How do you use long division to divide $\frac{2 x^{3} - 4 x^{2} - 15 x + 5}{{(x - 1)}^{2}}$ $(2x^3-4x^2-15x+5)/(x-1)^2$ ?

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How do you use long division to divide $\frac{2 x^{3} - 4 x^{2} - 15 x + 5}{{(x - 1)}^{2}}$ $(2x^3-4x^2-15x+5)/(x-1)^2$ ?

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Answer 1 · 2017-01-15T07:22:02

The answer is $= 2 x - \frac{(17 x - 5)}{(x^{2} - 2 x + 1)}$ $=2x-((17x-5))/((x^2-2x+1))$

Explanation:

Let 's do the long division

$a a a a$ $color(white)(aaaa)$ $2 x^{3} - 4 x^{2} - 15 x + 5$ $2x^3-4x^2-15x+5$ $a a a a$ $color(white)(aaaa)$ $∣$ $∣$ $x^{2} - 2 x + 1$ $color(red)(x^2-2x+1)$

$a a a a$ $color(white)(aaaa)$ $2 x^{3} - 4 x^{2} + 2 x$ $2x^3-4x^2+2x$ $a a a a a a a a a$ $color(white)(aaaaaaaaa)$ $∣$ $∣$ $2 x$ $color(blue)(2x)$

$a a a a a a a$ $color(white)(aaaaaaa)$ $0 - 0 - 17 x + 5$ $0-0-17x+5$

Therefore,

$\frac{2 x^{3} - 4 x^{2} - 15 x + 5}{x^{2} - 2 x + 1} = 2 x - \frac{(17 x - 5)}{(x^{2} - 2 x + 1)}$ $(2x^3-4x^2-15x+5)/(x^2-2x+1)=2x-((17x-5))/((x^2-2x+1))$

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How do you use long division to divide $\frac{2 x^{3} - 4 x^{2} - 15 x + 5}{{(x - 1)}^{2}}$ $(2x^3-4x^2-15x+5)/(x-1)^2$ ?

1 Answer

Explanation:

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How do you use long division to divide (2x^3-4x^2-15x+5)/(x-1)^22x3−4x2−15x+5(x−1)2(2x^3-4x^2-15x+5)/(x-1)^2?

1 Answer

Explanation:

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How do you use long division to divide $\frac{2 x^{3} - 4 x^{2} - 15 x + 5}{{(x - 1)}^{2}}$ $(2x^3-4x^2-15x+5)/(x-1)^2$ ?