Question

3x^4 - 4x^3 - 3x-1 ÷ x+1​

Answer 1

Answer:

$\dfrac{3x^4-4x^3-3x-1}{x+1}:\quad 3x^3-7x^2+7x-10+\dfrac{9}{x+1}$

Step-by-step explanation:

$\dfrac{3x^4-4x^3-3x-1}{x+1}$

$\mathrm{Divide}\:\dfrac{3x^4-4x^3-3x-1}{x+1}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}3x^4-4x^3-3x-1\mathrm{\:and\:the\:divisor\:}x+1\mathrm{\::\:}\dfrac{3x^4}{x}=3x^3$

$\mathrm{Quotient}=3x^3$

$\mathrm{Multiply\:}x+1\mathrm{\:by\:}3x^3:\:3x^4+3x^3$

$\mathrm{Subtract\:}3x^4+3x^3\mathrm{\:from\:}3x^4-4x^3-3x-1\mathrm{\:to\:get\:new\:remainder}$

$\mathrm{Remainder}=-7x^3-3x-1$

Therefore

$\dfrac{3x^4-4x^3-3x-1}{x+1}=3x^3+\dfrac{-7x^3-3x-1}{x+1}$

$=3x^3+\dfrac{-7x^3-3x-1}{x+1}$

$\mathrm{Divide}\:\dfrac{-7x^3-3x-1}{x+1}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-7x^3-3x-1\mathrm{\:and\:the\:divisor\:}x+1\mathrm{\::\:}\dfrac{-7x^3}{x}=-7x^2$

$\mathrm{Quotient}=-7x^2$

$\mathrm{Multiply\:}x+1\mathrm{\:by\:}-7x^2:\:-7x^3-7x^2$

$\mathrm{Subtract\:}-7x^3-7x^2\mathrm{\:from\:}-7x^3-3x-1\mathrm{\:to\:get\:new\:remainder}$

$\mathrm{Remainder}=7x^2-3x-1$

Therefore

$\dfrac{-7x^3-3x-1}{x+1}=-7x^2+\dfrac{7x^2-3x-1}{x+1}$

$=3x^3-7x^2+\dfrac{7x^2-3x-1}{x+1}$

$\mathrm{Divide}\:\dfrac{7x^2-3x-1}{x+1}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x^2-3x-1\mathrm{\:and\:the\:divisor\:}x+1\mathrm{\::\:}\dfrac{7x^2}{x}=7x$

$\mathrm{Quotient}=7x$

$\mathrm{Multiply\:}x+1\mathrm{\:by\:}7x:\:7x^2+7x$

$\mathrm{Subtract\:}7x^2+7x\mathrm{\:from\:}7x^2-3x-1\mathrm{\:to\:get\:new\:remainder}$

$\mathrm{Remainder}=-10x-1$

$\mathrm{Therefore}$

$\dfrac{7x^2-3x-1}{x+1}=7x+\dfrac{-10x-1}{x+1}$

$=3x^3-7x^2+7x+\dfrac{-10x-1}{x+1}$

$\mathrm{Divide}\:\dfrac{-10x-1}{x+1}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-10x-1\mathrm{\:and\:the\:divisor\:}x+1\mathrm{\::\:}\dfrac{-10x}{x}=-10$

$\mathrm{Quotient}=-10$

$\mathrm{Multiply\:}x+1\mathrm{\:by\:}-10:\:-10x-10$

$\mathrm{Subtract\:}-10x-10\mathrm{\:from\:}-10x-1\mathrm{\:to\:get\:new\:remainder}$

$\mathrm{Remainder}=9$

$\mathrm{Therefore}$

$\dfrac{-10x-1}{x+1}=-10+\dfrac{9}{x+1}$

$=3x^3-7x^2+7x-10+\dfrac{9}{x+1}$