How do you long divide $2 x^{3} - 3 x^{2} + 3 x - 4$ $2x^3- 3x^2+ 3x - 4$ by $x - 2$ $x-2$ ?

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How do you long divide $2 x^{3} - 3 x^{2} + 3 x - 4$ $2x^3- 3x^2+ 3x - 4$ by $x - 2$ $x-2$ ?

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Answer 1 · 2015-07-09T11:23:44

$(2 x^{3} - 3 x^{2} + 3 x - 4) \div (x - 2) = (2 x^{2} - x + 1) R : (- 2)$ $(2x^3-3x^2+3x-4) div (x-2) = (2x^2 - x + 1) R:(-2)$

Explanation:

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$[A]$ $color(red)("[A]")$ $x$ $x$ (the leading term of $(x - 2)$ $(x-2)$ goes into $2 x^{3}$ $2x^3$ (the leading term of $(2 x^{3} - 3 x^{2} + 3 x - 4)$ $(2x^3-3x^2+3x-4)$ , $(2 x^{2})$ $(2x^2)$ times.

$[B]$ $color(red)("[B]")$ Multiply $(x - 2)$ $(x-2)$ by $(2 x^{2})$ $(2x^2)$ and subtract from the dividend.

$[C]$ $color(red)("[C]")$ "Bring down" the next term ( $3 x$ $3x$ ); the first term of $(x - 2)$ $(x-2)$ goes into $(- x^{2} + 3 x)$ $(-x^2+3x)$ , $(- x)$ $(-x)$ times.

$[D]$ $color(red)("[D]")$ Multiply $(x - 2)$ $(x-2)$ by $(- x)$ $(-x)$ and subtract.

$[E]$ $color(red)("[E]")$ "Bring down" the final term ( $- 4$ $-4$ ); divide the first term of $(x - 2)$ $(x-2)$ (that is, $x$ $x$ ) into $(1 x - 4)$ $(1x-4)$ giving $1$ $1$ ; multiply $(x - 2)$ $(x-2)$ by $(1)$ $(1)$ ; and subtract giving the remainder.

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How do you long divide $2 x^{3} - 3 x^{2} + 3 x - 4$ $2x^3- 3x^2+ 3x - 4$ by $x - 2$ $x-2$ ?

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How do you long divide $2 x^{3} - 3 x^{2} + 3 x - 4$ $2x^3- 3x^2+ 3x - 4$ by $x - 2$ $x-2$ ?