How do you divide $(x^{3} - 3 x^{2} + 7 x + 2)$ $(x^3 − 3x^2 + 7x + 2)$ by $x - 1$ $x-1$ ?

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How do you divide $(x^{3} - 3 x^{2} + 7 x + 2)$ $(x^3 − 3x^2 + 7x + 2)$ by $x - 1$ $x-1$ ?

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Answer 1 · 2015-12-17T19:23:19

The answer will be : $x^{3} - 3 x^{2} + 7 x + 2 = (x - 1) (x^{2} - 2 x - 5) + 7$ $x^3 - 3x^2 + 7x + 2 = (x-1)(x^2 - 2x - 5) + 7$

Explanation:

You basically divide $x^{3} - 3 x^{2} + 7 x + 2$ $x^3 - 3x^2 + 7x + 2$ by $x - 1$ $x-1$ by using the euclidean method, just like you would do it if you were dividing a natural number a by another number b : you will here try to delete the 3rd degree terms, then the 2nd degree terms, then the 1st degree terms.

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How do you divide $(x^{3} - 3 x^{2} + 7 x + 2)$ $(x^3 − 3x^2 + 7x + 2)$ by $x - 1$ $x-1$ ?

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How do you divide (x3−3x2+7x+2) (x^3 − 3x^2 + 7x + 2) by x−1x-1?

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How do you divide $(x^{3} - 3 x^{2} + 7 x + 2)$ $(x^3 − 3x^2 + 7x + 2)$ by $x - 1$ $x-1$ ?