Question

Divide 9x ^ 3 + 18x ^ 2 - 13x + 5 by 3x - 1 using division and write the division in the form P = DQ + R

Answer 1

`(9x^3\:+\:18x\:^2\:-\:13x\:+\:5)/(3x-1)=9x^3+18x^2-13x+5`

Explanation:

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

`(p(x))/(d(x)) =q(x)+(r(x))/(d(x))`

and so that the degree of
`r(x)` is less than the degree of
`d(x)`.

To find
`(9 x^(3) + 18 x^(2) - 13 x + 5)/(3 x - 1)` you must:

`\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}9x^3+18x^2-13x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}(9x^3)/(3x)=3x^2`

`\mathrm{Quotient}=3x^2`

`\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}\\`

`\mathrm{Remainder}=21x^2-13x+5`

Therefore,

`(9x^3+18x^2-13x+5)/(3x-1)=3x^2+(21x^2-13x+5)/(3x-1)`

`\mathrm{Divide}\:(21x^2-13x+5)/(3x-1)`

`\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}21x^2-13x+54\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}(21x^2)/(3x)=7x\\\\\mathrm{Quotient}=7x`

`\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`

Therefore,

`(21x^2-13x+5)/(3x-1)=7x+(-6x+5)/(3x-1)\\\\(9x^3+18x^2-13x+5)/(3x-1)=3x^2+7x+(-6x+5)/(3x-1)`

`\mathrm{Divide}\:(-6x+5)/(3x-1)`

`\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}(-6x)/(3x)=-2\\\\\mathrm{Quotient}=-2`

`\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`

Therefore,

`(-6x+5)/(3x-1)=-2+(3)/(3x-1)\\\\(9x^3+18x^2-13x+5)/(3x-1)=3x^2+7x-2+(3)/(3x-1)`