:by\:}-2:\:-6x+2\\\mathrm{Subtract\:}-6x+2\mathrm{\:from\:}-6x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=3[/tex]

Therefore,

[tex]\frac{-6x+5}{3x-1}=-2+\frac{3}{3x-1}\\\\\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+7x-2+\frac{3}{3x-1}[/tex]

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